Section 1.A Exercises
Exercise 1.A.1
$$ 1/\left. (a + bi) \right. = c + di $$Suppose that $a$ and $b$ are real numbers, not both 0. Find real numbers $c$ and $d$ such that
We need to use the complex conjugate of $\left. (a + bi) \right.$ by multiplying it by $\left. (a - bi) \right.$, which gives $a^{2} - abi + abi - b^{2}i^{2} = a^{2} + b^{2}$.
$$ \frac{1}{a + bi}\frac{a - bi}{a - bi} = \frac{a - bi}{a^{2} + b^{2}} $$Now we have
$$ \frac{a - bi}{a^{2} + b^{2}} = c + di = \frac{a}{a^{2} + b^{2}} - \frac{b}{a^{2} + b^{2}}i $$Now, equating coefficients gives $c = a/\left. \left( a^{2} + b^{2} \right) \right.$ and $d = - b/\left. \left( a^{2} + b^{2} \right) \right.$.
Exercise 1.A.2
$$ \frac{- 1 + \sqrt{3}i}{2} $$Show that
is a cube root of 1 (meaning that its cube equals 1)
Let’s just brute force this. Multiply the following
$$ \left. \left( \frac{- 1 + \sqrt{3}i}{2} \right) \right.\left. \left( \frac{- 1 + \sqrt{3}i}{2} \right) \right. = \left. \left( - \frac{1}{2} - \frac{2\sqrt{3}}{4}i \right) \right. $$Now
$$ \left. \left( - \frac{1}{2} - \frac{2\sqrt{3}}{4}i \right) \right.\left. \left( \frac{- 1 + \sqrt{3}i}{2} \right) \right. = 1 $$Exercise 1.A.3
Find two distinct square roots of $i$.
We have that
$$ \begin{aligned} e^{i\pi/2} & = \cos\left. \left( \frac{- \pi}{2} \right) \right. + i\sin\left. \left( \frac{- \pi}{2} \right) \right.\quad\mathrm{\text{(Euler's formula)}} \\ & = \cos\left. \left( \frac{3\pi}{2} \right) \right. + i\sin\left. \left( \frac{3\pi}{2} \right) \right. \\ & = 0 + i\left. ( - 1) \right. \\ & = - i \end{aligned} $$Using the above, we have our two distinct square roots of $i$:
$$ \pm e^{i\pi/4} = \pm \left. \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right) \right. $$Exercise 1.A.4
Show that $\alpha + \beta = \beta + \alpha$ for all $\alpha,\beta \in \mathbb{C}$.
Let $\alpha = a + bi$ and $\beta = c + di$. Then the left-hand side of the equation gives
$$ \begin{aligned} \alpha + \beta & = \left. (a + bi) \right. + \left. (c + di) \right. \\ & = a + bi + c + di \\ & = \left. (a + c) \right. + \left. (b + d) \right.i \end{aligned} $$And the RHS gives
$$ \begin{aligned} \alpha + \beta & = \left. (c + di) \right. + \left. (a + bi) \right. \\ & = c + di + a + bi \\ & = \left. (c + a) \right. + \left. (d + b) \right.i \end{aligned} $$Thus $\alpha + \beta = \beta + \alpha$ via the above and via commutativity of the real numbers.
Exercise 1.A.5
Show that $\left. (\alpha + \beta) \right. + \lambda = \alpha + \left. (\beta + \lambda) \right.$ for all $\alpha,\beta,\lambda \in \mathbb{C}$.
Let $\alpha = a + bi$ and $\beta = c + di$ and $\lambda = e + fi$.
Then the left-hand side of the equation gives
$$ \begin{aligned} \left. (\alpha + \beta) \right. + \lambda & = \left. \left\lbrack \left. (a + bi) \right. + \left. (c + di) \right. \right\rbrack \right. + \left. (e + fi) \right. \\ & = a + bi + c + di + e + fi \\ & = \left. \left\lbrack \left. (a + c) \right. + e \right\rbrack \right. + \left. \left\lbrack \left. (b + d) \right.i + fi \right\rbrack \right. \end{aligned} $$Similarly, the RHS gives
$$ \begin{aligned} \alpha + \left. (\beta + \lambda) \right. & = \left. (a + bi) \right. + \left. \left\lbrack \left. (c + di) \right. + \left. (e + fi) \right. \right\rbrack \right. \\ & = a + bi + c + di + e + fi \\ & = \left. \left\lbrack a + \left. (c + e) \right. \right\rbrack \right. + \left. \left\lbrack bi + \left. (d + f) \right.i \right\rbrack \right. \end{aligned} $$The above and the associativity of addition of real numbers show that $\left. (\alpha + \beta) \right. + \lambda = \alpha + \left. (\beta + \lambda) \right.$.
Exercise 1.A.6
Show that $\left. (\alpha\beta) \right.\lambda = \alpha\left. (\beta\lambda) \right.$ for all $\alpha,\beta,\lambda \in \mathbb{C}$.
Let $\alpha = a + bi$ and $\beta = c + di$ and $\lambda = e + fi$.
Then the LHS gives
$$ \begin{aligned} \left. (e + fi) \right. & = \left. \left\lbrack \left. (ac - bd) \right. + \left. (ad + bc) \right.i \right\rbrack \right.\left. (e + fi) \right. \\ & = \left. \left\lbrack \left. (ac - bd) \right.e - \left. (ad + bc) \right.f \right\rbrack \right. + \left. \left\lbrack \left. (ac - bd) \right.f + \left. (ad + bc) \right.e \right\rbrack \right.i \\ & = \left. (ace - bde - adf - bcf) \right. + \left. (acf - bdf + ade + bce) \right.i \end{aligned} $$And the RHS gives
$$ \begin{aligned} \left. (a + bi) \right.\left. \left\lbrack \left. (c + di) \right.\left. (e + fi) \right. \right\rbrack \right. & = \left. (a + bi) \right.\left. \left\lbrack \left. (ce - df) \right. + \left. (cf + de) \right.i \right\rbrack \right. \\ & = \left. \left\lbrack a\left. (ce - df) \right. - b\left. (cf + de) \right. \right\rbrack \right. + \left. \left\lbrack a\left. (cf + de) \right. + b\left. (ce - df) \right. \right\rbrack \right.i \\ & = \left. (ace - adf - bcf - bde) \right. + \left. (acf + ade + bce - bdf) \right.i \end{aligned} $$The results above and the commutativity of addition of real numbers show that $\left. (\alpha\beta) \right.\lambda = \alpha\left. (\beta\lambda) \right.$.
Exercise 1.A.7
Show that for every $\alpha \in \mathbb{C}$, there exists a unique $\beta in\mathbb{C}$ such that $\alpha + \beta = 0$.
Let $\alpha = a + bi$, where $a,b \in \mathbb{R}$. Then, $\beta = - a - bi$ satisfies $\alpha + \beta = 0$. Now suppose that there exists some $B' \in \mathbb{C}$ such that $\alpha + \beta' = 0$. Then
$$ \begin{aligned} \beta & = \beta + \left. (\alpha + \beta') \right. \\ & = \left. (\beta + \alpha) \right. + \beta' \\ & = \beta' \end{aligned} $$The above equations show that $\beta' = \beta$.
Exercise 1.A.8
Show that for every $\alpha \in \mathbb{C}$ with $\alpha \neq 0$ there exists a unique $\beta \in \mathbb{C}$ such that $\alpha\beta = 1$
We already know that $\beta$ exists given the results of Exercise 1.A.1. We just need to show uniqueness now.
Suppose that there exists another $\beta' \in \mathbb{C}$ such that $\alpha\beta = 1$. Then
$$ \begin{aligned} \beta & = \beta\left. (\alpha\beta') \right. \\ & = \left. (\beta\alpha) \right.\beta' \\ & = \beta' \end{aligned} $$The above equations show that $\beta' = \beta$.
Exercise 1.A.9
Show that $\lambda\left. (\alpha + \beta) \right. = \lambda\alpha + \lambda\beta$ for all $\lambda,\alpha,\beta \in \mathbb{C}$.
Let $\alpha = a + bi$ and $\beta = c + di$ and $\lambda = e + fi$, where $a,b,c,d,e,f \in \mathbb{R}$.
Computing the LHS gives
$$ \begin{aligned} \lambda\left. (\alpha + \beta) \right. & = \left. (e + fi) \right.\left. \left\lbrack \left. (a + c) \right. + \left. (b + d) \right.i \right\rbrack \right. \\ & = ae + bei + ce + dei + afi - bf + cfi - df \end{aligned} $$Computing the RHS gives
$$ \begin{aligned} \lambda\alpha + \lambda\beta & = \left. (e + fi) \right.\left. (a + c) \right. + \left. (e + fi) \right.\left. (b + d) \right. \\ & = ae + bei + ce + dei + afi - bf + cfi - df \end{aligned} $$So the LHS equals the RHS.
Exercise 1.A.10
$$ \left. (4, - 3,1,7) \right. + 2x = \left. (5,9, - 6,8) \right. $$Find $x \in \mathbb{R}^{4}$ such that
$x = \left. \left( \frac{1}{2},6, - \frac{7}{2},\frac{1}{2} \right) \right.$
Exercise 1.A.11
$$ \lambda\left. (2 - 3i,5 + 4i, - 6 + 7i) \right. = \left. (12 - 5i,7 + 22i, - 32 - 9i) \right. $$Explain why there does not exist a $\lambda \in \mathbb{C}$ such that
For the first term, we have $\lambda\left. (2 - 3i) \right. = 12 - 5i \Rightarrow \lambda = 3 + 2i$.
But
$$ \begin{aligned} \lambda\left. (2 - 3i,5 + 4i, - 6 + 7i) \right. & = \left. (3 + 2i) \right.\left. (2 - 3i,5 + 4i, - 6 + 7i) \right. \\ & = \left. (12 - 5i,7 + 22i, - 32 + 9i) \right. \\ & \neq \left. (12 - 5i,7 + 22i, - 32 - 9i) \right. \end{aligned} $$Thus $\lambda$ does not exist.
Exercise 1.A.12
Show that $\left. (x + y) \right. + z = x + \left. (y + z) \right.$ for all $x,y,z \in \mathbb{F}^{n}$.
Let $x = \left. \left( x_{1},x_{2},\ldots,x_{n} \right) \right.,y = \left. \left( y_{1},y_{2},\ldots,y_{n} \right) \right.,z = \left. \left( z_{1},z_{2},\ldots,z_{n} \right) \right.$. Using Definition 1.12 (addition in $\mathbb{F}^{n}$), we compute the LHS:
$$ \begin{aligned} \left. (x + y) \right. + z & = \left. \left( x_{1} + y_{1},x_{2} + y_{2},\ldots,x_{n} + y_{n} \right) \right. + \left. \left( z_{1},z_{2},\ldots,z_{n} \right) \right. \\ & = \lbrack\left. \left( x_{1} + y_{1} \right) \right. + z_{1},\ldots,\left. \left( x_{n} + y_{n} \right) \right. + z_{n}) \end{aligned} $$Computing the RHS gives
$$ \begin{array}{rlr} x + \left. (y + z) \right. & = \left. \left( x_{1},x_{2},\ldots,x_{n} \right) \right. + \left. \left( y_{1} + z_{1},y_{2} + z_{2},\ldots,y_{n} + z_{n} \right) \right. & = \left. \left\lbrack x_{1} + \left. \left( y_{1} + z_{1} \right) \right.,\ldots,x_{n} + \left. \left( y_{n} + z_{n} \right) \right. \right\rbrack \right. \end{array} $$The equations above and the associativity of addition in $\mathbb{F}$ show that $\left. (x + y) \right. + z = x + \left. (y + z) \right.$.
Exercise 1.A.13
Show that $\left. (ab) \right.x = a\left. (bx) \right.$ for all $x \in \mathbb{F}^{n}$ and all $a,b \in \mathbb{F}$.
Let $x = \left. \left( x_{1},x_{2},\ldots,x_{n} \right) \right.$, where $x_{1},x_{2},\ldots,x_{n} \in \mathbb{F}$. Definition 1.17 (scalar multiplication in $\mathbb{F}^{n}$) shows that
$$ \left. (ab) \right.x = \left. \left\lbrack \left. (ab) \right.x_{1},\left. (ab) \right.x_{2},\ldots,\left. (ab) \right.x_{n} \right\rbrack \right. $$and
$$ \begin{aligned} a\left. (bx) \right. & = a\left. \left( bx_{1},bx_{2},\ldots,bx_{n} \right) \right. \\ & = \left. \left\lbrack a\left. \left( bx_{1} \right) \right.,a(bx_{2},\ldots,a\left. \left( bx_{n} \right) \right. \right\rbrack \right. \end{aligned} $$The equations above and the associativity of multiplication in $\mathbb{F}$ show that $\left. (ab) \right.x = a\left. (bx) \right.$.
