\documentclass[12pt]{math23a}
\problemset{2}
\year1999\month10\day13
\author{Li-Chung Chen}

\DeclareMathOperator{\Img}{Img}
\DeclareMathOperator{\Mat}{Mat}
\newcommand{\R}{{\mathbb R}}
\def\theenumii{\Alph{enumii}}

\begin{document}

\item[Themes]
\ \begin{itemize}
\item Theme 1 of assignment 2: Give an explicit example when you need to show
something exists (such as a counterexample).
(See 2.4.5(a), 2.5.3, 2.6.5(b).)

\item Theme 2 of assignment 2: Proofs using symbols are usually the cleanest.
(See 2.4.6(c), 2.4.6(d), 2.6.4(b).)
\end{itemize}

\item[2.4.5]
\ \begin{enumerate}
\item % (a)
We want to investigate the solutions for $a_1,a_2,a_3$ in the equation
	$$a_1 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} +
	a_2 \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} +
	a_3 \begin{bmatrix} 0 \\ 1 \\ \alpha \end{bmatrix} =
	\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$
We get three equations:
\begin{align*}
a_1 + a_2 &= 0 \\
a_1 + 2a_2 + a_3 &= 0 \\
a_2 + \alpha a_3 &= 0.
\end{align*}
The first two yield $a_1 = a_3$ and $a_2 = -a_3$, so the third equation
becomes $0 = -a_3 + \alpha a_3 = (\alpha - 1)a_3$.  At this point, many of
you said $\alpha - 1 = 0$ and $\alpha = 1$, without giving too much more
justification.  The reasoning should proceed as follows. 

Case 1: $\alpha \neq 1$.  Then $\alpha - 1 \neq 0$, and $a_3$ must be $0$.
But then $a_1 = 0$ and $a_2 = 0$.  Hence the only linear combination of
the three given vectors that gives 0 is the trivial linear combination.  This
means that the vectors are linearly independent.

Case 2: $\alpha = 1$.  Then $(a_1, a_2, a_3) = (1, -1, 1)$ is a solution, so
	$$ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} - 
	 \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} +
	 \begin{bmatrix} 0 \\ 1 \\ \alpha \end{bmatrix} =
	\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
and the vectors are linearly dependent.

Therefore, the vectors are linearly dependent if and only if $\alpha = 1$.

When you are asked to show the linear dependence of some vectors, it is
safest to provide some nontrivial linear combination of them that equals the
zero vector.

\item % (b)
Note that this problem should have asked for a plane \emph{through the origin},
because any three points in $\R^3$ are in a plane.

The only value of $\alpha$ to consider is $1$.  The equation of
a plane through the origin is $ax + by + cz = 0$, so we just plug in each of
the three vectors to get three equations:

\begin{align*}
a + b &= 0, \\
a + 2b + c &= 0, \\
b + c &= 0.
\end{align*}

Solving, we get $a = c$ and $b = -c$.  The equation $cx - cy + cz = 0$ gives
the same plane for any nonzero $c$, so pick $c = 1$ and get the equation
$x - y + z = 0$ (other answers are possible).  The three vectors automatically
satisfy this equation because that was how we got $a$, $b$, and $c$.
\end{enumerate}

\item[2.4.6]
\ \begin{enumerate}
\item % (a)
$v_1, \dots, v_k$ are linearly independent if and only if the only time
$c_1v_1 + c_2v_2 + \dots + c_kv_k = 0$ is when $c_1 = c_2 = \dots = c_k = 0$.
This is equivalent to saying that for any vector $w \in \R^n$, there is at
most one way of writing $w$ as a linear combination of $v_1,\dots,v_k$.  In
different situations, one formulation might be easier to use than the other.

$v_1, \dots, v_k$ span $\R^n$ if and only if every vector $w \in R^n$ can
be written as a linear combination of $v_1,\dots,v_k$.

$v_1, \dots, v_k$ form a basis of $\R^n$ if and only if they are linearly
independent and span $\R^n$.  

