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Section 6.5 #15: Determine whether the linear transformation is one-to-one and/or onto: $T \colon \mathbb{R}^{2 \times 1} \rightarrow \mathbb{R}^{2 \times 1}$ defined by $T \left( \begin{bmatrix} x \\ y \end{bmatrix} \right) = \begin{bmatrix} 2x-y \\ x+2y \end{bmatrix}$.
Solution: Is it one-to-one? Assume that $T\left( \begin{bmatrix} a \\ b \end{bmatrix} \right) = T\left( \begin{bmatrix} c \\ d \end{bmatrix} \right)$. We seek to determine whether or not $\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} c \\ d \end{bmatrix}$. To determine this, expand the equation $T\left( \begin{bmatrix} a \\ b \end{bmatrix} \right) = T\left( \begin{bmatrix} c \\ d \end{bmatrix} \right)$ to get $$\begin{bmatrix} 2a-b \\ a+2b \end{bmatrix} = \begin{bmatrix} 2c-d \\ c+2d \end{bmatrix}$$ Solve this equation for, say, $a$ and $b$ (could also solve for $c$ and $d$) by setting up the appropriate augmented matrix and putting it into reduced echelon form (or any way you choose to do it): compute $$\begin{bmatrix} 2 & -1 & 2c-d \\ 1 & 2 & c+2d \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & c \\ 0&1&d \end{bmatrix}.$$ Thus our solution shows $a=c$ and $b=d$, hence $\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} c \\ d \end{bmatrix}$. Therefore we have shown that $T$ is one-to-one.

To determine if $T$ is onto, pick an arbitrary $\begin{bmatrix} \alpha \\ \beta \end{bmatrix} \in \mathbb{R}^{2 \times 2}$. We ask the question "is there a $\begin{bmatrix} a \\ b \end{bmatrix}$ such that $T \left( \begin{bmatrix} a \\ b \end{bmatrix} \right) = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$?". Expand this equation by using the definition of $T$ to see $$\begin{bmatrix} 2a-b \\ a+2b \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}.$$ Solving this yields $$\begin{bmatrix} 2 & -1 & \alpha \\ 1 &2 & \beta \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & \frac{2\alpha+\beta}{5} \\ 0&1& \frac{2\beta-\alpha}{5} \end{bmatrix}.$$ This shows that there is such a $\begin{bmatrix} a \\ b \end{bmatrix}$. Therefore $T$ is onto.

Section 6.5 #16: Determine whether the linear transformation is one-to-one and/or onto: $T \colon \mathbb{R}^{2 \times 1} \rightarrow \mathscr{P}_2$ defined by $$T \left( \begin{bmatrix} a \\ b \end{bmatrix} \right) = (a-2b) + (3a+b)x + (a+b)x^2.$$ Solution: To check if $T$ is one-to-one, we assume $T \left( \begin{bmatrix} a \\ b \end{bmatrix} \right) = T \left( \begin{bmatrix} c \\ d \end{bmatrix} \right)$. Expanding this yields $$(a-2b)+(3a+b)x+(a+b)x^2 = (c-2d) + (3c+d)x + (c+d)x^2.$$ Solve this equation for $a$ and $b$: compute $$\begin{bmatrix}1&-2&c-2d \\ 3&1&3c+d \\ 1&1&c+d \end{bmatrix} \sim \begin{bmatrix} 1&0&c \\ 0&1&d \\ 0&0&0 \end{bmatrix}.$$ Therefore $a=c$ and $b=d$. Thus $T$ is one-to-one.

To check if $T$ is onto, let $\alpha x^2+\beta x+\gamma \in \mathscr{P}_2$ be arbitrary. We ask if there is a $\begin{bmatrix} a \\ b \end{bmatrix} \in \mathbb{R}^{2 \times 1}$ such that $T \left( \begin{bmatrix} a \\ b \end{bmatrix} \right) = \alpha x^2+\beta x+\gamma$. Expanding this equation yields $$(a-2b)+(3a+b)x+(a+b)x^2 = \gamma + \beta x + \alpha x^2.$$ Solve for $a,b,$ and $c$: compute $$\begin{bmatrix} 1&-2&\gamma \\ 3&1&\beta \\ 1&1&\alpha \end{bmatrix} \sim \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}.$$ From this we observe there is no solution and hence $T$ is not onto.