Exercise 1.A.14
Show that $1x = x$ for all $x \in \mathbb{F}^{n}$
Let $x = \left. \left( x_{1},x_{2},\ldots,x_{n} \right) \right.$, where $x_{1},x_{2},\ldots,x_{n} \in \mathbb{F}$. Definition 1.17 (scalar multiplication in $\mathbb{F}^{n}$), and the fact that 1 is the multiplicative identify of $\mathbb{F}$ shows that
$$ \begin{aligned} 1x & = \left. \left( 1x_{1},1x_{2},\ldots,1x_{n} \right) \right. \\ & = \left. \left( x_{1},x_{2},\ldots,x_{n} \right) \right. \\ & = x \end{aligned} $$Exercise 1.A.15
Show that $\lambda\left. (x + y) \right. = \lambda x + \lambda y$ for all $\lambda \in \mathbb{F}$ and all $x,y \in \mathbb{F}^{n}$.
Let $x = \left. \left( x_{1},x_{2},\ldots,x_{n} \right) \right.$, where $x_{1},x_{2},\ldots,x_{n} \in \mathbb{F}$. Definition 1.12 (addition in $\mathbb{F}^{n}$) and Definition 1.17 (scalar multiplication in $\mathbb{F}^{n}$) shows that
$$ \begin{aligned} \lambda\left. (x + y) \right. & = \lambda\left. \left( x_{1} + y_{1},x_{2} + y_{2},\ldots,x_{n} + y_{n} \right) \right. \\ & = \left. \left\lbrack \lambda\left. \left( x_{1} + y_{1} \right) \right.,\lambda\left. \left( x_{2} + y_{2} \right) \right.,\ldots,\lambda\left. \left( x_{n} + y_{n} \right) \right. \right\rbrack \right. \end{aligned} $$and
$$ \begin{aligned} \lambda x + \lambda y & = \left. \left( \lambda x_{1},\lambda x_{2},\ldots,\lambda x_{n} \right) \right. + \left. \left( \lambda y_{1},\lambda y_{2},\ldots,\lambda y_{n} \right) \right. \\ & = \left. \left( \lambda x_{1} + \lambda y_{1},\lambda x_{2} + \lambda y_{2},\ldots,\lambda x_{n} + \lambda y_{n} \right) \right. \end{aligned} $$The equations above and the distributive property of $\mathbb{F}$ show that $\lambda\left. (x + y) \right. = \lambda x + \lambda y$.
Exercise 1.A.16
Show that $\left. (a + b) \right.x = ax + bx$ for all $a,b \in \mathbb{F}$ and all $x \in \mathbb{F}^{n}$
Let $x = \left. \left( x_{1},x_{2},\ldots,x_{n} \right) \right.$, where $x_{1},x_{2},\ldots,x_{n} \in \mathbb{F}$. Definition 1.12 (addition in $\mathbb{F}^{n}$) and Definition 1.17 (scalar multiplication in $\mathbb{F}^{n}$) shows that
$$ \left. (a + b) \right.x = \left. \left\lbrack \left. (a + b) \right.x_{1},\left. (a + b) \right.x_{2},\ldots,\left. (a + b) \right.x_{n} \right\rbrack \right. $$and
$$ \begin{aligned} ax + bx & = \left. \left( ax_{1},ax_{2},\ldots,ax_{n} \right) \right. + \left. \left( bx_{1},bx_{2},\ldots,bx_{n} \right) \right. \\ & = \left. \left( ax_{1} + bx_{1},ax_{2} + bx_{2},\ldots,ax_{n} + bx_{n} \right) \right. \end{aligned} $$The equations above and the distributive property of $\mathbb{F}$ show that $\left. (a + b) \right.x = ax + bx$.
Section 1.B Examples
Example 1.22
$$ \mathbb{F}^{\infty} = \{\left. \left( x_{1},x_{2},\ldots \right) \right.:x_{j} \in \mathbb{F}\mathrm{\text{ for }}j = 1,2,\ldots\} $$$\mathbb{F}^{\infty}$ is defined to be the set of all sequences of elements of $\mathbb{F}$:
$$ \left. \left( x_{1},x_{2},\ldots \right) \right. + \left. \left( y_{1},y_{2},\ldots \right) \right. = \left. \left( x_{1} + y_{1},x_{2} + y_{2},\ldots \right) \right. $$Addition and scalar multiplication of $\mathbb{F}^{\infty}$ are defined as expected:
$$ \lambda\left. \left( x_{1},x_{2},\ldots \right) \right. = \left. \left( \lambda x_{1},\lambda x_{2},\ldots \right) \right. $$and
With these definitions $F^{\infty}$ becomes a vector_space over $\mathbb{F}$, as you should verify. The additive identity in this vector_space is the sequence of all 0’s.
Example 1.24: $\mathbb{F}^{S}$ is a vector_space
See this example.
Verify the following
If $S$ is a non-empty set, then $\mathbb{F}^{S}$ (with the operations of addition and scalar multiplication as defined above) is a vector_space over $\mathbb{F}$.
The additive identity of $\mathbb{F}^{S}$ is the function $0:S \rightarrow \mathbb{F}$ defined by
for all $x \in S$
- For $f \in \mathbb{F}^{S}$, the additive inverse of $f$ is the function $- f:S \rightarrow \mathbb{F}$ defined by
for all $x \in S$.
By definition,
$$ \mathbb{F}^{S} = \{ f:f \rightarrow \mathbb{R}:S \neq \varnothing\} $$With addition being defined on $\mathbb{F}^{S}$ as
$$ \left. (f + g) \right.\left. (x) \right. = f\left. (x) \right. + g\left. (x) \right. $$for all $x \in S$. By definition, this takes inputs in $S$ and produces two outputs, $f\left. (x) \right.$ and $g\left. (x) \right.$, in $F$, and since $F$ is closed under addition, $f\left. (x) \right. + g\left. (x) \right. \in \mathbb{F}$. Hence $f + g \in \mathbb{F}^{S}$.
Similarly, by definition, scalar multiplication is defined as
$$ \left. (\lambda f) \right.\left. (x) \right. = \lambda f\left. (x) \right. $$for $\lambda \in \mathbb{F}$ and $f \in \mathbb{F}^{S}$. By definition, $\lambda f$ takes inputs in $S$. Since $f$ maps $S$ to $\mathbb{F}$, $f\left. (x) \right. \in \mathbb{F}$. By closure of $\mathbb{F}$ under scalar multiplication, $\lambda f\left. (x) \right. \in \mathbb{F}$, so scalar multiplication likewise holds.
Commutativity: Let $f,g \in \mathbb{F}^{S}$. For any $x \in S$
$$ \left. (f + g) \right.\left. (x) \right. = f\left. (x) \right. + g\left. (x) \right. = g\left. (x) \right. + f\left. (x) \right. = \left. (g + f) \right.\left. (x) \right. $$so addition on $\mathbb{F}^{S}$ is commutative, which follows from commutativity under addition on $\mathbb{F}$.
Associativity of addition: This follows from associativity on $\mathbb{F}$. Let $f,g,h \in \mathbb{F}^{S}$. Then, for any $x \in S$,
$$ \begin{aligned} \left. \left( \left. (f + g) \right. + h \right) \right.\left. (x) \right. & = \left. (f + g) \right.\left. (x) \right. + h\left. (x) \right. \\ & = \left. \left( f\left. (x) \right. + g\left. (x) \right. \right) \right. + h\left. (x) \right. \\ & = f\left. (x) \right. + \left. \left( g\left. (x) \right. + h\left. (x) \right. \right) \right. \\ & = \left. \left( f + \left. (g + h) \right. \right) \right.\left. (x) \right. \end{aligned} $$so addition in $\mathbb{F}^{S}$ is associative.
Associativity of scalar multiplication: Let $\alpha,\beta \in \mathbb{F}$. For any $f \in \mathbb{F}^{S}$ and $x \in S$, we have
$$ \begin{aligned} \left. \left( \left. (\alpha\beta) \right.\left. (f) \right. \right) \right.\left. (x) \right. & = \left. (\alpha\beta) \right.f\left. (x) \right. \\ & = \alpha\left. \left( \beta f\left. (x) \right. \right) \right. \\ & = \alpha\left. \left( \left. (\beta f) \right.\left. (x) \right. \right) \right. \\ & = \left. \left( \alpha\left. (\beta f) \right. \right) \right.\left. (x) \right. \end{aligned} $$by associativity in $F$.
Additive identity: We consider the defined function $0\left. (x) \right. = 0$. For any $f \in \mathbb{F}^{S}$ and $x \in S$, we have
$$ \left. (f + 0) \right.\left. (x) \right. = f\left. (x) \right. + 0 = f\left. (x) \right. $$by the additive identity in $\mathbb{F}$. So $0\left. (x) \right.$ is the identity in $\mathbb{F}^{S}$.
Additive inverse: Given $f \in \mathbb{F}^{S}$, take $- f:S \rightarrow \mathbb{F}$ given by $\left. ( - f) \right.\left. (x) \right. = - f\left. (x) \right.$. By closure under scalar multiplication, $- f \in \mathbb{F}^{S}$. By the definition of addition, for any $x \in S$, we have
$$ \begin{aligned} \left. \left( f + \left. ( - f) \right. \right) \right.\left. (x) \right. & = f\left. (x) \right. + \left. ( - f) \right.\left. (x) \right. \\ & = f\left. (x) \right. + \left. \left( - f\left. (x) \right. \right) \right. \\ & = 0 \end{aligned} $$by the additive inverse axiom in $\mathbb{F}$.
Multiplicative identity: For any $f \in \mathbb{F}^{S}$ and $x \in S$, the definition of scalar multiplication gives
$$ \left. (1f) \right.\left. (x) \right. = 1f\left. (x) \right. = f\left. (x) \right. $$by the multiplicative identity axiom of $\mathbb{F}$.
Distributivity: Let $\lambda \in \mathbb{F}$ and $f,g \in \mathbb{F}^{S}$. For any $x \in S$, we have
$$ \begin{aligned} \left. \left( \lambda\left. (f + g) \right. \right) \right.\left. (x) \right. & = \lambda\left. \left( \left. (f + g) \right.\left. (x) \right. \right) \right. \\ & = \lambda\left. \left( f\left. (x) \right. + g\left. (x) \right. \right) \right. \\ & = \lambda f\left. (x) \right. \end{aligned} $$by distributivity in $\mathbb{F}$. So distributivity over addition of functions holds.
Distributivity over field addition: Taking $\alpha,\beta \in \mathbb{F}$ and $f \in \mathbb{F}^{S}$, for any $x \in S$, we have
$$ \left. \left( \left. (\alpha + \beta) \right.f \right) \right.\left. (x) \right. = \left. (\alpha + \beta) \right.f\left. (x) \right. = \alpha f\left. (x) \right. = \beta f\left. (x) \right. $$by distributivity in $\mathbb{F}$.
Thus, $\mathbb{F}^{S}$ is a vector space.
Section 1.B Exercises
Exercise 1.B.1
Prove that $- \left. ( - v) \right. = v$ for every $v \in V$.
Proof:
$$ \begin{aligned} \text{-} v\left. ( - v) \right. & = - \left. ( - v) \right. + 0 \\ & = - \left. ( - v) \right. + \left. \left\lbrack \left. ( - v) \right. + v \right\rbrack \right. \\ & = \left. \left\lbrack - \left. ( - v) \right. + \left. ( - v) \right. \right\rbrack \right. + v \\ & = 0 + v \\ & = v \end{aligned} $$Exercise 1.B.2
Suppose that $a \in \mathbb{F},v \in V$, and $av = 0$. Prove that $a = 0$ or $v = 0$.
Proof: proceed via proof_by_contradiction, and assume that $a \neq 0$ and $v \neq 0$. Then,
$$ \begin{aligned} v & = 1v \\ & = \left. \left( \frac{1}{a}a \right) \right.v \\ & = \frac{1}{a}\left. (av) \right. \\ & = 0 \end{aligned} $$But here $v = 0$, which contradicts our assumption that $v \neq 0$. Thus, $a = 0$ or $v = 0$.
Exercise 1.B.3
Suppose $v,w \in V$. Explain why there exists a unique $x \in V$ such that $v + 3x = w$.