\item % (b)
$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},
A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix},
A^2 = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}$, and
$A^3 = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}.$
%To see if these elements are linearly independent, we solve for
%$c_1,c_2,c_3,c_4$ in the equation $c_1 I + c_2 A + c_3 A^2 + c_4 A^3
%= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.  This is essentially
%four linear equations in terms of the $c_i$'s, and solving we get a
%nontrivial solution $(c_1,c_2,c_3,c_4) = (-3,-2,1,0)$.  Hence
%$I,A,A^2,A^3$ are not linearly independent.
%Here is another way to look at the problem.
We can regard $2 \times 2$ matrices as $4 \times 1$ vectors in $\R^4$ by
considering a matrix $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}
\end{bmatrix}$ as the vector $\begin{bmatrix} a_{11} \\ a_{12} \\ a_{21}
\\ a_{22} \end{bmatrix}$.  Then every linear combination
$c_1 I + c_2 A + c_3 A^2 + c_4 A^3$ of $I$, $A$,
$A^2$, and $A^3$ can be regarded as
$$c_1 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} +
c_2 \begin{bmatrix} 1 \\ 2 \\ 2 \\ 1 \end{bmatrix} +
c_3  \begin{bmatrix} 5 \\ 4 \\ 4 \\ 5 \end{bmatrix} +
c_4  \begin{bmatrix} 13 \\ 14 \\ 14 \\ 13 \end{bmatrix}
= \begin{bmatrix}
1 & 1 & 5 & 13 \\
0 & 2 & 4 & 14 \\
0 & 2 & 4 & 14 \\
1 & 1 & 5 & 13
\end{bmatrix}
\begin{bmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{bmatrix}.$$
Hence the span of $I$, $A$, $A^2$, and $A^3$ is the image of this
$4 \times 4$ matrix; the dimension of the span is $2$.  Elements of the
kernel correspond to linear combinations that give the zero element.
We can get many different nonzero elements in the kernel, such as
$\begin{bmatrix} -3 \\ -2 \\ 1 \\ 0 \end{bmatrix}$.  This means that
$-3I - 2A + A^2 + 0A^3 = 0$ and that $I,A,A^2,A^3$ are not linearly
independent.

If you did not see this connection between $\Mat(2,2)$ and $\R^4$, it is
perfectly okay.  It is acceptable to prove that $I$ and $A$ form a basis
for $V$, but then you need to prove that they are linearly independent
and span $V$.  One way to show they span $V$: Note that
$A^2 = 2A + 3I$ and $A^3 = 7A + 6I$.  So if an element $B$ is in $V$,
then for some $c_1,c_2,c_3,c_4$, $B = c_1 I + c_2 A + c_3 A^2 + c_4 A^3
= c_1 I + c_2 A + c_3 (2A + 3I) + c_4 (7A + 6I)
= (c_1 + 3c_3 + 6c_4)I + (c_2 + 2c_3 + 7c_4)A$.  Hence any element in $V$
is a linear combination of $I$ and $A$.

\item % (c)
A good way to start is to try to figure out what $W$ looks like.  So let us
take a matrix $B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.  Then $B$ is 
in $W$ if and only if $AB = BA$, which is the same as
$$\begin{bmatrix} a+2c & b+2d \\ 2a+c & 2b+d \end{bmatrix} = 
\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}
\begin{bmatrix} a & b \\ c & d \end{bmatrix}=
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}=
\begin{bmatrix} a+2b & 2a+b \\ c+2d & 2c+d \end{bmatrix}.$$
This is equivalent to the four equations
\begin{align*}
a+2c &= a+2b\\
b+2d &= 2a+b\\
2a+c &= c+2d\\
2b+d &= 2c+d.
\end{align*}

Solving the equations, we get $a=d$ and $b=c$.  Therefore, these are necessary
and sufficient conditions for $B$ to be in $W$.  Hence
$$W = \left\{\begin{bmatrix}a & b \\ b & a \end{bmatrix}: a,b\in \R\right\}.$$
In this form, it is easy to show that $W$ is a subspace of $\Mat(2,2)$, and that
$W$ has dimension $2$ (due to the basis $\left\{
\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix},
\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}\right\}$).

It is also possible to show that $W$ is a subspace without characterizing $W$
as above (though it is harder to find $W$'s dimension without characterizing
$W$).  To show that $W$ is a subspace of $\Mat(2,2)$, there are three things to 
show:

(i) $W$ is nonempty. (Almost all of you forgot about this.) \\
(ii) For any matrices $B_1$ and $B_2$ in $W$, $B_1 + B_2$ is also in $W$. \\
(iii) For any real number $a$ and matrix $B$ in $W$, $aB$ is also in $W$.

