\documentclass{amsart}
\usepackage{amsmath,amssymb}

\DeclareMathOperator{\im}{Im}

\newcommand{\R}{{\mathbb R}}

\title{Math 23A SOLUTION SET 3}
\author{George Kirkup}

\begin{document}
\maketitle
\begin{enumerate}

\item
$$0v = (0 + 0)v = 0v + 0v$$
Therefore
$$0v + (-(0v)) = 0v + 0v + (-(0v))$$
and hence
$$0 = 0v$$

\item
\begin{enumerate}
\item
Suppose
$$a_1 l_1 + \dots + a_r l_r + b_1 v_1 + \dots + b_m v_m = 0$$
for some $\{a_i\}$ and $\{b_j\}$.  Since $F(0) = 0$, we have
$$F(a_1 l_1 + \dots + a_r l_r + b_1 v_1 + \dots + b_m v_m) = 0.$$
Expanding by linearity we have
$$a_1 F(l_1) + \dots + a_r F(l_r) + b_1 F(v_1) + \dots + b_m F(v_m) = 0.$$
Since $F(l_i) = 0$ and $F(v_j) = w_j$ this reduces to
$$ b_1 w_1 + \dots + b_m w_m = 0.$$
Since $\{w_i\}$ are linearly independent this means that $b_i = 0$ for all $i$.
That leaves us with $$a_1 l_1 + \dots + a_r l_r = 0,$$ but since $\{l_i\}$ are
linearly independent, $a_i = 0$ for all $i$.
Therefore $\{l_1, \ldots , l_r ,v_1, \ldots , v_m\}$ are linearly independent.

\item
Given $v \in V$, $F(v) \in W$ so we can write 
$$F(v) = a_1 w_1 + \dots + a_m w_m$$ since $\{w_i\}$ spans $W$.  Therefore 
$$F(v - (a_1 v_1 + \dots + a_m v_m)) = F(v) - (a_1 w_1 + \dots + a_m w_m) = 0$$
so $v - (a_1 v_1 + \dots + a_m v_m)$ is in the kernel of $F$.  Since $\{l_i\}$
spans $\ker F$ we can write
$$v - (a_1 v_1 + \dots + a_m v_m) = b_1 l_1 + \dots + b_r l_r$$ and hence
$$v = a_1 v_1 + \dots + a_m v_m + b_1 l_1 + \dots + b_r l_r$$ so $v$ is in the
span of $\{l_1, \ldots , l_r, v_1, \ldots , v_m\}$.

\item
$\{l_1, \ldots , l_r\}$ a basis for $L$ means that they are linearly independent
and span $L$ and $\{w_1, \ldots , w_r\}$ a basis for $W$ means that they are
linearly independent and span $W$.  Combining parts (a) and (b) we clearly have
that $\{l_1, \ldots , l_r, v_1, \ldots , v_m\}$ is a basis of $V$.

\end{enumerate}

\item
Claim:  $v^1 = \frac{e^1 + e^2}{2}$ and $v^2 = \frac{e^1 - e^2}{2}$
\begin{proof}
It is clear that $v^1$ and $v^2$ are indeed elements of $V^*$ so we just check
that they have the desired values on $v_1$ and $v_2$.
$$
v^1 (v_1) = \frac{e^1 + e^2}{2} (e_1 + e_2) $$
$$          = \frac{e^1(e_1 + e_2) + e^2(e_1 + e_2)}{2}$$
$$          = \frac{(1 + 0) + (0 + 1)}{2}$$
$$	  = 1.$$
Likewise,
$$
v^1 (v_2) = \frac{e^1 + e^2}{2} (e_1 - e_2)$$
$$          = \frac{e^1(e_1 - e_2) + e^2(e_1 - e_2)}{2}$$
$$          = \frac{(1 - 0) + (0 - 1)}{2}$$
$$          = 0$$
The other calculations are similar and hence will not be carried out.
\end{proof}

\item
Since $\im F$ is a subspace of $\R$ its dimension is either $1$ or $0$.  However,
it is not $0$ because the image of $e_1$, the first standard basis vector, is
$1$.  Therefore, $\dim (\im F) = 1$.  Since $\dim (\R^n) = n$, the dimension
formula gives us $\dim (\ker F) = n-1$.

\item
\begin{enumerate}
\item
$$T(a p_1 + p_2) = (x^2 + 1) (a p_1 + p_2)'' - (x)(a p_1 + p_2)' + (2)(a p_1 +
p_2)$$
By well-known propertiesof the derivative we have
$$
T(a p_1 + p_2) = (x^2 + 1) (a p_1'' + p_2'') - (x)(a p_1' + p_2') + (2)(a p_1 +
p_2)$$
$$= (x^2 + 1) (a p_1'') + (x^2 + 1) p_2'' - (x)(a p_1') - (x)p_2' + (2)(a
p_1) + (2) p_2$$
$$= a((x^2 + 1)p_1'' - (x) p_1' + (2) p_1) + (x^2 + 1)p_2'' - (x) p_2' + 
(2) p_2$$
$$= a(T(p_1)) + T(p_2)$$


\item
We evaluate the map 
$\Phi_{\{p\}}^{-1} \circ T \circ \Phi_{\{p\}}$ 
on the standard basis.  
$$
\Phi_{\{p\}}^{-1} \circ T \circ \Phi_{\{p\}}(e_1) = \Phi_{\{p\}}^{-1}
\circ T (1)$$
$$= \Phi_{\{p\}}^{-1} (2)$$
$$= 2e_1$$

$$\Phi_{\{p\}}^{-1} \circ T \circ \Phi_{\{p\}}(e_2) =
\Phi_{\{p\}}^{-1} \circ T (x)$$
$$= \Phi_{\{p\}}^{-1} (x)$$
$$= e_2$$

Similarly $\Phi_{\{p\}}^{-1} \circ T \circ \Phi_{\{p\}}(e_3) = 2e_1 +
2e_3.$
Thus the columns of the matrix are as desired.

\item
Extending $T$ to an operator from $P_3$ to $P_3$ it is clear that the first
three columns are the same (with $0$s in the last entry) as above.
$T(x^3) = (x^2 + 1)(6x) - x(3x^2) + 2(x^3)$ so the third column is
$$\begin{matrix}
0 \\ 6 \\ 0 \\ 5
\end{matrix}$$
I will now treat the general case.
If $T$ is a map from $P_n$ to $P_n$ then we evaluate $T(x^j)$ for all $j \leq
n$
$$
T(x^j) = (x^2 + 1)(j(j-1)x^{j-2}) - x(j(x^{j-1})) + 2(x^j)$$
$$= (j(j-1) - j + 2)x^j + (j(j-1))x^{j-2}$$
$$= (j^2 - 2j + 2)x^j + (j^2 - j)x^{j-2}$$

Since the $k^{th}$ column is $T(x^{k-1})$, we have the following matrix:
$$\begin{pmatrix}
2 & 0 & 2 & 0      & \ldots      & 0      & 0  \\
0 & 1 & 0 & \ddots & 0           & 0      & 0  \\ 
0 & 0 & 2 & \ddots & k^2 - 3k +2 & \ddots & \vdots \\
0 & 0 & 0 & \ddots & 0           & \ddots & 0  \\
0 & 0 & 0 & \ddots & k^2 -4k +5  & \ddots & n^2 - n \\
0 & 0 & 0 & \ddots & 0           & \ddots & 0  \\ 
0 & 0 & 0 & \ldots & 0           & 0      & n^2-2n+2 \\
\end{pmatrix}$$
\end{enumerate}
\end{enumerate}
\end{document}