We claim that $x = \left. (1/3) \right.\left. (w - v) \right.$ satisfies $v + 3x = w$.
Proof:
$$ \begin{aligned} v + 3x & = v + 3\frac{1}{3}\left. (w - v) \right. \\ & = v + \left. (w - v) \right. \\ & = w \end{aligned} $$This implies existence. Now we need to show uniqueness.
Suppose that there exists another $x' \in V$ such that $v + 3x' = w$. Then,
$$ \begin{aligned} 0 & = w - w \\ & = \left. (v + 3x) \right. - \left. (v + 3x') \right. \\ & = v + 3x - v - 3x' \\ & = 3x - 3x' \end{aligned} $$Thus
$$ \begin{aligned} x' & = x' + \frac{1}{3} \times 0 \\ & = x' + \frac{1}{3}\left. (3x - 3x') \right. \\ & = x' + x - x' \\ & = x \end{aligned} $$which shows uniqueness.
Exercise 1.B.4
The empty set is not a vector space. The empty set fails to satisfy only one of the requirements listed in 1.19. Which one?
The empty set fails to satisfy the additive identity requirement. This is because there does not exist an element $0 \in \varnothing$.
Exercise 1.B.5
$$ 0v = 0\mathrm{\text{ for all }}v \in V $$Show that in the definition of a vector_space (1.19), the additive inverse condition can be replaced with the condition that
Here the 0 on the left side is the number 0, and the 0 on the right side is the additive identity of $V$. (The phrase “a condition can be replaced” in a definition means that the collection of objects satisfying the definition is unchanged if the original condition is replaced with the new condition.)
The original definition of the additive inverse from 1.19 states that there exists an element $0 \in V$ such that $v + 0 = v$ for all $v \in V$.
We already showed that $0v = 0$ for all $v \in V$ in Theorem 1.29. Now we assume that $0v = 0$ for all $v \in V$ and then show the additive inverse condition.
$$ \begin{aligned} v + \left. ( - 1) \right.v & = 1v + \left. ( - 1) \right.v \\ & = \left. \left\lbrack 1 + \left. ( - 1) \right. \right\rbrack \right.v \\ & = 0v \\ & = 0 \end{aligned} $$So the new definition implies the original definition.
Exercise 1.B.6
$$ \begin{array}{r} t\infty = \{ - \infty,\mathrm{\text{ if }}t < 0, \\ 0,\mathrm{\text{ if }}t = 0, \\ \infty,\mathrm{\text{ if }}t > 0,\quad t\left. ( - \infty) \right.\{\infty,\mathrm{\text{ if }}t < 0, \\ 0,\mathrm{\text{ if }}t = 0, \\ \text{-} \infty,\mathrm{\text{ if }}t > 0, \end{array} $$Let $\infty$ and $- \infty$ denote two distinct objects, neither of which is in $\mathbb{R}$. Define an addition and scalar multiplication on $\mathbb{R} \cup \{\infty\} \cup \{ - \infty\}$ as you could guess from the notation. Specifically, the sum and product of two real numbers is as usual, and for $t \in \mathbb{R}$ define
where
$t + \infty = \infty + t = \infty$,
$t + \left. ( - \infty) \right. = \left. ( - \infty) \right. + t = - \infty$,
$\infty + \infty = \infty$,
$\left. ( - \infty) \right. + \left. ( - \infty) \right. = - \infty$,
$\infty + \left. ( - \infty) \right. = 0$.
Now, is $R \cup \{\infty\} \cup \{ - \infty\}$ a vector_space over $\mathbb{R}$? Explain.
This is not a vector_space over $\mathbb{R}$. Consider the distributive properties in 1.19. If this is a vector_space over $\mathbb{R}$, we will have
$$ \begin{aligned} \infty & = \left. \left\lbrack 2 + \left. ( - 1) \right. \right\rbrack \right.\infty \\ & = 2\infty + \left. ( - 1) \right.\infty \\ & = \infty + \left. ( - \infty) \right. \\ & = 0 \end{aligned} $$Hence for any $t \in \mathbb{R}$, one has
$$ \begin{aligned} t & = 0 + t \\ & = \infty + t \\ & = \infty \\ & = 0 \end{aligned} $$Which is a contradiction since the zero vector is unique. See here.
Section 1.C Examples
Example 1.C.35: subspace
(a)
$$ \{\left. \left( x_{1},x_{2},x_{3},x_{4} \right) \right. \in \mathbb{F}^{4}:x_{3} = 5x_{4} + b\} $$If $b \in \mathbb{F}$, then
is a subspace of $\mathbb{F}^{4}$ if and only if $b = 0$.
Since $\left. (0,0,0,0) \right. \in U$, then for $\left. (0,0,b,0) \right.$ to be an element of $U$, $b$ must be zero.
(b)
The set of continuous real-valued functions on the interval $\left. \lbrack 0,1\rbrack \right.$ is a subspace of $\mathbb{R}^{\left. \lbrack 0,1\rbrack \right.}$.
Claim: The sum of two continuous functions $f$ and $g$ is continuous.
Proof: Let $\epsilon > 0$ be arbitrary. There exists a $\delta > 0$ such that whenever $|x - c| < \delta$, it follows that $\left| \left. (f + g) \right.\left. (x) \right. - \left. (f + g) \right.\left. (c) \right. \right| < \epsilon$.
For $\epsilon/2$, we can find a $\delta_{1}$ such that $\left| f\left. (x) \right. - f\left. (c) \right. \right| < \epsilon/2$ whenever $|x - c| < \delta_{1}$.
Similarly, for $\epsilon/2$, we can find a $\delta_{2}$ such that $\left| g\left. (x) \right. - g\left. (c) \right. \right| < \epsilon/2$ whenever $|x - c| < \delta_{2}$.
Now, using the triangle inequality, we have
$$ \begin{aligned} \left| \left. (f + g) \right.\left. (x) \right. - \left. (f + g) \right.\left. (c) \right. \right| & = \left| f\left. (x) \right. + g\left. (x) \right. - f\left. (c) \right. - g\left. (c) \right. \right| \\ & = |\left. \left\lbrack f\left. (x) \right. - f\left. (c) \right. \right\rbrack \right. + \left. \left\lbrack g\left. (x) \right. - g\left. (c) \right. \right\rbrack \right. \\ & \leq \left| f\left. (x) \right. - f\left. (c) \right. \right| + \left| g\left. (x) \right. - g\left. (c) \right. \right| \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & = \epsilon \end{aligned} $$Now, take $\delta = \min\{ d_{1},d_{2}\}$, so that whenever $|x - c| < \delta$, it follows that $\left| \left. (f + g) \right.\left. (x) \right. - \left. (f + g) \right.\left. (c) \right. \right| < \epsilon$, as desired.
Thus we have shown that the subspace is closed under addition. A similar argument can be used to show it is closed under scalar multiplication. Also, the zero function is a continuous function. Thus, this is a subspace.
(c)
The set of differentiable real-valued functions on $\mathbb{R}$ is a subspace $\mathbb{R}^{\mathbb{R}}$.
Proof: Let $\mathbb{D}$ denote the set of all differentiable functions on $\mathbb{R}$. It is clear that $\mathbb{D}$ inherits most of the required properties from the vector_space of all functions on $\mathbb{R}$, so we only need to verify that $\mathbb{D}$ is a subspace. Let $f\left. (x) \right.$ and $g\left. (x) \right.$ be in $\mathbb{D}$, and let $c$ be an arbitrary real number. By the Algebraic_Differentiability_Theorem (Abbott — Theorem 5.2.4 part (i)), we know that
$$ \left. (f + g) \right.' = f' + g' $$To show this formally, we write
$$ \begin{aligned} \left. (f + g) \right.' & = \lim\limits_{x \rightarrow c}\frac{\left. (f + g) \right.\left. (x) \right. - \left. (f + g) \right.\left. (c) \right.}{x - c} \\ & = \lim\limits_{x \rightarrow c}\frac{f\left. (x) \right. + g\left. (x) \right. - f\left. (c) \right. + g\left. (c) \right.}{x - c} \\ & = \lim\limits_{x \rightarrow c}\frac{\left. \left\lbrack f\left. (x) \right. - f\left. (c) \right. \right\rbrack \right. + \left. \left\lbrack g\left. (x) \right. - g\left. (c) \right. \right\rbrack \right.}{x - c} \\ & = \lim\limits_{x \rightarrow c}\frac{f\left. (x) \right. - f\left. (c) \right.}{x - c} + \lim\limits_{x \rightarrow c}\frac{g\left. (x) \right. - g\left. (c) \right.}{x - c} \\ & = f' + g' \end{aligned} $$Again by the Algebraic_Differentiability_Theorem (Abbott — Theorem 5.2.4 part (ii)), we know that
$$ \left. (cf) \right.' = c \cdot f' $$Which means that $\mathbb{D}$ is closed under vector addition and scalar multiplication and is thus a subspace.
(d)
The set of differentiable real-valued functions $f$ on the interval $\left. (0,3) \right.$ such that $f'\left. (2) \right. = b$ is a subspace of $\mathbb{R}^{\left. (0,3) \right.}$ if and only if $b = 0$.
(e)
The set of all sequences of complex numbers with limit 0 is a subspace of $\mathbb{C}^{\infty}$.
See: https://math.stackexchange.com/a/3777128
Let $U$ be the set of all complex sequences $\left. \left( z_{n} \right) \right._{n = 0}^{\infty}$ such that
$$ \lim\limits_{n \rightarrow \infty}z_{n} = 0 $$Let us now show that $U$ is a subspace.
\(1\) Remember that the additive identity of our vector_space $\mathbb{C}^{\infty}$ is the sequence whose terms are all zero: $\left. (0,0,0,\ldots) \right.$. It is indeed the case that the limit of this sequence is zero, so it is an element of $U$.
\(2\) Now take two sequences $\left. \left( z_{n} \right) \right._{n = 0}^{\infty}$ and $\left. \left( w_{n} \right) \right._{n = 0}^{\infty}$, both in $U$. Then
$$ \begin{aligned} \lim\limits_{n \rightarrow \infty}\left. \left( z_{n} + w_{n} \right) \right. & = \lim\limits_{n \rightarrow \infty}z_{n} + \lim\limits_{n \rightarrow \infty}w_{n} \\ & = 0 + 0 \\ & = 0 \end{aligned} $$So the sequence $\left. \left( z_{n} + w_{n} \right) \right._{n = 0}^{\infty}$ is in $U$. This shows that $U$ is closed under vector addition.
\(3\) Now let $\lambda$ be an arbitrary complex number. We see that
$$ \begin{aligned} \lim\limits_{n \rightarrow \infty}\lambda z_{n} & = \lambda\lim\limits_{n \rightarrow \infty}z_{n} \\ & = \lambda \cdot 0 \\ & = 0 \end{aligned} $$So the sequence $\left. \left( \lambda z_{n} \right) \right._{n = 0}^{\infty}$ is in $U$. This shows that $U$ is closed under scalar multiplication.
So we can therefore conclude that $U$ is a subspace of $\mathbb{C}^{\infty}$
Example 1.C.37
$$ U = \{\left. (a,0,0) \right. \in \mathbb{F}^{3}:a \in \mathbb{F}\}\quad\mathrm{\text{and}}\quad W = \{\left. (0,b,0) \right. \in \mathbb{F}^{3}:b \in \mathbb{F}\} $$Suppose $U$ is the set of all elements of $\mathbb{F}^{3}$ whose second and third coordinates equal 0, and $W$ is the set of all elements of $\mathbb{F}^{3}$ whose first and third coordinates equal 0:
$$ U + W = \{\left. (x,y,0) \right.:x,y \in \mathbb{F}\} $$Then
We must show that $U + W = S$, where $S = \{\left. (x,y,0) \right.:x,y \in \mathbb{F}\}$.
Showing $\subseteq$: Let $u = \left. (x,0,0) \right. \in U$ and $w = \left. (0,y,0) \right. \in W$. Then
$$ u + w = \left. (x + 0,0 + y,0 + 0) \right. $$is in $S$ because $x + 0,0 + y,0 + 0$ are all elements of $\mathbb{F}$.