For (i), it suffices to note that the zero matrix satisfies $AB = BA$ and is
in $W$.  For (ii), suppose $B_1$ and $B_2$ are in $W$.  Then
$A(B_1 + B_2) = AB_1 + AB_2 = B_1A + B_2A = (B_1+B_2)A$ (here we use the
fact that $B_1$ and $B_2$ are in $W$).  Thus $B_1+B_2$ is in $W$ by definition.
For (iii), suppose $a$ is real and $B$ is in $W$.  Then
$A(aB) = a(AB) = a(BA) = B(aA)$ (using the fact that $AB = BA$), so $aB$ is
in $W$ by definition.

\item % (d)
To show that $V \subset W$, we need to show that any matrix $B$ in $V$ is
also in $W$, i.e. $AB = BA$.  Now if $B$ is in $V$, then $B$ can be written
as a linear combination $c_1 I + c_2 A + c_3 A^2 + c_4 A^3$.  This means
that $AB = A(c_1 I + c_2 A + c_3 A^2 + c_4 A^3)
= A(c_1I) + A(c_2A) + A(c_3A^2) + A(c_4A^3)
= c_1A + c_2A^2 + c_3A^3 + c_4A^4$.  On the other hand,
$BA = (c_1 I + c_2 A + c_3 A^2 + c_4 A^3)A
= (c_1 I)A + (c_2 A)A + (c_3 A^2)A + (c_4 A^3)A
= c_1A + c_2A^2 + c_3A^3 + c_4A^4$.  Therefore, $AB = BA$ and $B$ is in $W$.
If you have already found a basis for $V$, it is also possible to do this problem 
by showing that every basis element of $V$ is in $W$.
(Why does this imply that $V \subset W$?)

The facts that $\dim V = 2 = \dim W$ and $V \subset W$ imply that
$V = W$.  Alternatively, you can show that $W \subset V$ (for which it suffices
to show that each basis element of a basis for $W$ is in $V$).
\end{enumerate}

\item[2.4.8]
The two different kinds of coordinates must specify the same vector, so
$xe_1 + xe_2 = uv_1 + vv_2$.  This means that
$\begin{bmatrix}x \\ y \end{bmatrix} =
\begin{bmatrix}u+v \\ u+3v \end{bmatrix}$.
\begin{itemize}
\item
To go from the $\{v_1,v_2\}$ coordinates to the $\{e_1,e_2\}$
coordinates, we start with $u$ and $v$ and must solve for $x$ and $y$
in terms of $u$ and $v$.  The above equation immediately gives $x = u+v$ and
$y = u+3v$.  Thus we go from $(u,v)$ to $(u+v,u+3v)$.
\item
To go from the $\{e_1,e_2\}$ coordinates to the $\{v_1,v_2\}$
coordinates, we start with $x$ and $y$ and must solve for $u$ and $v$ in terms of
$x$ and $y$.  The above vector equation gives $u = \frac{3x-y}{2}$ and
$v = \frac{-x+y}{2}$.  Thus we go from $(x,y)$ to
$(\frac{3x-y}{2},\frac{-x+y}{2})$.
\end{itemize}

The vector $\begin{bmatrix}3 \\ -5 \end{bmatrix}$ is just $(3,-5)$ in
the $\{e_1,e_2\}$ coordinates (i.e. $x$ and $y$), so we use the second
set of equations above to get $(\frac{3(3)-(-5)}{2},\frac{-3+(-5)}{2})
= (7,-4)$ in the $\{v_1,v_2\}$ coordinates.

\item[2.4.9]
\ \begin{enumerate}
\item % (a)
First suppose that $v_1,\dots,v_n$ are linearly independent.  Recall that
a function is one-to-one when different elements in the domain get mapped
to different elements in the range.  So suppose
$\begin{psmallmatrix} a_1 \\ \vdots \\ a_n \end{psmallmatrix}$ and
$\begin{psmallmatrix} b_1 \\ \vdots \\ b_n \end{psmallmatrix}$ are distinct
elements in $\R^n$.  They get mapped to
$P_{\{v\}}\begin{psmallmatrix} a_1 \\ \vdots \\ a_n \end{psmallmatrix}
= \sum a_iv_i$ and
$P_{\{v\}}\begin{psmallmatrix} b_1 \\ \vdots \\ b_n \end{psmallmatrix}
= \sum b_iv_i$.  If $\sum a_iv_i = \sum b_iv_i$, then they represent two distinct
linear combination representations of the same element.  But this cannot
happen by the linear independence of $v_1,\dots,v_n$.  Therefore,
$\sum a_iv_i \neq \sum b_iv_i$, and indeed different elements get mapped
to different elements.  This means that $P_{\{v\}}$ is one-to-one.