Section 6.5 #21: Determine whether $V$ and $W$ are isomorphic. If they are, given an explicit isomorphism $T \colon V \rightarrow W$: $V=D_3$ (the diagonal $3\times 3$ real-valued matrices) and $W=\mathbb{R}^{3 \times 1}$.
Solution: The diagonal $3 \times 3$ matrices are of the form $\begin{bmatrix} a & 0 & 0 \\ 0&b&0 \\ 0&0&c \end{bmatrix}$. Therefore we may use the following set as a basis: $\left\{ \begin{bmatrix} 1 & 0 & 0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}, \begin{bmatrix} 0&0&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix}, \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&1 \end{bmatrix} \right\}$. Therefore the dimension of the diagonal $3 \times 3$ matrices is $3$. It is well-known that the dimension of $\mathbb{R}^{3 \times 1}$ is $3$. Therefore these vector spaces are isomorphic. Define an isomorphism $T \colon D_3 \rightarrow \mathbb{R}^{3 \times 1}$ by $$T \left( \begin{bmatrix} a&0&0 \\ 0&b&0 \\ 0&0&c \end{bmatrix} \right) = \begin{bmatrix} a \\ b \\ c \end{bmatrix} .$$ Such a function is "obviously" an isomorphism, but it can be checked by showing $T$ is one-to-one, onto, and a linear transformation (good practice!).

Section 6.5 #25: Determine whether $V$ and $W$ are isomorphic. If they are, give an explicit isomorphism $T \colon V \rightarrow W$: $V=\mathbb{C}$ and $W=\mathbb{R}^{2 \times 1}$.
Solution: How this problem is interpreted may change the answer:
Interpretation 1: If we are comparing the spaces $(V,\mathbb{F}_1)=(\mathbb{C},\mathbb{C})$ and $(W,\mathbb{F}_2)=(\mathbb{R}^{2 \times 1},\mathbb{C})$. Then the question makes no sense because $(\mathbb{R}^{2 \times 1},\mathbb{C})$ is not closed under scalar multiplication. So it is not a vector space.

Interpretation 2: If we are comparing $(V,\mathbb{F}_1)=(\mathbb{C},\mathbb{C})$ and $(W,\mathbb{F}_2)=(\mathbb{R}^{2\times 1},\mathbb{R})$, then the problem is that the fields do not match. Mismatched fields make checking if $T$ is a linear transformation troublesome: $T(\alpha \vec{u} + \beta \vec{v})=\alpha T(\vec{u})+\beta T(\vec{v})$.

Interpretation 3: We are comparing $(V,\mathbb{F}_1)=(\mathbb{C},\mathbb{R})$ and $(W,\mathbb{F})=(\mathbb{R}^{2 \times 1},\mathbb{R})$. In this case, we can give a coherent answer. We write $\mathbb{C}= \left\{ a+bi \colon a, b \in \mathbb{R}\right\}=\mathrm{span}\{1,i\}$ (this is a basis!), and so its dimension is $2$. On the other hand, $\mathbb{R}^{2 \times 1} = \mathrm{span} \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\}$ (this is a basis!) so its dimension is $2$. Therefore the two vector spaces are isomorphic. An isomorphism $T \colon \mathbb{C} \rightarrow \mathbb{R}^{2 \times 1}$ may be defined by $$T(a+bi) = \begin{bmatrix} a \\ b \end{bmatrix}.$$

Section 6.5 #28: Show that $T \colon \mathscr{P}_2 \rightarrow \mathscr{P}_2$ defined by $T(p(x))=p(x)+p'(x)$ is an isomorphism.
Solution: We must show that $T$ is one-to-one, is onto, and is a linear transformation. To show $T$ is onto, assume that $T(p)=T(q)$ for $p=ax^2+bx+c$ and $q=dx^2+ex+f$. Then the equation $T(p)=T(q)$ becomes $$(ax^2+bx+c)+(2ax+b)=(dx^2+ex+f)+(2dx+e).$$ Combining like-terms on each side yields $$(a)x^2+(b+2a)x+(c+b) = (d)x^2+(e+2d)x+(f+e)$$ Let us solve for $a,b,$ and $c$: compute $$\begin{bmatrix} 1&0&0&d \\ 2&1&0&e+2d \\ 0&1&1&f+e \end{bmatrix} \sim \begin{bmatrix} 1&0&0&d \\ 0&1&0&e \\ 0&0&1&f \end{bmatrix}.$$ This shows that $a=d$, $b=e$, and $c=f$, proving that $T$ is one-to-one.