Now, showing $\supseteq$: Let $s = \left. (x,y,0) \right.$. We need to find $u \in W$ and $w \in W$ such that $s = u + w$. Consider $u = \left. (x,0,0) \right.$ and $w = \left. (0,y,0) \right.$. Then $u \in W$ and $w \in W$ and
$$ \begin{aligned} s & = \left. (x,y,0) \right. \\ & = \left. (x,0,0) \right. + \left. (0,y,0) \right. \\ & = u + w \in U + W \end{aligned} $$Example 1.C.38
$$ U + W = \{\left. (x,x,y,z) \right. \in \mathbb{F}^{4}:x,y,z \in \mathbb{F}\} $$Suppose that $U = \{\left. (x,x,y,y) \right. \in \mathbb{F}^{4}:x,y \in \mathbb{F}\}$ and $W = \{\left. (x,x,x,y) \right. \in \mathbb{F}^{4}:x,y \in \mathbb{F}\}$. Then
as you should verify.
See https://math.stackexchange.com/questions/2397605/sum-of-vector-subspaces
Recall that if we want to show that two sets $A$ and $B$ are equal, then we need to show $A \subseteq B$ and $B \subseteq A$.
We must show that $U + W = S$, where $S = \{\left. (x,x,y,z) \right. \in \mathbb{F}^{4}:x,y,z \in \mathbb{F}\}$.
Showing $\subseteq$: Let $u = \left. (x,x,y,y) \right. \in U$ and $w = \left. (a,a,a,b) \right. \in W$. Then
$$ u + w = \left. (x + a,x + a,y + a,y + b) \right. $$is in $S$ because $x + a,x + a,y + a,y + b$ are all elements of $\mathbb{F}$.
Showing $\supseteq$: Let $s = \left. (x,x,y,z) \right. \in S$. We need to find $u \in U$ and $w \in W$ such that $s = u + w$. Consider $u = \left. (x - y,x - y,0,0) \right. \in U$ and $w = \left. (y,y,y,z) \right. \in W$. Then,
$$ \begin{aligned} s & = \left. (x,x,y,z) \right. \\ & = \left. (x - y,x - y,0,0) \right. + \left. (y,y,y,z) \right. \\ & = u + w \in U + W \end{aligned} $$Therefore $U + W = \{\left. (x,x,y,z) \right. \in \mathbb{F}^{4}:x,y,z \in \mathbb{F}\}$.
Theorem 1.39
Suppose $U_{1},\ldots,U_{m}$ are subspaces of $V$. Then $U_{1} + \cdots + U_{m}$ is the smallest subspace of $V$ containing $U_{1},\ldots,U_{m}$.
See here.
Claim: $U_{1} + \cdots + U_{m}$ is a subspace containing $U_{1},\ldots,U_{m}$, and if $W \subseteq V$ is any subspace with $U_{1},\ldots,U_{m} \subseteq W$, then we already have $U_{1} + \cdots + U_{m} \subseteq W$.
Proof: It is easy to see that $0 \in U_{1} + \cdots + U_{m}$ and that $U_{1} + \cdots + U_{m}$ is closed under addition and scalar multiplication. Thus Theorem 1.34 implies that $U_{1} + \cdots + U_{m}$ is a subspace of $V$.
Now, suppose that $W \subseteq V$ is some subspace containing $U_{1},\ldots,U_{m}$. For all $u_{1} \in U_{1},\ldots,u_{m} \in U_{m}$, we then have $u_{1},\ldots,u_{m} \in W$. Because $W$ is a subspace it follows that $u_{1} + \cdots + u_{m} \in W$ (because $W$ is closed under finite sums). Therefore
$$ U_{1} + \cdots U_{n} = \{ u_{1} + \cdots + u_{n}|u_{1} \in U_{1},\ldots,u_{n} \in U_{n}\} \subseteq W $$Example 1.C.41
$$ U = \{\left. (x,y,0) \right. \in \mathbb{F}^{3}:x,y \in \mathbb{F}\}\quad\mathrm{\text{and}}\quad W = \{\left. (0,0,z) \right. \in \mathbb{F}^{3}:z \in \mathbb{F}\}. $$Suppose $U$ is the subspace of $\mathbb{F}^{3}$ of those vectors whose last coordinate equals 0, and $W$ is the subspace of $\mathbb{F}^{3}$ of those vectors whose first two coordinates equal 0:
Then $\mathbb{F}^{3} = U \oplus W$.
See here.
Let $v \in \mathbb{F}^{3}$. Then we can write $v = \left. (x,y,z) \right.$ for $x,y,z, \in \mathbb{F}$. Let $u = \left. (x,y,0) \right. \in U$ and $w = \left. (0,0,z) \right. \in W$. Then $v = u + w$, so $v \in U + W$. Now let $u' \in U$ and $w' \in W$ be arbitrary such that $u' + w' = v$. So, $u' = \left. (a,b,0) \right.$ and $w' = \left. (0,0,c) \right.$ for $a,b,c \in \mathbb{F}$. It follows that
$$ \begin{aligned} u' + w' & = \left. (a,b,0) \right. + \left. (0,0,c) \right. \\ & = \left. (a,b,c) \right. \\ & = v \\ & = \left. (x,y,0) \right. + \left. (0,0,z) \right. \\ & = \left. (x,y,z) \right. \end{aligned} $$See also here.
Example 1.C.42
$$ \mathbb{F}^{n} = U_{1} \oplus \cdots \oplus U_{n} $$Suppose $U_{j}$ is the subspace of $\mathbb{F}^{n}$ of those vectors whose coordinates are all 0, except possible in the $j^{\mathrm{\text{th}}}$ (thus, for example, $U_{2} = \{\left. (0,x,0,\ldots,0) \right. \in \mathbb{F}^{n}:x \in \mathbb{F}\}$). Then
Section 1.C Exercises
Exercise 1.C.1
For each of the following subsets of $\mathbb{F}^{3}$, determine whether it is a subspace of $\mathbb{F}^{3}$:
(a)
$\{\left. \left( x_{1},x_{2},x_{3} \right) \right. \in \mathbb{F}^{3}:x_{1} + 2x_{2} + 3x_{3} = 0\}$
Let $U = \{\left. \left( x_{1},x_{2},x_{3} \right) \right. \in \mathbb{F}^{3}:x_{1} + 2x_{2} + 3x_{3} = 0\}$.
See https://www.youtube.com/watch?v=pUkYjBw6qJw
Proving the three properties of subspaces:
\(1\) Notice that $0 + 2\left. (0) \right. + 3\left. (0) \right. = 0$. So, $\left. (0,0,0) \right. \in U$.
\(2\) Take any $\left. \left( x_{1},x_{2},x_{3} \right) \right. \in U$ and $\left. \left( y_{1},y_{2},y_{3} \right) \right. \in U$. Then
$$ x_{1} + 2x_{2} + 3x_{3} = 0\quad\mathrm{\text{ and }}y_{1} = 2y_{2} + 3y_{3} = 0 $$Then
$$ \left. \left( x_{1} + 2x_{2} + 3x_{3} \right) \right. + \left. \left( y_{1} + 2y_{2} + 3y_{3} \right) \right. = 0 + 0 = 0 $$From which we can conclude that $\left. \left( x_{1} + y_{1},x_{2} + y_{2},x_{3} + y_{3} \right) \right. \in U$.
\(3\) Take any $a \in \mathbb{F}$ and $\left. \left( x_{1},x_{2},x_{3} \right) \right. \in U$. This means that $x_{1} + 2x_{2} + 3x_{3} = 0$. Then
$$ \begin{aligned} \left. \left( ax_{1} \right) \right. + 2\left. \left( ax_{2} \right) \right. + 3\left. \left( ax_{3} \right) \right. & = a\left. \left( x_{1} + 2x_{2} + 3x_{3} \right) \right. \\ & = a \cdot 0 \\ & = 0 \end{aligned} $$Thus this is a subspace.
(b)
$\{\left. \left( x_{1},x_{2},x_{3} \right) \right. \in \mathbb{F}^{3}:x_{1} + 2x_{2} + 3x_{3} = 4\}$
Of course this is not a subspace.. This fails part (1) of Theorem 1.34 since $0 + 2\left. (0) \right. + 3\left. (0) \right. \neq 4$.
(c)
$\{\left. \left( x_{1},x_{2},x_{3} \right) \right. \in \mathbb{F}^{3}:x_{1}x_{2}x_{3} = 0\}$
Let $U = \{\left. \left( x_{1},x_{2},x_{3} \right) \right. \in \mathbb{F}^{3}:x_{1}x_{2}x_{3} = 0\}$.
$U$ is not a subspace of $\mathbb{F}^{3}$. Take $u = \left. (0,1,1) \right. \in U$ and $w = \left. (1,1,0) \right. \in U$. Then, $u + w = \left. (1,2,1) \right. \notin U$. So this fails the condition of closure under vector addition.
(d)
$\{\left. \left( x_{1},x_{2},x_{3} \right) \right. \in \mathbb{F}^{3}:x_{1} = 5x_{3}\}$
Let $U = \{\left. \left( x_{1},x_{2},x_{3} \right) \right. \in \mathbb{F}^{3}:x_{1} = 5x_{3}\}$
1) Take any $x_{1} = x_{2} = x_{3} = 0$. Then $x_{1} = 5x_{3}$ gives $0 = 0$. So, $\left. (0,0,0) \right. \in U$.
2) Take any $x = \left. \left( x_{1},x_{2},x_{3} \right) \right. \in U$ and $y = \left. \left( y_{1},y_{2},y_{3} \right) \right. \in U$. Then $x + y$ gives
$$ x + y = \left. \left( x_{1} + y_{1},x_{2} + y_{2},x_{3} + y_{3} \right) \right. $$and
$$ x_{1} + y_{1} = 5x_{3} + 5y_{3} = 5\left. \left( x_{3} + y_{3} \right) \right. $$so $U$ is closed under vector addition.
3) Choose any $a \in \mathbb{F}$ and $x = \left. \left( x_{1},x_{2},x_{3} \right) \right. \in U$. Then, $ax = a\left. \left( 5x_{3} \right) \right. = 5\left. \left( ax_{3} \right) \right.$. So, $U$ is closed under scalar multiplication.
Exercise 1.C.2
Verify all the assertions in Example 1.35.
This was worked through previously. See above.
Exercise 1.C.3
Show that the set of differentiable real-valued functions $f$ on the interval $\left. ( - 4,4) \right.$ such that $f'\left. ( - 1) \right. = 3f\left. (2) \right.$ is a subspace of $\mathbb{R}^{\left. ( - 4,4) \right.}$.
See https://math.stackexchange.com/a/3471524
Let $X = \{ f \in \mathbb{R}^{\left. ( - 4,4) \right.}:f\mathrm{\text{ is differentiable and }}f'\left. ( - 1) \right. = 3f\left. (2) \right.\}$.
Prove the three properties of subspaces.
\(1\) The zero vector of the vector_space $\mathbb{R}^{\left. ( - 4,4) \right.}$ is the function $0:\left. ( - 4,4) \right. \rightarrow \mathbb{R}$ defined by $0\left. (x) \right. = 0$. The zero function is differentiable and satisfies that $0'\left. (1) \right. = 3 \cdot 0\left. (2) \right. = 0$. So $0 \in X$.
\(2\) Choose $f,g \in X$. We know that $\left. (f + g) \right.$ is differentiable by the Algebraic_Differentiability_Theorem (Abbott — Theorem 5.2.4 part i), and $\left. (f + g) \right.'\left. ( - 1) \right. = f'\left. ( - 1) \right. + g'\left. ( - 1) \right. = 3f\left. (2) \right. + 3g\left. (2) \right. = 3\left. \left\lbrack f\left. (2) \right. + g\left. (2) \right. \right\rbrack \right. = 3\left. \left\lbrack \left. (f + g) \right.\left. (2) \right. \right\rbrack \right.$. Thus $f + g \in X$.
\(3\) Choose $a \in \mathbb{F}$ and $f \in X$. By the Algebraic_Differentiability_Theorem (Abbott — Theorem 5.2.4 part ii), we know that $af$ is differentiable, and that $\left. (af) \right.'\left. ( - 1) \right. = a\left. \left( f'\left. ( - 1) \right. \right) \right. = a\left. \left\lbrack 3f\left. (2) \right. \right\rbrack \right. = 3\left. \left\lbrack \left. (af) \right.\left. (2) \right. \right\rbrack \right.$. Thus $af \in X$.
Exercise 1.C.4
Suppose $b \in \mathbb{R}$. Show that the set of continuous real-valued functions $f$ on the interval $\left. \lbrack 0,1\rbrack \right.$ such that $\int_{0}^{1}f = b$ is a subspace of $\mathbb{R}^{\left. \lbrack 0,1\rbrack \right.}$ if and only if $b = 0$.