Next suppose that $P_{\{v\}}$ is one-to-one.  One way to prove that
$v_1,\dots,v_n$ are linearly independent is to show that for any
$w$ in $\R^m$, $w$ can be written as a linear combination of
$v_1,\dots,v_n$ in at most one way.  We proceed by contradiction and suppose
that this is not true.  Then there exists a $w$ in $\R^m$ that can be written
in two different ways as a linear combation, say as
$\sum a_iv_i$ and $\sum b_iv_i$.  But $\sum a_iv_i =
P_{\{v\}}\begin{psmallmatrix} a_1 \\ \vdots \\ a_n \end{psmallmatrix}
$ and $\sum b_iv_i
= P_{\{v\}}\begin{psmallmatrix} b_1 \\ \vdots \\ b_n \end{psmallmatrix}
$, so the two different vectors
$\begin{psmallmatrix} a_1 \\ \vdots \\ a_n \end{psmallmatrix}$ and
$\begin{psmallmatrix} b_1 \\ \vdots \\ b_n \end{psmallmatrix}$ get mapped
to the same element by $P_{\{v\}}$, contradicting the fact that $P_{\{v\}}$
is one-to-one.  Therefore, $v_1,\dots,v_n$ are indeed linearly independent.

\item % (b)
First suppose that $v_1,\dots,v_n$ span $\R^m$.  Then for any $w$ in $\R^m$,
it is possible to write $w$ as $\sum c_iv_i$ for some $c_1,\dots,c_n$.  But
then $w = \sum c_iv_i =
P_{\{v\}}\begin{psmallmatrix} c_1 \\ \vdots \\ c_n \end{psmallmatrix}$.
What we just showed is that any element in $\R^m$ is in the image of
$P_{\{v\}}$.  Therefore, $P_{\{v\}}$ is onto.

Next suppose that $P_{\{v\}}$ is onto.  Then for any $w$ in $\R^m$, there
exists element 
$\begin{psmallmatrix} a_1 \\ \vdots \\ a_n \end{psmallmatrix}$ in
$\R^n$ such that
$w = P_{\{v\}}\begin{psmallmatrix} a_1 \\ \vdots \\ a_n \end{psmallmatrix}$
.  But this means exactly that $w = \sum c_iv_i$ and that $w$ is in the
span of $v_1,\dots,v_n$.  Because $w$ is an arbitrary element in $\R^m$,
we have just shown that $v_1,\dots,v_n$ span $\R^m$.

Note that it is possible to avoid a two-part proof by going from 
``$v_1,\dots,v_n$ span $\R^m$'' to ``$P_{\{v\}}$ is onto'' by a series of
if and only if statements.  However, this method requires \emph{all}
deductions to be if and only if, and it often gets confusing which things
you are assuming.  Hence it is usually best to write out the two directions
separately.

\item % (c)
Here is an example of a proof that avoids being two-part.  Note that it is
extremely short; long proofs are difficult to do in one part.
\begin{align*}
v_1,\dots,v_n \text{ is a basis of } \R^m
&\iff v_1,\dots,v_n \text{ are linearly independent and } 
v_1,\dots,v_n \text{span}
\R^m \\
&\iff P_{\{v\}} \text{ is one-to-one and } v_1,\dots,v_n 
\text{ span } \R^m \text{(by part (a))} \\ 
&\iff P_{\{v\}} \text{ is one-to-one and } P_{\{v\}}
\text{ is onto (by part (b)).} 
\end{align*}
\end{enumerate}

\item[2.5.2]
\ \begin{enumerate}
\item % (a)
A basis for the kernel is
$\big\{
\begin{bmatrix}
-1 \\ 1 \\ 0
\end{bmatrix},
\begin{bmatrix}
-3 \\ 0 \\ 1
\end{bmatrix}
\big\}$.  A basis for the image is
$\big\{
\begin{bmatrix}
1 \\ 2
\end{bmatrix}
\big\}$.  Note that many solutions are possible for all parts
of this problem.  Feel free to ask a CA any questions you have about row
reduction, but row reduction should not be necessary for the remainder of
the course.