Now to check if $T$ is onto, let $\alpha x^2 + \beta x + \gamma \in \mathscr{P}_2$. We seek a $ax^2+bx+c \in \mathscr{P}_2$ such that $T(ax^2+bx+c)=\alpha x^2+\beta x + \gamma$. Simplifying the left-hand side yields $$(ax^2+bx+c)+(2ax+b) = \alpha x^2 + \beta x + \gamma,$$ and combining like-terms yields $$(a)x^2 + (2a+b)x + (b+c) = \alpha x^2 + \beta x + \gamma.$$ Therefore we solve for $a,b,$ and $c$: compute $$\begin{bmatrix} 1&0&0&\alpha \\ 2&1&0&\beta \\ 0&1&1&\gamma \end{bmatrix} \sim \begin{bmatrix} 1&0&0&\alpha \\ 0&1&0&\beta-2\alpha \\ 0&0&1&2\alpha-\beta+\gamma \end{bmatrix}$$ Therefore $T$ is onto.

Now we check that $T$ is a linear transformation. Let $p=ax^2+bx+c$ and $q=dx^2+ex+f$ and let $\alpha$ and $\beta$ be scalars. Compute $$\begin{array}{ll} T(\alpha p + \beta q) &= T((\alpha a + \beta d)x^2 + (\alpha b + \beta e)x + (\alpha c + \beta f)) \\ &=[(\alpha a + \beta d)x^2 + (\alpha b + \beta e)x + (\alpha c + \beta f) ] + [2(\alpha a + \beta d)x + (\alpha b + \beta e)] \\ &= \alpha [(ax^2+bx+c) + (2a+b)] + \beta [(cx^2+dx+f)+(2dx+e)] \\ &= \alpha T(p) + \beta T(q), \end{array}$$ showing that $T$ is a linear transformation.

Problem A: ("Laguerre polynomials") Let $T_n \colon \mathscr{P}_2 \rightarrow \mathscr{P}_2$ be defined by $T_n(p)=xp''+(1-x)p'+np$ for $n=0,1,2$. Find $\mathrm{ker}(T_n)$ for $n=0,1,2$.
Solution:
$\mathbf{\underline{n=0}}$
By definition, $T_0(p) = xp'' + (1-x)p'$. A polynomial $p=ax^2+bx+c$ is in $\mathrm{ker}(T_0)$ provided that $T_0(ax^2+bx+c)=0$, i.e. $$x(2a) + (1-x)(2ax+b) = 0.$$ Combining like-terms on the left yields $$-2a x^2 + (-b) x + b=0.$$ Thus we have $-2a=0$, $4a-b=0$, and $b=0$. Hence $a=b=0$. Therefore we have shown $$\mathrm{ker}(T_0)=\{ax^2+bx+c \colon a=b=0, c \in \mathbb{R}\} = \{c \colon c \in \mathbb{R}\} = \mathrm{span}\{1\}.$$
$\mathbf{\underline{n=1}}$
By definition, $T_1(p)=xp''+(1-x)p'+p$. If $p=ax^2+bx+c \in \mathrm{ker}(T_1)$ then $T(p)=0$, i.e. $$(-2a)x + (1-x)(2ax+b) + (ax^2+bx+c)=0.$$ Combining like-terms on the left yields $$(-a)x^2+0x+(b+c)=0.$$ This leads to the equations $-a=0$ and $b+c=0$. Hence $a=0$ and $b=-c$. Therefore $$\mathrm{ker}(T_1) = \{ax^2+bx+c \colon a,b,c \in \mathbb{R}, a=0, b=-c \} = \{ -cx+c \colon c \in \mathbb{R}\} = \mathrm{span}\{-x+1\}.$$ $\mathbf{\underline{n=2}}$
By definition, $T_2(p)=xp''+(1-x)p'+2p$. Plugging in $p=ax^2+bx+c$ yields $$x(2a)+(1-x)(2ax+b) + 2(ax^2+bx+c)=0.$$ Combining like-terms yields $$(4a+b)x+(2c+b)=0.$$ This yields $4a+b=0$ and $2c+b=0$. Hence $c=-\frac{b}{2}$ and $a=-\frac{b}{4}$. Thus $$\mathrm{ker}(T_2) = \{ ax^2+bx+c \colon a,b,c \in \mathbb{R}, c=-\frac{b}{2}, a=-\frac{b}{4}\} = \left\{-\frac{b}{4}x^2 + bx - \frac{b}{2} \colon b \in \mathbb{R} \right\}=\mathrm{span}\left\{-\frac{x^2}{4} + x - \frac{1}{2} \right\}.$$