Let X = $\{ f \in \mathbb{R}^{\left. \lbrack 0,1\rbrack \right.}:f\mathrm{\text{ is continuous and }}\int_{0}^{1}f = b\}$.
\(1\) Certainly, $0 \in X$ since the zero function $0\left. (x) \right. = 0$ is continuous and $\int_{0}^{1}0\left. (x) \right.dx = 0$.
\(2\) Choose $f,g \in X$. It follows by the Algebraic_Continuity_Theorem (Abbott – Theorem 4.3.4 part ii) that $f + g$ is continuous. By the Properties_of_the_Integral (Abbott – Theorem 7.4.2 part i), we know that $\left. (f + g) \right.$ is integrable. So we have $\int_{0}^{1}f\left. (x) \right.dx + \int_{0}^{1}g\left. (x) \right. = 0 + 0 = 0$.
\(3\) Choose $a \in \mathbb{F}$ and $f \in X$. Again by the Algebraic_Continuity_Theorem (Abbott — Theorem 7.4.2 part i), $af$ is continuous. Then, $af$ is also integrable by the Properties_of_the_Integral (Abbott — Theorem 7.4.2 part ii) So we have $\int_{0}^{1}0\left. (x) \right.dx = 0 = b$.
Exercise 1.C.5
Is $\mathbb{R}^{2}$ a subspace of the complex vector_space $\mathbb{C}^{2}$?
If $\mathbb{R}^{2}$ is a subspace of the complex vector_space $\mathbb{C}^{2}$, then
$$ i\left. (1,1) \right. = \left. (i,i) \right. \in \mathbb{R}^{2} $$which is a contradiction. Thus, $\mathbb{R}^{2}$ is not a subspace of the complex vector_space $\mathbb{C}^{2}$.
Exercise 1.C.6
(a)
Is $\{\left. (a,b,c) \right. \in \mathbb{R}^{3}:a^{3} = b^{3}\}$ a subspace of $\mathbb{R}^{3}$?
Because $a^{3} = b^{3}$ if and only if $a = b \in R$,
$$ \{\left. (a,b,c) \right. \in \mathbb{R}^{3}\} = \{\left. (a,b,c) \right. \in \mathbb{R}^{3}:a = b\} $$is obviously a subspace of $\mathbb{R}^{3}$ by similar arguments in Exercise 1.C.1 and 1.C.2.
(b)
Is $\{\left. (a,b,c) \right. \in \mathbb{C}^{3}:a^{3} = b^{3}\}$ a subspace of $\mathbb{C}^{3}$?
Note that
$$ x = \left. \left( 1,\frac{- 1 + \sqrt{3}i}{2},0 \right) \right. \in \{\left. (a,b,c) \right. \in \mathbb{C}^{3}:a^{3} = b^{3}\} $$and
$$ y = \left. \left( 1,\frac{- 1 - \sqrt{3}i}{2},0 \right) \right. \in \{\left. (a,b,c) \right. \in \mathbb{C}^{3}:a^{3} = b^{3}\} $$However
$$ x + y = \left. (2, - 1,0) \right. \notin \{\left. (a,b,c) \right. \in \mathbb{C}^{3}:a^{3} = b^{3}\} $$This implies that $\{\left. (a,b,c) \right. \in \mathbb{C}^{3}:a^{3} = b^{3}\}$ is not closed under addition and thus is not a subspace of $\mathbb{C}^{3}$.
Exercise 1.C.7
Give an example of a nonempty subset $U$ of $\mathbb{R}^{2}$ such that $U$ is closed under addition and under taking additive inverses (meaning $- u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbb{R}^{2}$.
Let $U = \{\left. (x,y) \right. \in \mathbb{R}^{2}:x,y, \in \mathbb{Z}\}$. $U$ is not empty. If $\left. \left( x_{1},y_{1} \right) \right. \in A$ and $\left. \left( x_{2},y_{2} \right) \right. \in A$, then $x_{1},x_{2},y_{1},y_{2} \in \mathbb{Z}$. Hence $x_{1} + x_{2}$ and $y_{1} + y_{2}$ are integers. This means that $\left. \left( x1 + x_{2},y_{1} + y_{2} \right) \right. = \left. \left( x_{1},y_{1} \right) \right. + \left. \left( x_{2},y_{2} \right) \right. \in U$, and that $U$ is closed under addition. Similarly, since $\left. \left( - x_{1}, - y_{1} \right) \right. \in U$, it follows that $U$ is closed under additive inverses. However, $U$ is not closed under sclar multiplication since $\left. (1,1) \right. \in A$ while $\frac{1}{2}\left. (1,1) \right. \notin U$. Hence $U$ is not a subspace of $\mathbb{R}^{2}$.
Exercise 1.C.8
Give an example of a nonempty subset $U$ of $\mathbb{R}^{2}$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbb{R}^{2}$.
Let $U = \{\left. (x,y) \right. \in \mathbb{R}^{2}:x = 0\mathrm{\text{ or }}y = 0\}$. $U$ is not empty. If $\left. (x,0) \right. \in U$, then for any $\lambda \in \mathbb{R}$, we have
$$ \lambda\left. (x,0) \right. = \left. (\lambda x,0) \right. \in U $$Similarly, $\lambda\left. (0,y) \right. \in U$. Hence $U$ is closed under sclar multiplication. However, $\left. (1,0) \right.,\left. (0,1) \right. \in U$ while $\left. (1,1) \right. = \left. (1,0) \right. + \left. (0,1) \right. \notin U$. This implies that $U$ is not closed under addition, and thus is not a subspace of $\mathbb{R}^{2}$.
Exercise 1.C.9
A function $f:\mathbb{R} \rightarrow \mathbb{R}$ is called periodic if there exists a positive number $p$ such that $f\left. (x) \right. = f\left. (x + p) \right.$ for all $x \in \mathbb{R}$. Is the set of periodic functions from $\mathbb{R}$ to $\mathbb{R}$ a subspace of $\mathbb{R}^{\mathbb{R}}$? Explain.
Let $f\left. (x) \right. = \cos\left. (\pi x) \right. + \cos\left. (x) \right.$. $f\left. (x) \right.$ is not periodic.
Exercise 1.C.10
Suppose $U_{1}$ and $U_{2}$ are subspaces of $V$. Prove that the intersection $U_{1} \cap U_{2}$ is a subspace of V.
Proving the three properties of subspaces:
\(1\) By definition, $0 \in U_{1}$ and $0 \in U_{2}$, thus $0 \in U_{1} \cap U_{2}$.
\(2\) If $x \in U_{1} \cap U_{2}$ and $y \in U_{1} \cap U_{2}$, then $x \in U_{1}$ and $y \in U_{1}$, thus $x + y \in U_{1}$ because $U_{1}$ is closed under addition (since we are given that $U_{1}$ is a subspace ). Similarly for $x + y \in U_{2}$. Therefore $x + y \in U_{1} \cap U_{2}$.
\(3\) If $x \in U_{1} \cap U_{2}$, then $x \in U_{1}$. Then for any $\lambda \in \mathbb{F}$, we have $\lambda x \in U_{1}$ since $U_{1}$ is closed under scalar multiplication (since we are given that $U_{1}$ is a subspace ). Similarly, $\lambda x \in U_{2}$. Therefore $\lambda x \in U_{1} \cap U_{2}$.
Exercise 1.C.11
Prove that the intersection of every collection of subspaces of $V$ is a subspace of $V$.
Let $U_{i}$ be subspaces of $V$, where $i \in I$. We claim that $\bigcap_{i \in I}U_{i}$ is a subspace of $V$.
Proving the three properties of subspaces:
\(1\) By definition, $0 \in U_{i}\forall i \in I$, thus $0 \in \bigcap_{i \in I}U_{i}$.
\(2\) If $x \in \bigcap_{i \in I}U_{i}$ and $y \in \bigcap_{i \in I}U_{i}$, then for any given $i \in I$, we have $x \in U_{i}$ and $y \in U_{i}$, thus $x + y \in U_{i}$ since $U_{i}$ is closed under addition because we are given that $U_{i}$ is a subspace. Thus $x + y \in \bigcap_{i \in I}U_{i}$.
\(3\) If $x \in \bigcap_{i \in I}U_{i}$ for any given $i \in I$, then $x \in U_{i}$. Then for any $\lambda \in \mathbb{F}$, we have $\lambda x \in U_{i}$ since $U_{i}$ is closed under scalar multiplication. Thus $\lambda x \in \bigcap_{i \in I}U_{i}$.
Exercise 1.C.12
Prove that the union two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.
See: here.
Let $U$ and $W$ be subspaces of $V$.
$\left. ( \Rightarrow ) \right.$ Claim: If $U \subseteq W$ or $W \subseteq U$, then $U \cup W$ is a subspace of $V$.
Proof: Assume $U \cup W$ is a subspace. Proceed via proof_by_contradiction and assume that $U \nsubseteq W$ and $W \nsubseteq U$. This means that there are elements $u \in U\backslash W$ and $winW\backslash U$. Now, since $U \cup W$ is a subspace, it is closed under addition. Thus we have $u + w \in U + W$. It follows that
$$ u + w \in U\quad\mathrm{\text{or}}\quad u + w \in W $$Suppose that $u + w \in U$. Then we write
$$ w = \left. (u + w) \right. - u $$Since both $u$ and $u + w$ are elements of the subspace $U$, their difference $u = \left. (u + w) \right. - w$ is also in $U$. However, this contradicts the choice of $w \in W\backslash U$. The same is true for the other case, $u + w \in W$. By contradiction, then, we have either $U \subseteq W$ or $W \subseteq U$.
$\left. ( \Leftarrow ) \right.$ Claim: $U \cup W$ is a subspace of $V$ if $U \subseteq W$ $W \subseteq U$.
Proof: Assume $U \subseteq W$, in which case $U \cup W = W$. Since $W$ is a subspace, it follows that $U \cup W$ is also a subspace.
Exercise 1.C.13
Prove that the union of three subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces contains the other two. [This exercise is surprisingly harder than the previous exercise, possibly because this exercise is not true if we replace $\mathbb{F}$ with a field containing only two elements.]
Exercise 1.C.14
Verify the assertion in Example 1.38.
This has been worked through above.
Exercise 1.C.15
Suppose $U$ is a subspace of $V$. What is $U + U$?
$U$ is a subspace of $V$ and thus closed under addition.
\(1\) For any $x,y \in U$, since $U$ is closed under addition, we have $x + y \in U$, i.e. $U + U \subseteq U$.
\(2\) If $x \in U$, then $x = x + 0 \in U + U$, thus $U \subseteq U + U$.
Having shown these two things, it follows that $U + U = U$.
Exercise 1.C.16
Is the operation of addition on the subspaces of $V$ commutative? In other words, if $U$ and $W$ are subspaces of $V$, is $U + W = W + U$?
For $x \in U$ and $y \in W$, because addition in a vector_space $V$ is commutative, we have $x + y = y + x \in W + U$. This implies that $U + W \subseteq W + U$. Similarly, $W + U \subseteq U + W$. Thus $U + W = W + U$.
Exercise 1.C.17
$$ \left. \left( U_{1} + U_{2} \right) \right. + U_{3} = U_{1} + \left. \left( U_{2} + U_{3} \right) \right.? $$Is the operation of addition on the subspaces of V associative? In other words, if $U_{1},U_{2},U_{3}$ are subspaces of $V$, is
In some vector_space $V$, we have $\left. (x + y) \right. + z = x + \left. (y + z) \right.$. Let $xi \in U_{i}$ for $i = 1,2,3$, then
$$ \left. \left( x_{1} + x_{2} \right) \right. + x_{3} = x_{1} + \left. \left( x_{2} + x_{3} \right) \right. \in U_{1} + \left. \left( U_{2} + U_{3} \right) \right. $$Since every element in $\left. \left( U_{1} + U_{2} \right) \right. + U_{3}$ can be expressed in the form $\left. \left( x_{1} + x_{2} \right) \right. + x_{3}$, it follows that $\left. \left( U_{1} + U_{2} \right) \right. + U_{3} \subseteq U_{1} + \left. \left( U_{2} + U_{3} \right) \right.$. Similarly, $U_{1} + \left. \left( U_{2} + 3 \right) \right. \subseteq \left. \left( U_{1} + U_{2} \right) \right. + U_{3}$. Hence $\left. \left( U_{1} + U_{2} \right) \right. + U_{3} = U_{1} + \left. \left( U_{2} + U_{3} \right) \right.$.
Exercise 1.C.18
Does the operation of addition on the subspaces of $V$ have an additive identity? Which subspaces have additive inverses?