\item % (b)
A basis for the kernel is
$\big\{
\begin{bmatrix}
-\frac{1}{3} \\ -\frac{4}{3} \\ 1
\end{bmatrix}
\big\}$.
A basis for the image is
$\big\{
\begin{bmatrix}
1 \\ -1 \\ -1
\end{bmatrix},
\begin{bmatrix}
2 \\ 1 \\ 4
\end{bmatrix}
\big\}$.

\item % (c)
A basis for the kernel is
$\big\{
\begin{bmatrix}
1 \\ -2 \\ 1
\end{bmatrix}
\big\}$.
A basis for the image is
$\big\{
\begin{bmatrix}
1 \\ 1 \\ 2
\end{bmatrix},
\begin{bmatrix}
1 \\ 2 \\ 3
\end{bmatrix}
\big\}$.
\end{enumerate}

\item[2.5.3]
False.  Many of you saw that $\Img g \subset \ker f$ by noting that
$f(g(x)) = (f \circ g)(x) = 0$ for any $x$ and hence for any element 
$g(x)$ in Img $g$.  However, showing that $f$ is $0$ on $\Img g$ is
\emph{not} the same thing as showing that $f$ is $0$ \emph{only} 
on $\Img g$.  Many of you noted this and stopped.  This is not good
enough, because sometimes the two things \emph{are} the same!  For 
example, suppose $m$ = 0 and that the domain of $f$ is just $\{0\}$.
Then $\Img g$ and $\ker f$ are both always $\{0\}$ (check this).

Therefore, to disprove something, you must give an explicit counterexample.
Many are possible for this problem.  One is to let $n = m = k = 1$ and
let $f$ and $g$ be the constant $0$ map, mapping everything to $0$.
Clearly $f\circ g = 0$.  But Img $g = \{0\}$ and ker $f = \R^1
\neq \{0\}$.

\item[2.5.4]
\ \begin{enumerate}
\item % (a)
Some of you were confused by the identification of $P_2$ as $\R^3$.
What it means practically is that the coefficients of a polynomial
``represent'' the polynomial in some ways.  However, polynomials are still
polynomials---the identification as $\R^3$ is just a convenient way of
looking at things.  A similar situation is in 2.4.6(b) where $\Mat(2,2)$ is
identified as $\R^4$ but is still in actuality the set of $2 \times 2$
matrices.

On to the problem.  We see what $T$ does to a general polynomial
$a + bx + cx^2$ in $P_2$:
\begin{align*}
T(\text{the polynomial }a+bx+cx^2)
&= x(\text{the derivative of the polynomial }a+bx+cx^2)\\
&+ x^2(\text{the second derivative of the polynomial }a+bx+cx^2)\\
& = x(b+2cx) + x^2(2c) \\
& = bx + 4cx^2.
\end{align*}
(You do not have to write the text in the above equations; they are
just there for explanatory purposes.)  In the same way that we think
of $a+bx+cx^2$ as $\begin{bmatrix} a \\ b \\ c \end{bmatrix}$,
we can think of $bx+4cx^2$ as
$\begin{bmatrix} 0 \\ b \\ 4c \end{bmatrix}$.  The array
for $T$ is just the $3 \times 3$ array $A$ for which
$A\begin{bmatrix}a \\ b \\ c \end{bmatrix}
= \begin{bmatrix}0 \\ b \\ 4c \end{bmatrix}$.
The answer is
$\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 4
\end{bmatrix}$.

\item % (b)
If we use the above matrix, then a basis for the image is
$ \big\{
\begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix},
\begin{bmatrix}
0 \\ 0 \\ 4
\end{bmatrix}
\big\}$, which really \emph{means}
$\{x, 4x^2\}$.  A basis for the kernel is
$ \big\{
\begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix}
\big\}$, which really \emph{means} $\{1\}$.  Different answers are
possible for this problem.
\end{enumerate}

\item[2.5.7]
A basis for the image of $A$ is
$$ \left\{
\begin{bmatrix}
1 \\ 2 \\ 4
\end{bmatrix},
\begin{bmatrix}
1 \\ -1 \\ 1
\end{bmatrix}
\right\}.$$
A basis for the kernel of $A$ is
$$ \left\{
\begin{bmatrix}
-1 \\ -2 \\ 1 \\ 0 \\ 0
\end{bmatrix},
\begin{bmatrix}
-\frac{10}{3} \\ -\frac{8}{3} \\ 0 \\ 1 \\ 0
\end{bmatrix},
\begin{bmatrix}
-1 \\ -1 \\ 0 \\ 0 \\ 1
\end{bmatrix}
\right\}.$$
Thus $\dim(\Img) + \dim(\ker) = 2 + 3 = 5,$ which is the dimension of
the domain, $\R^5$.

