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Section 6.1 #46: Let $(V,\mathbb{F})$ be a vector space with subspace $(U,\mathbb{F})$ and $(W,\mathbb{F})$. Prove that $(U \cap W,\mathbb{F})$ is a subspace of $(V,\mathbb{F})$.
Solution: Let $\vec{x}, \vec{y} \in U \cap W$. Then, in particular, $\vec{x} \in U$ & $\vec{x} \in W$ and $\vec{y} \in U$ & $\vec{y} \in W$. Since $(U,\mathbb{F})$ is a vector space, $\vec{x} \in U$, and $\vec{y} \in U$, then we may conclude that $\vec{x}+\vec{y} \in U$. Since $(W,\mathbb{F})$ is a vector space, $\vec{x} \in W$, and $\vec{y} \in W$, then we may conclude that $\vec{x}+\vec{y} \in W$. Therefore $\vec{x} + \vec{y} \in U$ and $\vec{x}+\vec{y} \in W$. Therefore $\vec{x}+\vec{y} \in U \cap W$.

Let $\alpha \in \mathbb{F}$ be a scalar. Then since $(U,\mathbb{F})$ is a vector space, $\vec{x} \in U$ implies $\alpha \vec{x} \in U$. But also $(W,\mathbb{F})$ is a vector space and so $\vec{x} \in W$ implies $\alpha \vec{x} \in W$. Therefore $\alpha \vec{x} \in U$ and $\alpha \vec{x} \in W$. Therefore $\alpha \vec{x} \in U \cap W$.

Therefore by Theorem 6.2, we may conclude that $(U \cap W, \mathbb{F})$ is a subspace of $(V,\mathbb{F})$. $\blacksquare$

Section 6.1 #48: Let $(V,\mathbb{F})$ be a vector space with subspace $(U,\mathbb{F})$ and $(W,\mathbb{F})$. We define the sum of $U$ and $W$ to be the set $$U+W = \{ \vec{u} + \vec{w} \colon \vec{u} \in U, \vec{w} \in W \}.$$ a.) If $(V,\mathbb{F})=\left( \mathbb{R}^{3 \times 1}, \mathbb{R} \right)$, $U$ is the x-axis, and $W$ is the $y$-axis, then what is $U+W$?
b.) If $(U,\mathbb{F})$ and $(W,\mathbb{F})$ are subspaces of $(V,\mathbb{F})$, prove that $(U+W,\mathbb{F})$ is a subspace of $(V,\mathbb{F})$.
Solution: Solution of a.): $U$ is comprised of all vectors in $\mathbb{R}^{3 \times 1}$ of the form $\begin{bmatrix} a \\ 0 \\ 0 \end{bmatrix}$ and $W$ is comprised of all vectors in $\mathbb{R}^{3 \times 1}$ of the form $\begin{bmatrix} 0 \\ b \\ 0 \end{bmatrix}$. Thus $U+W$ is all vectors in $\mathbb{R}^{3 \times 1}$ of the form $$\begin{bmatrix} a \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ b \\ 0 \end{bmatrix} = \begin{bmatrix} a \\ b \\ 0 \end{bmatrix}.$$ This describe all vectors that comprise the $xy$-plane.

Solution of b.): Let $\vec{x}, \vec{y} \in U+W$. Then we may write $\vec{x}=u_1+w_1$ and $\vec{y} = u_2+w_2$ where $u_1, u_2 \in U$ and $w_1, w_2 \in W$. Consider the sum $$\vec{x}+\vec{y} = (u_1+w_1) + (u_2+w_2) = (u_1+u_2) + (w_1+w_2).$$ Since $(U,\mathbb{F})$ is a vector space and $u_1, u_2 \in U$, then we may conclude that $u_1+u_2 \in U$. Since $(W,\mathbb{F})$ is a vector space and $w_1, w_2 \in W$, then we may conclude that $w_1+w_2 \in W$. Therefore $\vec{x}+\vec{y} \in U+W$ since $u_1+u_2 \in U$ and $w_1+w_2 \in W$.