Claim: The subspace $\{ 0\}$ is the additive identity, i.e. $U + \{ 0\} = U$ for all subspaces $U$.
Proof: (1) Choose $v \in U + \{ 0\}$. There exists a $u \in U$ such that $v = u + 0$, and this shows that $v = u \in U$. Thus $U + \{ 0\} \subseteq U$.
\(2\) Now, choose $u \in U$. Then, $u = u + 0$ shows that $u \in U + \{ 0\}$. Thus $U \subseteq U + \{ 0\}$.
\(1\) and (2) above imply that $U + \{ 0\} = U$.
We claim that for all subspaces $U$ and $W$, $U + W = \{ 0\}$ implies $U = W = \{ 0\}$, which means that the subspaces that have additive inverses are only $\{ 0\}$ itself.
Choose subspaces $U$ and $W$. Assume $U + W = \{ 0\}$. Theorem 1.39 of Axler shows that $U \subseteq U + W$ and $W \subseteq U + W$. Thus, $U \subseteq \{ 0\}$ and $W \subseteq \{ 0\}$. Note that the only two subsets of $\{ 0\}$ are $\varnothing$ and $\{ 0\}$. Also, the premise that $U$ and $W$ are subspaces implies that $0 \in U$ and $0 \in W$, thus $U = W = \{ 0\}$.
Exercise 1.C.19
$$ U_{1} + W = U_{2} + W $$Prove or give a counterexample: if $U_{1}$, $U_{2}$, $W$ are subspaces of $V$ such that
then $U_{1} = U_{2}$.
Counterexample: Define the following:
$V = \mathbb{R}^{2}$
$U_{1} = \mathbb{R}^{2}$
$U_{2} = \{\left. (x,0) \right.:x \in \mathbb{R}\}$
$W = \{\left. (0,x) \right.:x \in \mathbb{R}\}$
Then $U_{1} + W = U_{2} + W$, but $U_{1} \neq U_{2}$.
Exercise 1.C.20
$$ U = \{\left. (x,x,y,y) \right. \in \mathbb{F}^{4}:x,y \in \mathbb{F}\} $$Suppose
Find a subspace $W$ of $\mathbb{F}^{4}$ such that $\mathbb{F}^{4} = U \oplus W$
Restating Theorem 1.C.45
Suppose $U$ and $W$ are subspaces of $V$. Then $U + W$ is a direct sum if and only if $U \cap W = \{ 0\}$.
Let $W = \{\left. (0,b,0,d) \right. \in \mathbb{F}^{4}:b,d \in \mathbb{F}\}$. For any $\left. (a,b,c,d) \right. \in \mathbb{F}^{4}$, we have
$$ \left. (a,b,c,d) \right. = \left. (x,x,y,y) \right. + \left. (0,b - x,0,d - y) \right. \in U + W $$since $\left. (x,x,y,y) \right. \in U$ and $\left. (0,b - x,0,d - y) \right. \in W$. So we have $\mathbb{F}^{4} = U + W$.
Moreover, if $\left. (a,b,c,d) \right. \in U \cap W$, then $a = b$ and $c = d$ since $\left. (a,b,c,d) \right. \in U$.
Similarly, since $\left. (a,b,c,d) \right. \in W$, we have $a = 0$ and $b = 0$. Therefore, $a = b = 0$ and $c = d = 0$, thus $\left. (a,b,c,d) \right. = \left. (0,0,0,0) \right.$. It follows that $U \cap W = \{ 0\}$. Hence $F = U \oplus W$ by Theorem 1.4.5.
Exercise 1.C.21
$$ U = \{\left. (x,y,x + y,x - y,2x) \right. \in \mathbb{F}^{5}:x,y \in F\} $$Suppose
Find a subspace W of $\mathbb{F}^{5}$ such that $\mathbb{F}^{5} = U \oplus W$.
Let $W = \{\left. (0,0,c,d,e) \right. \in \mathbb{F}^{5}:c,d,e \in \mathbb{F}\}$. For any $\left. (a,b,c,d,e) \right. \in \mathbb{F}^{5}$, we have
$$ \left. (a,b,c,d,e) \right. = \left. (x,y,x + y,x - y,2x) \right. + \left. (0,0,c - x - y,d - x + y,e - 2x) \right. \in U + W $$Since $\left. (x,y,x + y,x - y,2x) \right. \in U$ and $\left. (0,0,c - x - y,d - x + y,e - 2x) \right. \in W$, $\mathbb{F}^{5} = U + W$.
Finally, since $U \cap W = 0$, it follows from Theorem 1.45 that $\mathbb{F}^{5} = U \oplus W$.
Exercise 1.C.22
$$ U = \{\left. (x,y,x + y,x - y,2x) \right. \in \mathbb{F}^{5}:x,y \in F\} $$Suppose
Find three subspaces $W_{1},W_{2},W_{3}$ of $\mathbb{F}^{5}$, none of which equals $\{ 0\}$, such that $\mathbb{F}^{5} = U \oplus W_{1} \oplus W_{2} \oplus W_{3}$.
Let $W_{1} = \{\left. (0,0,a,0,0) \right. \in \mathbb{F}^{5}:a \in \mathbb{F}\}$ and
Let $W_{2} = \{\left. (0,0,0,b,0) \right. \in \mathbb{F}^{5}:b \in \mathbb{F}\}$ and
Let $W_{3} = \{\left. (0,0,0,0,c) \right. \in \mathbb{F}^{5}:c \in \mathbb{F}\}$ and
By the same arguments in Exercises 1.C.20/21,
$$ \mathbb{F}^{5} = U \oplus W1 \oplus W2 \oplus W3 $$Exercise 1.C.23
$$ V = U_{1} \oplus W\quad\mathrm{\text{and}}\quad V = U_{2} \oplus W. $$Prove or give a counterexample: if $U_{1},U_{2},W$ are subspaces of $V$ such that
then $U_{1} = U_{2}$.
Counterexample: Let
$V = \mathbb{R}^{2}$,
$U_{1} = \{\left. (x,0) \right. \in \mathbb{R}^{2}:x \in \mathbb{R}\}$,
$U_{2} = \{\left. (0,y) \right. \in \mathbb{R}^{2}:y \in \mathbb{R}\}$,
$W = \{\left. (a,a) \right. \in \mathbb{R}^{2}:a \in \mathbb{R}\}$,
Then by the same argument in Exercise 1.C.21, we have
$$ V = U_{1} \oplus W\quad\mathrm{\text{and}}\quad U_{2} \oplus W $$but $U_{1} \neq U_{2}$.
Exercise 1.C.24
$$ f\left. ( - x) \right. = f\left. (x) \right. $$A function $f:\mathbb{R} \rightarrow \mathbb{R}$ is called even if
$$ f\left. ( - x) \right. = - f\left. (x) \right. $$for all $x \in \mathbb{R}$. A function $f:\mathbb{R} \rightarrow \mathbb{R}$ is called odd if
for all $x \in \mathbb{R}$. Let $U_{e}$ denote the set of real-valued functions on $\mathbb{R}$ and let $U_{o}$ denote the set of real-valued functions on $\mathbb{R}$. Show that $\mathbb{R}^{\mathbb{R}} = U_{e} \oplus U_{o}$.
Given any $f \in \mathbb{R}^{\mathbb{R}}$, define $f_{e}\left. (x) \right.$ to be $\left. \left\lbrack f\left. (x) \right. + f\left. ( - x) \right. \right\rbrack \right./2$ and define $f_{o}\left. (x) \right.$ to be $\left. \left\lbrack f\left. (x) \right. - f\left. ( - x) \right. \right\rbrack \right./2$ for all $x \in \mathbb{R}$. Then $f_{e},f_{o} \in \mathbb{R}^{\mathbb{R}}$. Morever, for all $x \in \mathbb{R}$, we have
$$ f_{e}\left. ( - x) \right. = \frac{f\left. ( - x) \right. + f\left. (x) \right.}{2} = \frac{f\left. (x) \right. + f\left. ( - x) \right.}{2} = f_{e}\left. (x) \right. $$and
$$ f_{o}\left. ( - x) \right. = \frac{f\left. ( - x) \right. - f\left. (x) \right.}{2} = - \frac{f\left. (x) \right. - f\left. ( - x) \right.}{2} = f_{o}\left. (x) \right. $$hence $f = f_{e} + f_{0} \in U_{e} + U_{o}$. Since we can choose any $f$ arbitrarily, one has $\mathbb{R}^{\mathbb{R}} = U_{e} + U_{o}$.
By Theorem 1.45, to show $\mathbb{R}^{\mathbb{R}} = U_{e} \oplus U_{o}$, it sufficies to prove that $U_{e} \cap U_{o} = \{ 0\}$. Let $f \in U_{e} \cap U_{o}$, then $f\left. (x) \right. = f\left. ( - x) \right.$ since $f \in U_{e}$ and $f\left. (x) \right. = - f\left. ( - x) \right.$ since $f \in U_{o}$ for all $x \in \mathbb{R}$. Sum up $f\left. (x) \right. = f\left. ( - x) \right.$ and $f\left. (x) \right. = - f\left. ( - x) \right.$, we have $f\left. (x) \right. = 0$ for all $x \in \mathbb{R}$. Hence $f = 0$, which implies $U_{e} \cap U_{o} = \{ 0\}$.
Section 2.A Examples
Example 2.A.4
$$ \left. (17, - 4,2) \right. = 6\left. (2,1, - 3) \right. + 5\left. (1, - 2,4) \right. $$$\left. (17, - 4,2) \right.$ is a linear_combination of $\left. (2,1,3) \right.,\left. (1, - 2,4) \right.$ because
$$ \left. (17, - 4,5) \right. = a_{1}\left. (2,1, - 3) \right. + a_{2}\left. (1, - 2,4) \right. $$However, $\left. (17, - 4,5) \right.$ is not a linear_combination of $\left. (2,1, - 3) \right.,\left. (1, - 2,4) \right.$ because there do not exist numbers $a_{1},a_{2} \in \mathbb{F}$ such that
$$ \begin{aligned} 2a_{1} + a_{2} & = 17 \\ a_{1} - 2a_{2} & = - 4 \\ \text{-} 3a_{1} + 4a_{2} & = 5 \end{aligned} $$In other words, the system of equations
has no solutions ()as you should verify).
If we take the above system of equations and put it into reduced row echelon form, we get
$$ \begin{pmatrix} 2 & 1 & 17 \\ 1 & - 2 & - 4 \\ - 3 & 4 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$Since the last row has a nonzero number in the last column and 0’s in all other columns, we know that this system of equations has no solution.
Theorem 2.7: span is the smallest containing subspace
The span of a list of vectors in $V$ is the smallest subspace of $V$ containing all the vectors in the list.
See this example.
Also see this video.
Let $L = \left. \left( v_{1},\ldots,v_{m} \right) \right.$ be a list of vectors in $V$.
\(1\) We will show that $\text{span }\left. (L) \right.$ is a subspace of $V$. The additive identity is in $\text{span }\left. (L) \right.$, because
$$ 0 = 0v_{1} + \cdots + 0v_{m} $$Also, $\text{span }\left. (L) \right.$ is closed under addition, because
$$ \left. \left( a_{1}v_{1} + \cdots + a_{m}v_{m} \right) \right. + \left. \left( c_{1}v_{1} + \cdots + c_{m}v_{m} \right) \right. = \left. \left( a_{1} + c_{1} \right) \right.v_{1} + \cdots + \left. \left( a_{m} + c_{m} \right) \right.v_{m} $$which of course is an element of $\text{span }\left. (L) \right.$.
Furthermore, $\text{span }\left. (L) \right.$ is closed under scalar multiplication, because
$$ \lambda\left. \left( a_{1}v_{1} + \cdots + a_{m}v_{m} \right) \right. = \lambda a_{1}v_{1} + \cdots + \lambda a_{m}v_{m} $$which again is an element of $\text{span }\left. (L) \right.$. So, $\text{span }\left. (L) \right.$ is a subspace.
\(2\) Here, we want to show that $\text{span }\left. (L) \right.$ contains $v_{1},\ldots,v_{m}$. We wish to express $v_{j}$ as a linear_combination of the vectors $v_{1},\ldots,v_{m}$. So suppose that
$$ v_{j} = a_{1}v_{1} + \cdots + a_{j}v_{j} + \cdots + a_{m}v_{m} $$Then, we can let $a_{j} = 1$ and let all other $a$ terms be zero, so we then get $v_{j}$ back. Which means we can get any of the individual vectors back. So we have shown that $\text{span }\left. (L) \right.$ contains each $v_{j}$.