A basis for the image of $B$ is
$$ \left\{
\begin{bmatrix}
2 \\ 2
\end{bmatrix},
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
\right\}.$$
A basis for the kernel of $B$ is
$$\left\{
\begin{bmatrix}
-\frac{3}{4} \\ -\frac{3}{2} \\ 1 \\ 0 \\ 0
\end{bmatrix},
\begin{bmatrix}
-\frac{5}{2} \\ -1 \\ 0 \\ 1 \\ 0
\end{bmatrix},
\begin{bmatrix}
-\frac{3}{4} \\ -\frac{1}{2} \\ 0 \\ 0 \\ 1
\end{bmatrix}
\right\}.$$
Thus $\dim(\Img) + \dim(\ker) = 2 + 3 = 5$, which is the dimension of
the domain, $\R^5$.

Many answers are possible for both parts.

\item[2.5.9]
This is probably the most challenging problem on the set, so please do
not be discouraged if you did not solve it.  One hint in the probem is
that you need to show the unique existence of something.  This often means
that there is a bijective map around, because for any element $y$ in the
range of the map, there exists a unique element $x$ in the domain that is
mapped to $y$.  We now define this map.  Let $P_{2k-1}$ be the set of
polynomials of degree at most $2k-1$.  Note that $P_{2k-1}$ is a
$2k$-dimensional vector space with basis $\{1,x,x^2,\dots,x^{2k-1}\}$.
Now we define map
$T\from P_{2k-1} \to \R^{2k}$ by
$T(p) = \begin{bmatrix}
p(1) \\ p(2) \\ \vdots \\ p(k) \\
p'(1) \\ p'(2) \\ \vdots \\ p'(k)
\end{bmatrix}$.  Because the derivative is a linear map, it can be
shown that $T$ is linear.  We are almost done, because if we can show
that $T$ is bijective, then for any $a_1,\dots,a_k,b_1,\dots,b_k$,
there exists a unique $p$ in $P_{2k-1}$ such that
$T(p) = \begin{bmatrix} a_1 \\ \vdots \\ a_k \\ b_1 \\ \vdots \\ b_k
\end{bmatrix}$, \ie $p(i) = a_i$ and $p'(i) = b_i$ for $i=1,\dots,k$.

Because the domain and range of $T$ have the same dimension, $T$ is
bijective if and only if it is one-to-one, which occurs if and only if
the kernel of $T$ is $\{0\}$ (check this).  So we have boiled down the
problem into showing that $0$ is the only element in the kernel.  Suppose
$p$ is in the kernel.  Then $T(p) = 0$, so $0 = p(1) = p(2) = \dots
= p(k) = p'(1) = p'(2) = \dots = p'(k)$.  By the hint in the book, 
$p(x) = (x-1)^2 q_1(x)$ for some polynomial $q_1$.  Now
$q_1(i) = q_1'(i) = 0$ for $i=2,3,\dots,k$ because
$p(i) = p'(i) = 0$ for $i=2,3,\dots,k$ (why does it follow?).  So by
the hint, $q_1(x) = (x-2)^2 q_2(x)$ for some polynomial $q_2$.  Hence
$p(x) = (x-1)^2(x-2)^2 q_2(x)$.  We can
keep doing this and eventually get
$p(x) = (x-1)^2(x-2)^2\dots(x-k)^2 q_k(x)$.  But the right hand side
is at least degree $2k$ if $q_k \neq 0$.  Because deg$(p(x)) < 2k$, this
must mean that $q_k = 0$.  Hence $p = 0$, and the only element that can
be in the kernel is $0$.

\item[2.6.4]
\ \begin{enumerate}
\item % (a)
A lot of you confused $\cL(\Mat(n,n),\Mat(n,n))$ with
$\Mat(n,n)$.  In this problem, $n \times n$ matrices are being
regarded as \emph{vectors}, and $L$ contains $\emph{maps}$ that take
matrix \emph{vectors} to matrix \emph{vectors}.  Like in 2.4.6(b), we
can regard $\Mat(n,n)$ as $\R^{n^2}$ ($\Mat(n,n)$ just has the coordinates
arranged in a square).  Hence we are really dealing with linear functions
from $\R^{n^2}$ to $\R^{n^2}$.  From this it is easy to check that
$\cL(\Mat(n,n),\Mat(n,n))$ is
a vector space, so I will not do it here.  One final observation is that
linear maps from $\R^{n^2}$ to $\R^{n^2}$ correspond to
$n^2 \times n^2$ matrices.  Hence the dimension of $L$ is $(n^2)(n^2)
= n^4$.