Let $\alpha \in \mathbb{F}$ be a scalar. Consider $$\alpha \vec{x} = \alpha (u_1+w_1) = \alpha u_1 + \alpha w_1.$$ Since $(U,\mathbb{F})$ is a vector space, we may conclude that $\alpha u_1 \in U$. Since $(W,\mathbb{F})$ is a vector space, we may conclude that $\alpha w_1 \in W$. Therefore we may conclude that $\alpha \vec{x} \in U+W$ since $\alpha u_1 \in U$ and $\alpha w_1 \in W$. Therefore by Theorem 6.2, $(U+W,\mathbb{F})$ is a subspace of $(V,\mathbb{F})$. $\blacksquare$

Section 6.1 #50: Let $(W,\mathbb{F})$ be a subspace of a vector space $(V,\mathbb{F})$. Prove that $(\Delta,\mathbb{F})$ where $\Delta=\{(\vec{w},\vec{w}) \colon \vec{w} \in W\}$ is a subspace of $V \times V$.
Solution: By Exercise 49, we know that $(V \times V, \mathbb{F})$ is a vector space (with componentwise operations). Let $\vec{x}, \vec{y} \in \Delta$ and write $\vec{x}=(w_1,w_1)$ and write $\vec{y}=(w_2,w_2)$. Then consider $$\vec{x}+\vec{y}=(w_1,w_1)+(w_2,w_2)=(w_1+w_2,w_1,w_2).$$ We see that $\vec{x}+\vec{y} \in \Delta$ because its first and second components are the same.

Let $\alpha \in \mathbb{F}$ be a scalar. Consider $$\alpha \vec{x} = \alpha (w_1,w_1) = (\alpha w_1, \alpha w_1).$$ We may conclude that $\alpha \vec{x} \in \Delta$ since its first and second components are the same.

Therefore by Theorem 6.2, we may conclude that $(\Delta, \mathbb{F})$ is a subspace of $V \times V$. $\blacksquare$

Section 6.1 #53: Let $p(x)=1-2x$, $q(x)=x-x^2$, and $r(x)=-2+3x+x^2$. Determine whether $s(x)=3-5x-x^2$ lies in $\mathrm{span}\{p,q,r\}$.
Solution: This question is asking us to solve the following equation: $$(*) \hspace{35pt} c_1 p(x) + c_2 q(x) + c_3 r(x) = s(x),$$ or in other words $$c_1 (1-2x) + c_2 (x-x^2) + c_3 (-2+3x+x^2) = 3-5x-x^2.$$ Combining like terms on the left yields $$(c_1-2c_3) + (-2c_1 + c_2 + 3c_3)x +(-c_2 + c_3)x^2 = 3-5x-x^2.$$ This leads us to the following system of equations:
$$\left\{ \begin{array}{llll} c_1 & & -2c_3 & = 3 \\ -2c_1 & + c_2 & +3c_3 &= -5\\ & -c_2 & +c_3 & = -1 \\ \end{array} \right.$$ Set up the augmented matrix and getting reduced echelon form yields $$\begin{bmatrix} 1&0&-2&3 \\ -2&1&3&-5 \\ 0&-1&1&-1 \end{bmatrix} \sim \ldots \sim \begin{bmatrix} 1&0&-2&3 \\ 0&1&-1&1 \\ 0&0&0&0 \end{bmatrix}.$$ Therefore we will have a solution, since the reduced echelon form encodes the system of equations $$\left\{ \begin{array}{ll} c_1 -2c_3 = 3 \\ c_2-c_3 = 1 \\ 0=0, \end{array} \right.$$ implying we have a free variable $c_3$. Pick a value for $c_3$, say $c_3=0$, to get a particular solution of $(*)$: $$\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}.$$ note: you can check to see your answer works: plug in $c_1=3$, $c_2=1$, and $c_3=0$ into (*) and observe that it works

Section 6.1 #54: Let $p(x)=1-2x$, $q(x)=x-x^2$, and $r(x)=-2+3x+x^2$. Determine whether $s(x)=1+x+x^2$ lies in $\mathrm{span}\{p,q,r\}$.
Solution: This question is asking us to solve the following equation: $$(*) \hspace{35pt} c_1 p(x) + c_2 q(x) + c_3 r(x) = s(x),$$ or in other words $$c_1 (1-2x) + c_2 (x-x^2) + c_3 (-2+3x+x^2) = 1+x+x^2.$$ Combining like terms on the left yields $$(c_1-2c_3) + (-2c_1 + c_2 + 3c_3)x +(-c_2 + c_3)x^2 =1+x+x^2.$$ This leads us to the following system of equations:
$$\left\{ \begin{array}{llll} c_1 & & -2c_3 &=1 \\ -2c_1 & +c_2 & +3c_3 &= 1 \\ & -c_2 &+ c_3 &= 1. \end{array} \right.$$ Set up an augmented matrix and put it into reduced echelon form to get $$\begin{bmatrix} 1 & 0 & -2 & 1 \\ -2 & 1 & 3 & 1 \\ 0 & -1 & 1 &1 \end{bmatrix} \sim \ldots \sim \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0&1&-1&0 \\ 0&0&0&1 \end{bmatrix}.$$ This shows that there is no solution to $(*)$, since the last row of the reduced echelon form matrix encodes the equation $0=1$, which is always false. Therefore we must conclude that $s$ is not in the span of $p,q,$ and $r$.