\(3\) Suppose that $W$ is a subspace containing $v_{1},\ldots,v_{m}$. Then we show that $W$ contains $\text{span }\left. (L) \right.$. Let $u \in \text{ span }\left. (L) \right.$. Then $u = a_{1}v_{1} + \cdots + a_{m}v_{m}$.[1] Since $W$ is a subspace, it is closed under scalar multiplication and addition, thus it contains $u$, so we know that $u \in W$. Thus, $u \in \text{ span }\left. (L) \right. \Rightarrow u \in W$ and thus $\text{span }\left. (L) \right. \subseteq W$.
Definition 2.11
$$ p\left. (z) \right. = a_{0} + a_{1}z + a_{2}z^{2} + \cdots + a_{m}z^{m} $$A function $p:\mathbb{F} \rightarrow \mathbb{F}$ is called a polynomial with coefficients in $\mathbb{F}$ if there exist $a_{0},\ldots,a_{m} \in \mathbb{F}$ such that
for all $z \in \mathbb{F}$.
$\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ is the set of all polynomials with coefficients in $\mathbb{F}$.
With the usual operations of addition and scalar multiplication, $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ is a vector_space over $\mathbb{F}$, as you should verify. In other words, $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ is a subspace of $\mathbb{F}^{\mathbb{F}}$, the vector_space of functions from $\mathbb{F}$ to $\mathbb{F}$.
See this example.
Instead of checking all ten vector_space axioms, we only need to check that $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ is a subspace. To determine this, we need to check that $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ : (1) contains the zero vector, (2) is closed under addition, (3) is closed under scalar multiplication.
\(1\) For all $a \in R$, let $a = 0 \Rightarrow p\left. (z) \right. = 0$.
\(2\) For $b_{0},b_{1},\ldots,b_{m} \in \mathbb{F}$ and $c_{0},c_{1},\ldots,c_{m} \in \mathbb{F}$, we have that
$$ \left. \left( b_{0} + b_{1}z + \cdots + b_{m}z^{m} \right) \right. + \left. \left( c_{0} + c_{1}z + \cdots c_{m}z^{m} \right) \right. = \left. \left( b_{0} + c_{0} \right) \right. + \left. \left( b_{1} + c_{1} \right) \right.z + \cdots + \left. \left( a_{m} + b_{m} \right) \right.z^{m} $$Which is an element of $p\left. (z) \right. = a_{0} + a_{1}z + a_{2}z^{2} + \cdots + a_{m}z^{m}$.
\(3\) For some arbitrary scalar $d \in \mathbb{F}$, $b \cdot \left. \left( a_{0} + a_{1}z + \cdots + a_{m}z^{m} \right) \right. = ba_{0} + ba_{1}z + \cdots + ba_{m}z^{m}$, which again is an element of $p\left. (z) \right. = a_{0} + a_{1}z + a_{2}z^{2} + \cdots + a_{m}z^{m}$.
Thus, $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ is a subspace.
Example 2.14
$\mathcal{P}_{m}\left. \left( \mathbb{F} \right) \right.$ is a finite-dimensional vector_space for each non-negative integer $m$.
Example 2.16
Show that $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ is infinite-dimensional.
Solution: Consider any list of elements of $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$. Let $m$ denote the highest degree of the polynomials in this list. Then every polynomial in the span of this list has degree at most $m$. Thus $z^{m + 1}$ is not in the span of our list. Hence no list spans $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$. Thus $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ is infinite-dimensional.
Example 2.18
A list $v$ of one vector $v \in V$ is linearly_independent if and only if $v \neq 0$.
A list of two vectors in $V$ is linearly_independent if and only if neither vector is a scalar multiple of the other.
$\left. (1,0,0,0) \right.,\left. (0,1,0,0) \right.,\left. (0,0,1,0) \right.$ is linearly_independent in $\mathbb{F}^{4}$.
The list $1,z,\ldots,z^{m}$ is linearly_independent in $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ for each non-negative integer $m$.
If some vectors are removed from a linearly_independent list, the remaining list is also linearly_independent, as you should verify.
See this example.
Proceed via proof_by_contrapositive: if the vectors of a list are linearly_independent, then when we add a vector, the new list will still be linearly_dependent.
It should be immediately obvious then that you can append any list of vectors, finite or infinite, even uncountable, to a linearly_dependent set and the result is still a linearly_dependent set. That’s because once you have a nontrivial linear_combination summing to zero, you can add any additional vectors into that sum with all additional coefficients identically zero; this means that (1) the sum is still zero, and (2) the linear_combination is still non-trivial.
Example 2.20: linearly_dependent lists
Part 1
$$ 2\left. (2,3,1) \right. + 3\left. (1, - 1,2) \right. + \left. ( - 1) \right.\left. (7,3,8) \right. = \left. (0,0,0) \right. $$$\left. (2,3,1) \right.,\left. (1, - 1,2) \right.,\left. (7,3,8) \right.$ is linearly_dependent in $\mathbb{F}^{3}$ because
Part 2
The list $\left. (2,3,1) \right.,\left. (1, - 1,2) \right.,\left. (7,3,c) \right.$ is linearly_dependent in $\mathbb{F}^{3}$ if and only if $c = 8$, as you should verify.
We need to solve the following system of equations
$$ \begin{aligned} x\left. (2,3,1) \right. + y\left. (1, - 1,2) \right. + z\left. (7,3,c) \right. & = \left. (0,0,0) \right. \\ \left. (2x + y + 7z,3x - y + 3z,x + 2y + cz) \right. & = \left. (0,0,0) \right. \end{aligned} $$Let us isolate $y$ in the first element of the above list, which gives $y = - 7z - 2x$. Substituting this in the second element of the above list gives $3x - \left. ( - 7z - 2x) \right. + 3z = 0$. Then, solving for $x$ gives $x = - 2z$.
Then, plugging $x$ back into either equation and solving for $y$ gives $y = - 3z$.
Then, plugging $x$ and $y$ into the third element of the above list gives $- 2z + 2\left. ( - 3z) \right. + c*z = 0$, where solving for $c$ gives $c = 8$.
Part 3
If some vector in a list of vectors in V is a linear_combination of the other vectors, then the list is linearly_dependent. (Proof: After writing one vector in the list as equal to a linear_combination of the other vectors, move that vector to the other side of the equation, where it will be multiplied by -1.)
Part 4
Every list of vectors in $V$ containing the 0 vector is linearly_dependent. (This is a special case of the previous bullet point.)
Section 2.A Exercises
Exercise 2.A.1 (alternative span from an existing span)
$$ v_{1} - v_{2},v_{2} - v_{3},v_{3} - v_{4},v_{4} $$Suppose $v_{1},v_{2},v_{3},v_{4}$ spans $V$. Prove that the list
also spans $V$.
We need to show that $v_{1},v_{2},v_{3},v_{4}$ can be expressed as a linear_combination of $v_{1} - v_{2},v_{2} - v_{3},v_{3} - v_{4},v_{4}$.
$$ \begin{aligned} v_{1} & = \left. \left( v_{1} - v_{2} \right) \right. + \left. \left( v_{2} - v_{3} \right) \right. + \left. \left( v_{3} - v_{4} \right) \right. + v_{4} \\ v_{2} & = \left. \left( v_{2} - v_{3} \right) \right. + \left. \left( v_{3} - v_{4} \right) \right. + v_{4} \\ v_{3} = \left. \left( v_{3} - v_{4} \right) \right. + v4 \\ v_{4} = v_{4} \end{aligned} $$Exercise 2.A.2 (examples of linearly_independent lists)
Verify the assertions in Example 2.18.
(a)
A list $v$ of one vector $v \in V$ is linearly_independent if and only if $v \neq 0$.
$\left. ( \Rightarrow ) \right.$: Assume that $v \in V$ is linearly_independent. Then, $v \neq 0$, otherwise we have that $1v = v = 0$, which means that $v$ is linearly_dependent, which is a contradiction.
$\left. ( \Leftarrow ) \right.$: Conversely, assume that $v \neq 0$. By Exericse 1.B.2, if $v \neq 0$, then $av = 0$ implies that $a = 0$, so $v \in V$ must be linearly_independent.
(b)
A list of two vectors in $V$ is linearly_independent if and only if neither vector is a scalar multiple of the other.
$\left. ( \Rightarrow ) \right.$: Assume that $v_{1},v_{2} \in V$ are linearly_independent. Then, by definition, neither vector is a scalar multiple of the other. Otherwise, without loss of generality, we can assume that $v_{1} = cv_{2}$, then $1v_{1} + \left. ( - c) \right.v_{2} = 0$. It follows that $v_{1} \in V$, $v_{2} \in V$ is linearly_dependent, which is a contradiction.
$\left. ( \Leftarrow ) \right.$: Conversely, if $v_{1} \in V$ and $v_{2} \in V$ are linearly_dependent, then there exist $a_{1}$ and $a_{2}$ such that $a_{1}v_{1} + a_{2}v_{2} = 0$ where $a_{1}$ and $a_{2}$ are not both zero. Without loss of generality, we can assume that $a_{1} \neq 0$. Then, $a_{1}v_{2} + a_{2}v_{2} = 0$ means that $v_{1} = - \left. \left( a_{2}/a_{1} \right) \right.v_{2}$, which is a contradiction since neither vector is a scalar multiple of the other.
(c)
$\left. (1,0,0,0) \right.,\left. (0,1,0,0) \right.,\left. (0,0,1,0) \right.$ is linearly_independent in $\mathbb{F}^{4}$.
If there exist $x,y,z \in \mathbb{F}$ such that
$$ x\left. (1,0,0,0) \right. + y\left. (0,1,0,0) \right. + z\left. (0,0,1,0) \right. = 0 $$then it means $\left. (x,y,z,0) \right. = \left. (0,0,0,0) \right.$. Hence $x = y = z = 0$, and it follows that the list is linearly_independent in $\mathbb{F}^{4}$.
Put differently, consider the matrix form
$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$Since the coefficient matrix is non-singular, the only solution is $x = y = z = 0$.
(d)
The list $1,z,\ldots,z^{m}$ is linearly_independent in $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ (see polynomial) for each non-negative integer $m$.
To say that “the list $1,z,\ldots,z^{m}$ is linearly_independent in $\mathcal{P}\left. \left( \mathbb{F} \right) \right.$ (see polynomial) for each non-negative integer $m$” means that the only choice of $a_{1},\ldots,a_{m} \in \mathbb{F}$ that makes $1 + a_{1}z + \cdots + a_{m}z^{m} = 0$ is $a_{1} = \cdots = a_{m} = 0$.
From page 31 of Axler, we have the conclusion that “the coefficients of a polynomial are uniquely determined by the polynomial.” Then, by the definition of linearly_independent and using a similar method as in part c above, we can prove this case.
Another proof: consider the function
$$ f\left. (z) \right. = x_{0} + x_{1}z + \cdots + x_{N - 1}^{N - 1} $$where $N - 1 \geq 0$ and $x_{0},x_{1},\ldots,x_{N - 1} \in \mathbb{F}$. Suppose that $f\left. (z) \right. = 0$ for all $z \in \mathbb{F}$. Then, in particular, $f\left. (1) \right. = 0,f\left. (2) \right. = 0,\ldots,f\left. (N) \right. = 0$. In matrix form
$$ \begin{bmatrix} 1 & 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 1 & 1 & \cdots & 1 \\ 1 & 3 & 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & N & N^{2} & N^{3} & \cdots & N^{N - 1} \\ & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & & \end{bmatrix}\begin{bmatrix} x_{0} \\ x_{1} \\ \vdots \\ x_{N - 1} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} $$Notice that the above coefficient matrix is a Vandermonde matrix. Since all the numbers $1,2,\ldots,N$ are distinct, its determinant is non-zero. Therefore the only solution is $x_{0} = x_{1} = \cdots = x_{N - 1} = 0$.
Exercise 2.A.3 (a linearly_independent list)
$$ \left. (3,1,4) \right.,\left. (2, - 3,5) \right.,\left. (5,9,t) \right. $$Find a number $t$ such that
is not linearly_independent (i.e., is linearly_dependent) in $\mathbb{R}^{3}$.
Similar to Example 2.18 part 2, we set
$$ x\left. (3,1,4) \right. + y\left. (2, - 3,5) \right. + z\left. (5,9,t) \right. = \left. (0,0,0) \right. $$So we have
$$ \left. (3x + 2y + 5z,x - 3y + 9z,4x + 5y + zt) \right. = \left. (0,0,0) \right. $$Let’s look for a convenient variable to solve for. In the second element of the above list, solve for $x$: $x - 3y + 9z = 0 \Rightarrow x = 3y - 9z$.