\item % (b)
Let $A$ be any given $n \times n$ matrix.  To show that $L_A$ is linear,
we need to show that
\begin{itemize}
\item
for any $B_1$ and $B_2$ in $\Mat(n,n)$,
$L_A(B_1 + B_2) = L_A(B_1) + L_A(B_2)$
\item
for any real number $c$ and any $B$ in $\Mat(n,n)$,
$L_A(cB) = cL_A(B)$.
\end{itemize}
For the former, $L_A(B_1 + B_2) = A(B_1 + B_2) = AB_1 + AB_2
= L_A(B_1) + L_A(B_2)$.  For the latter, $L_A(cB) = A(cB) = cAB
= cL_A(B)$.  Hence $L_A$ is linear.  The proof of the linearity of
$R_A$ is similar.

It is also correct to do this problem by comparing the entries of both
sides of the equations.

\item % (c)
Many of you wrote that $\{L_A: A \in \Mat(n,n)\}$ has dimension $n^2$
because it corresponds to the $n^2$-dimensional vector space $\Mat(n,n)$.
However, you also need to show that correspondence between the two spaces is linear.
Otherwise, it is conceivable that the two spaces have different dimensions.
(For example, there exists non-linear bijective maps from $\R^2$ to $\R$.)  Here
is how the proof can be made rigorous.  Denote
$W = \{L_A: A \in \Mat(n,n)\}$.  Then we can define a map
$T\from \Mat(n,n) \to W$ by $T(A) = L_A$.  We must then check the
following:

\begin{itemize}
\item
$T$ is linear.
\item
$T$ is one-to-one.
\item
$T$ is onto.
\end{itemize}

It is not difficult to show that $T$ is linear.  It is obvious that $T$ is
onto (by how $W$ is defined).  The only hard part is to show that $T$ is
one-to-one.  Recall that this is equivalent to showing that ker$(T) = \{0\}$.
So given a nonzero $n \times n$ matrix $A$, we must show that $T(A) = L_A$ is
nonzero, meaning that $L_A$ is not the zero map.  Let $I$ be the $n \times n$
identity matrix.  Then $L_A(I) = AI = A$, which is nonzero by assumption.
Hence $L_A$ is not the zero map, and indeed ker$(T) = \{0\}$.

A similar proof is used to show that the
space of transformations of the form $R_A$ also has dimension $n^2$.

\item % (d)
We try to find the general form of all transformations of the form
$L_A + R_B$.  Let $A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and
$B = \begin{bmatrix}e & f \\ g & h\end{bmatrix}$.  Then
$L_A + R_B$ takes
$$\begin{bmatrix}x_{11} & x_{12} \\ x_{21} & x_{22}
\end{bmatrix} \text{ to }
\begin{bmatrix}
(a+e)x_{11} + gx_{12} + bx_{21} & fx_{11} + (a+h)x_{12} + bx_{22} \\
cx_{11} + (d+e)x_{21} + gx_{22} & cx_{12} + fx_{21} + (d+h)x_{22}
\end{bmatrix}.$$
Let us now regard $2 \times 2$ matrices 
$\begin{bmatrix}x_{11} & x_{12} \\ x_{21} & x_{22}
\end{bmatrix}$
as vectors 
$\begin{bmatrix}x_{11} \\ x_{12} \\ x_{21} \\ x_{22}
\end{bmatrix}$
in $\R^4$.  Then the matrix representation of $L_A + R_B$ is
$\begin{bmatrix}
a+e & g & b & 0 \\  
f & a+h & 0 & b \\
c & 0 & d+e & g \\
0 & c & f & d+h
\end{bmatrix}$.  It is clear that no matter what values 
$a,b,c,d,e,f,g,h$ are, this matrix can never equal
$\begin{bmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}$.
Hence this represents a linear transformation $T_0$ that cannot be
written as $L_A + R_B$.  Explicitly,
$T_0\begin{bmatrix}
x_{11} & x_{12} \\ x_{21} & x_{22}
\end{bmatrix} = 
\begin{bmatrix}
x_{22} & 0 \\ 0 & 0
\end{bmatrix}$.
\end{enumerate}
Many other answers are possible.