Section 6.1 #61: Is $(\mathscr{P}_2,\mathbb{R})$ spanned by $\{1+x, x+x^2, 1+x^2 \}$?
Solution: We are being asked whether or not $$\mathscr{P}_2 = \mathrm{span}\left\{ 1+x, x+x^2, 1+x^2 \right\}.$$ Clearly, $$\mathrm{span}\left\{ 1+x, x+x^2, 1+x^2 \right\} \subset \mathscr{P}_2.$$ We must check whether or not $\mathscr{P}_2 \subset \mathrm{span}\left\{ 1+x, x+x^2, 1+x^2 \right\}$. To do so, let $q \in \mathscr{P}_2$ with $q(x)=c+bx+ax^2$. We must decide whether or not the following vector equation has a solution: $$(*) \hspace{35pt} c_1 (1+x) + c_2(x+x^2) + c_3 (1+x^2) = c + bx + ax^2.$$ Combining like terms on the left-hand side gives us $$(c_1+c_3)+(c_1+c_2)x+(c_2+c_3)x^2 = c+bx+ax^2.$$ This is equivalent to asking to solve the following system of equations: $$\left\{\begin{array}{llll} c_1 & &+c_3 &= c \\ c_1 & +c_2 & &=b \\ & c_2 & +c_3 &= a. \end{array} \right.$$ To solve us, set up an augmented matrix and put it into reduced echelon form: compute $$\begin{bmatrix} 1&0&1&c \\ 1&1&0&b \\ 0&1&1&a \end{bmatrix} \sim \ldots \sim \begin{bmatrix} 1 & 0 &0 & -\frac{a}{2}+\frac{b}{2}+\frac{c}{2} \\ 0&1&0&\frac{a}{2}+\frac{b}{2}-\frac{c}{2} \\ 0&0&1&\frac{a}{2}-\frac{b}{2}+\frac{c}{2} \end{bmatrix}.$$ Therefore we see that $(*)$ has the following solution: $$\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} -\frac{a}{2}+\frac{b}{2}+\frac{c}{2} \\ \frac{a}{2}+\frac{b}{2}-\frac{c}{2} \\ \frac{a}{2}-\frac{b}{2}+\frac{c}{2} \end{bmatrix},$$ and so we can conclude that $\mathscr{P}_2 \subset \mathrm{span}\left\{ 1+x, x+x^2, 1+x^2 \right\}$, allowing us to conclude that $\mathscr{P}_2 = \mathrm{span}\left\{ 1+x, x+x^2, 1+x^2 \right\}$.

Section 6.1 #62: Is $(\mathscr{P}_2,\mathbb{R})$ spanned by $\{1+x+2x^2, 2+x+2x^2, -1+x+2x^2\}$?
Solution: We are being asked whether or not $$\mathscr{P}_2 = \mathrm{span} \{ 1+x+2x^2, 2+x+2x^2, -1+x+2x^2 \}.$$ Clearly, $$\mathrm{span} \{ 1+x+2x^2, 2+x+2x^2, -1+x+2x^2 \} \subset \mathscr{P}_2.$$ We must check whether or not $\mathscr{P}_2 \subset \mathrm{span} \{ 1+x+2x^2, 2+x+2x^2, -1+x+2x^2 \}$. To do so, let $q \in \mathscr{P}_2$ with $q(x)=c+bx+ax^2$. We must decide whether or not the following vector equation has a solution: $$(*) \hspace{35pt} c_1 (1+x+2x^2) + c_2 (2+x+2x^2) + c_3 (-1+x+2x^2) = c + bx + ax^2.$$ Combining like terms on the left-hand side yields $$(c_1 + 2c_2 - c_3)+(c_1 + c_2 +c_3)x+(2c_1 + 2c_2 + 2c_3)x^2 = c+bx+ax^2.$$ This is equivalent to the following system of equations: $$\left\{ \begin{array}{llll} c_1 & +2c_2 & -c_3 &= c \\ c_1 &+c_2 &+c_3 &= b \\ 2c_1 &+2c_2 &+2c_3 &= a. \end{array} \right.$$ In order to solve this system, we write an augmented matrix and put it into reduced echelon form: compute $$\begin{bmatrix} 1&2&-1&c \\ 1&1&1&b \\ 2&2&2&a \end{bmatrix} \sim \ldots \sim \begin{bmatrix} 1&0&3&0 \\ 0&1&-2&0 \\ 0&0&0&1 \end{bmatrix}.$$ We conclude from this reduced echelon form that there is no solution to $(*)$ (for all possible $q \in \mathscr{P}_2$), and so we may conclude that $$\mathscr{P}_2 \neq \mathrm{span} \{ 1+x+2x^2, 2+x+2x^2, -1+x+2x^2 \}.$$