Plugging $x = 3y - 9z$ into the first element of the list and solving for $y$ gives $y = 2z$.
Finally, plugging both $x = 3y - 9z$ and $y = 2z$ into the third element of the list and solving for $t$ gives $t = 2$.
Exercise 2.A.4 (conditions to linear independence)
Verify the assertion in the second bullet point in Example 2.20.
This was worked through above.
Exercise 2.A.5 (difference between $\mathbb{C}$ and $\mathbb{R}$ as scalar fields)
(a)
Show that if we think of $\mathbb{C}$ as a vector_space over $\mathbb{R}$, then the list $\left. (1 + i,1 - i) \right.$ is linearly_independent.
For $x,y \in \mathbb{R}$ and $x\left. (1 + i) \right. + y\left. (1 - i) \right. = 0$, we have
$$ x\left. (1 + i) \right. + y\left. (1 - i) \right. = x + ix + y - iy = \left. (x + y) \right. + \left. (x - y) \right.i = 0 $$From the above, since $x + y = 0$ and $x - y = 0$, it follows that $x = 0$ and $y = 0$. Thus the list $\left. (1 + i,1 - i) \right.$ linearly_independent over $\mathbb{R}$.
(b)
Show that if we think of $\mathbb{C}$ as a vector_space over $\mathbb{C}$, then the list $\left. (1 + i,1 - i) \right.$ is linearly_dependent.
For $\left. (a + bi) \right.,\left. (c + di) \right. \in \mathbb{C}$, where $a,b,c,din\mathbb{R}$, we have
$$ \left. (a + bi) \right.\left. (1 + i) \right. + \left. (c + di) \right.\left. (1 - i) \right. = \left. (a - b + c + d) \right. + \left. (a + b - c + d) \right.i = 0 $$From the above, since $\left. (a - b + c + d) \right. = 0$ and $\left. (a + b - c + d) \right. = 0$, it follows that $a = b = c = d = 0$. Thus the list $\left. (1 + i,1 - i) \right.$ linearly_independent over $\mathbb{C}$.
Exercise 2.A.6 (constructing another independent list from a list)
$$ v_{1} - v_{2},v_{2} - v_{3},v_{3} - v_{4},v_{4} $$Suppose $v_{1},v_{2},v_{3},v_{4}$ is linearly_independent in $V$. Prove that the list
is also linearly_independent.
Let $x,y,z,w \in \mathbb{F}$ such that
$$ x\left. \left( v_{1} - v_{2} \right) \right. + y\left. \left( y_{2} - v_{3} \right) \right. + z\left. \left( v_{3} - v_{4} \right) \right. + wv_{4} = 0 $$Then,
$$ \begin{aligned} x\left. \left( v_{1} - v_{2} \right) \right. + y\left. \left( y_{2} - v_{3} \right) \right. + z\left. \left( v_{3} - v_{4} \right) \right. + wv_{4} & = xv_{1} - xv_{2} + yv_{2} - yv_{3} + zv_{3} - zv_{4} + wv_{4} \\ & = xv_{1} + \left. (y - x) \right.v_{2} + \left. (z - y) \right.v_{3} + \left. (w - z) \right.v_{4} = 0 \end{aligned} $$Because $v_{1},v_{2},v_{3},v_{4}$ are all assumed to be linearly_independent, then $x = 0,y - x = 0,z - y = 0,w - z = 0$, which implies that $x = y = z = w = 0$. Thus the list
$$ v_{1} - v_{2},v_{2} - v_{3},v_{3} - v_{4},v_{4} $$is linearly_independent in $V$.
Exercise 2.A.7 (constructing another independent list from a list)
$$ 5v_{1} - 4v_{2},v_{2},v_{3},\ldots,v_{m} $$Prove or give a counterexample: If $v_{1},v_{2},\ldots,v_{m}$ is a linearly_independent list of vectors in $V$, then
This is true. For $a_{1},\ldots,a_{m} \in \mathbb{F}$ we have
$$ a_{1}\left. \left( 5v_{1} - 4v_{2} \right) \right. + a_{2}v_{2} + a_{3}v_{3} + \cdots + a_{m}v_{m} = 0 $$so
$$ 5a_{1}v_{1} + \left. \left( a_{2} - 4a_{1} \right) \right.v_{2} + a_{3}v_{3} + \cdots + a_{m}v_{m} = 0 $$Because we are given that $v_{1},\ldots,v_{m}$ is a linearly_independent list of vectors in $V$, it follows that
$$ 5a_{1} = 0,a_{2} - 4a_{1} = 0,a_{3} = \cdots = a_{m} = 0 $$so it must be that $a_{1} = a_{2} = \cdots = a_{m} = 0$, hence the list of vectors
$$ 5v_{1} - 4v_{2},v_{2},v_{3},\ldots,v_{m} $$is also linearly_independent.
Exercise 2.A.8 ()linearly_independent list multiplied by scalar)
Prove or give a counterexample: If $v_{1},\ldots,v_{m}$ is a linearly_independent list of vectors in $V$ and $\lambda \in \mathbb{F}$ with $\lambda \neq 0$, then $\lambda v_{1},\lambda v_{2},\ldots,\lambda v_{m}$ is linearly_independent.
This is true. For $a_{1},\ldots,a_{m} \in \mathbb{F}$ such that
$$ a_{1}\left. \left( \lambda v_{1} \right) \right. + a_{2}\left. \left( \lambda v_{2} \right) \right. + \cdots + a_{m}\left. \left( \lambda v_{m} \right) \right. = 0 $$we have
$$ \left. \left( a_{1}\lambda \right) \right.v_{1} + \left. \left( a_{2}\lambda \right) \right.v_{2} + \cdots + \left. \left( a_{m}\lambda \right) \right.v_{m} = 0 $$Since $v_{1},v_{2},\ldots,v_{m}$ is linearly_independent, it follows that
$$ a_{1}\lambda = a_{2}\lambda = \cdots = a_{m}\lambda $$Since $\lambda = 0$, we can conclude that $a_{1} = a_{2} = \cdots = a_{m} = 0$, hence $\lambda v_{1},\lambda v_{2},\ldots,\lambda v_{m}$ is linearly_independent.
Exercise 2.A.9 (Addition of two linearly_independent lists)
Prove or give a counterexample: if $v_{1},\ldots,v_{m}$ and $w_{1},\ldots,w_{m}$ are linearly_independent lists of vectors in $V$, then $v_{1} + w_{1},\ldots,v_{m} + w_{m}$ is linearly_independent.
Counterexample: let $w_{i} = - v_{i}$. Then, if $v_{1},v_{2},\ldots,v_{m}$ is linearly_independent, we have $w_{1},w_{2},\ldots,w_{m}$ is linearly_independent by Exercise 2.A.8. However, $v_{1} + w_{1} = 0,v_{2} + w_{2} = 0,v_{m} + w_{m} = 0$ is linearly_dependent.
Exercise 2.A.10 (Adding a vector to a linearly_independent list)
Suppose $v_{1},\ldots,v_{m}$ is linearly_independent in $V$ and $w \in V$. Prove that if $v_{1} + w,\ldots,v_{m} + w$ is linearly_dependent then $w \in \text{ span }\left. \left( v_{1},\ldots,v_{m} \right) \right.$ (see span).
Assume that $v_{1} + w,\ldots,v_{m} + w$ is linearly_dependent. Then there exist $a_{1},\ldots,a_{m} \in \mathbb{F}$, not all 0, such that
$$ a_{1}\left. \left( v_{1} + w \right) \right. + a_{2}\left. \left( v_{2} + w \right) \right. + a_{3}\left. \left( v_{3} + w \right) \right. + \cdots + a_{m}\left. \left( v_{m} + w \right) \right. = 0 $$So we have
$$ \begin{aligned} a_{1}\left. \left( v_{1} + w \right) \right. + a_{2}\left. \left( v_{2} + w \right) \right. + a_{3}\left. \left( v_{3} + w \right) \right. + \cdots + a_{m}\left. \left( v_{m} + w \right) \right. & = a_{1}v_{1} + \cdots + a_{m}v_{m} + \left. \left( a_{1} + \cdots + a_{m} \right) \right.w = 0 \\ & = \left. \left( a_{1}v_{1} + a_{1}w \right) \right. + \left. \left( a_{2}v_{2} + a_{2}w \right) \right. + \cdots + \left. \left( a_{m}v_{m} + a_{m}w \right) \right. \\ & = a_{1}v_{1} + \cdots + a_{m}v_{m} + \left. \left( a_{1} + \cdots + a_{m} \right) \right.w \\ & = 0 \end{aligned} $$If $a_{1} + \cdots + a_{m} = 0$, then we get $a_{1}v_{1} + \cdots + a_{m}v_{m} = 0$, from which we can deduce that $a_{i} = 0$. Hence $a_{1} + \cdots + a_{m} \neq 0$. It follows that
$$ w = - \frac{1}{a_{1} + \cdots + a_{m}}\left. \left( a_{1}v_{1} + \cdots + a_{m}v_{m} \right) \right. \in \text{ span }\left. \left( v_{1},\ldots,v_{m} \right) \right. $$$\square$
Exercise 2.A.11 (concatenating a vector to a linearly_independent list)
$$ w \notin \text{ span }\left. \left( v_{1},\ldots,v_{m} \right) \right. $$Suppose $v_{1},\ldots,v_{m}$ is linearly_independent in $V$ and $w \in V$. Show that $v_{1},\ldots,v_{m},w$ is linearly_independent if and only if
See span.
Also, see this video, particularly at the 5m30s mark.
It is equivalent to show that $v_{1},\ldots,v_{m}$ is linearly_dependent if and only if
$$ w \in \text{ span }\left. \left( v_{1},\ldots,v_{m} \right) \right. $$()$\Rightarrow$): Assume that $v_{1},\ldots,v_{m},w$ is linearly_dependent. Then there exist $a_{1},\ldots,a_{m},b \in \mathbb{F}$, not all 0, such that
$$ a_{1}v_{1} + \cdots + a_{m}v_{m} + bw = 0 $$If $b = 0$, we get $a_{1}v_{1} + \cdots + a_{m}v_{m} = 0$, which implies that $v_{1},\ldots,v_{m}$ is linearly_dependent, which contradicts the supposition in the problem description that $v_{1},\ldots,v_{m}$ is linearly_independent. Thus, it must be that $b \neq 0$ and the above equation implies that
$$ w = - \frac{1}{b}\left. \left( a_{1}v_{1} + a_{2}v_{2} + \cdots + a_{m}v_{m} \right) \right. \in \text{ span }\left. \left( v_{1},\ldots,v_{m} \right) \right. $$()$\Leftarrow$): Assume that $w \in \text{ span }\left. \left( v_{1},\ldots,v_{m} \right) \right.$. Then there exist $a_{1},\ldots,a_{m} \in \mathbb{F}$ such that
$$ w = a_{1}v_{1} + \ldots + a_{m}v_{m} \Rightarrow a_{1}v_{1} + \cdots + a_{m}v_{m} - w = 0 $$Again, our supposition in the problem description states that $v_{1},\ldots,v_{m}$ is linearly_independent, which implies that $a_{1}v_{1} + \cdots + a_{m}v_{m} = 0$, so we can conclude that $v_{1},\ldots,v_{m},w$ is linearly_dependent.
Exercise 2.A.12 ()linearly_independent lists in $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$)
Explain why there does not exist a list of six polynomials that is linearly_independent in $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$.
From page 31 of Axler, we know that the list $1,z,z^{2},z^{3},z^{4}$ spans $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$, which has length 5. By Theorem 2.23 (length of linearly_independent list $\leq$ length of spanning list), the length of every linearly_independent list of vectors in $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$ is therefore less than or equal to 5. Thus there does not exist a list of 6 polynomials that is linearly_independent in $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$.
Exercise 2.A.13 (possible spans of $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$)
Explain why no list of four polynomials spans $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$.
By Exercise 2.A.2(d), we know that the list $\left. \left( 1,z,z^{2},z^{3},z^{4} \right) \right.$ is linearly_independent in $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$, which has length 5. By Theorem 2.23, we know that the length of every spanning list of $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$ is greater than or equal to 5. Thus no list of 4 polynomials spans $\mathcal{P}_{4}\left. \left( \mathbb{F} \right) \right.$.