\item[2.6.5]
\ \begin{enumerate}
\item % (a)
$W$ is a subspace of $V$ if and only if
\begin{description}
\item[(i)] $W$ is nonempty. (Almost all of you forgot about this.)
\item[(ii)] For any vectors $w_1$ and $w_2$ in $W$, $w_1 + w_2$ is also in $W$.
\item[(iii)] For any real number $a$ and vector $w$ in $W$, $aw$ is also in $W$.
\end{description}

\item % (b)
\ \begin{description}
\item[(i)]
Let $W_1$ denote the set in question.  It is not a subspace of $V$.  One
way to see this is that $W_1$ does not satisfy additive closure.  First of
all, if $W_1$ is empty then it is automatically not a subspace.  But even
if $W_1$ is nonempty (say it contains $f$), we know that $f+f = 2f$ is not
in $W_1$ because $2f = f+f = (f'(x)+1)+(f'(x)+1) = (2f)'(x)+2$, which
cannot possibly be the function $(2f)'(x)+1$ (just try plugging in $x=1/2$).

What many of you did contains a subtle fallacy.  Many of you wrote that
if $f,g \in W_1$ then $f(x)+g(x) = f'(x)+g'(x)+2 \neq f'(x)+g'(x)+1$, and
hence additive closure is violated.  Well, it is only violated if we can
find an actual example of a violation.  An analogous example is the
statement ``For any apple that is an orange, the apple likes to shop.''  It
is absurd that any apple likes to shop, but since we cannot find an apple
that is an orange, the above statement is actually true.

Hence you need to exhibit some $f$ and $g$ in $W_1$ (or deduce that some 
exists, like I did above) in order to have the additive closure violated.
The function $f(x) = 1$ is a \emph{specific} function you could have 
used.

Despite what I just said, it is actually a good technique to see if additive
closure is satisfied by using $f$ and $g$.  This way, you discovered that
$f+g$ is not in $W_1$ if $f'(x)+g'(x)+2$ and $f'(x)+g'(x)+1$ differ.  Often
it will not be so obvious why they might differ.  Then you need to find
specific $f$ and $g$ in $W_1$ \emph{such that} the two quantities (or in
this case functions) differ.

\item[(ii)]
Let $W_2$ denote the set in question.  It is a subspace of $V$.  Let us
check the three criteria of a subspace (see 2.6.5(a) above):

\begin{itemize}
\item $W_2$ is nonempty because it contains the function $f(x) = 0$.
\item Suppose $f(x)$ and $g(x)$ are in $W_2$.  Then
$(f+g)(x) = f(x) + g(x) = xf'(x) + xg'(x) = x(f'(x)+g'(x))
= x((f+g)'(x))$.  Hence $(f+g)(x)$ is also in $W_2$.
\item Suppose $a$ is a real number and $f(x)$ is in $W_2$.  Then
$(af)(x) = af(x) = a(xf'(x)) = x(af)'(x)$.  Hence $(af)(x)$ is also
in $W_2$.
\end{itemize}

\item[(iii)]
Let $W_3$ denote the set in question.  It is not a subspace of $V$.
We note that the function $h(x) = \frac{1}{4}x^2$ is in $W_3$.  However,
$(2h)(x)$ turns out not to be in $W_3$: $((2h)'(x))^2
= (x)^2 = x^2$, while $(2h)(x) = \frac{1}{2}x^2$, a different function
from $x^2$.  Hence $W_3$ violates additive closure.

Many of you wrote that 
if $f,g \in W_3$ then $((f+g)'(x))^2 = (f'(x)+g'(x))^2
= (f'(x))^2 + f'(x)g'(x) + (g'(x))^2 \neq 
(f'(x))^2 + (g'(x))^2 = f(x) + g(x)$,
and hence additive closure is violated.  Well, the $\neq$ becomes $=$ if
$f'(x)g'(x)$ is identically $0$.  This can happen if $f(x)=g(x)=0$: you
can check that such $f$ and $g$, as well as $f+g$, are in $W_3$.  This is
situation where we need to find actual $f$ and $g$ such that
$f'(x)g'(x)$ is not identically $0$.
  \end{description}
 \end{enumerate}
\end{document}
