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\begin{document}


\noindent
{\bf Solutions to HW 6.}

\medskip
  
\begin{enumerate}
\item[1.]
 Let $A$ be the $4\times 4$ matrix whose entries are all $1$'s,
  and let $T(x)=Ax$. 
\begin{enumerate}
\item[(a)] Find a basis for the image of $T$. 
\item[(b)] Find a basis for the kernel of $T$.
\item[(c)] What is the rank of $T$? 
(Recall that $\textrm{rank}(T) = \dim(\textrm{im}(T))$.)
\end{enumerate}
\end{enumerate}

\medskip

\color{blue}
{\bf Solution.}
The first step is to put $A$ in RRE form:
\[
{\tiny
\left[
\begin{array}{cccc}
1 & 1 & 1 & 1\\
1 & 1 & 1 & 1\\
1 & 1 & 1 & 1\\
1 & 1 & 1 & 1\\
\end{array}
\right]
{\overset{\hphantom{AAAA}}{\xrightarrow{\hspace*{1.2cm}}}}
\left[
\begin{array}{cccc}
\color{purple} 1 & 1 & 1 & 1\\
\color{purple}0 & 0 & 0 & 0\\
\color{purple}0 & 0 & 0 & 0\\
\color{purple}0 & 0 & 0 & 0\\
\end{array}
\right].
}
\]

Part (a): There is a \color{purple}unique pivot column \color{blue}
in the RRE form, the first column,
so the first column of the original matrix
$A$ constitutes a $1$-vector
basis for the image (=column space).
I write 
$
{\mathcal I} =
{\tiny
  \left(\begin{bmatrix}1\\1\\1\\1\end{bmatrix}\right)
    }$
  for this basis.

Part (b): From the RRE form obtained in
Part (a), we find that the free variables are $x_2, x_3, x_4$,
and the general solution to the homogeneous system in vector form is
\[
{\tiny
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} = 
\begin{bmatrix}-x_2-x_3-x_4\\x_2\\x_3\\x_4\end{bmatrix} = 
x_2\begin{bmatrix}-1\\1\\0\\0\end{bmatrix} +
x_3\begin{bmatrix}-1\\0\\1\\0\end{bmatrix} +
x_4\begin{bmatrix}-1\\0\\0\\1\end{bmatrix}.
}
\]
According to our algorithm, ${\mathcal K} =
{\tiny
  \left(\begin{bmatrix}-1\\1\\0\\0\end{bmatrix},
\begin{bmatrix}-1\\0\\1\\0\end{bmatrix},
  \begin{bmatrix}-1\\0\\0\\1\end{bmatrix}\right)
    }$
  is a basis for the kernel.

  Part (c): The rank of $T$ is the dimension of the image of $T$,
  which is the size of $\mathcal I$, namely it is $1$.
\color{black}

\medskip

\begin{enumerate}
\item[2.]
Let $A$ be the matrix for differentiation
  $D\colon P_3\to P_3  \colon f(x)\mapsto f'(x)$
  relative to the ordered basis $(1, x, x^2, x^3)$,
and let $T(x)=Ax$. 
\begin{enumerate}
\item[(a)] Find a basis for the image of $T$. 
\item[(b)] Find a basis for the kernel of $T$. 
\item[(c)] What is the rank of $T$?  
\end{enumerate}
\end{enumerate}
\medskip

\color{blue}
{\bf Solution.} 
We know the matrix for $A$, from HW5(1),
so let's put it in RRE form:
\[
{\tiny
\left[
\begin{array}{cccc}
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 3\\
0 & 0 & 0 & 0\\
\end{array}
\right]
{\overset{\hphantom{AAAA}}{\xrightarrow{\hspace*{1.2cm}}}}
\left[
\begin{array}{cccc}
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\\
\end{array}
\right].
}
\]

Part (a): The last three columns are the pivot columns,
so a basis for the image consists of the last three
columns of the original matrix.
I write 
${\mathcal I} =
{\tiny
  \left(\begin{bmatrix}1\\0\\0\\0\end{bmatrix},
  \begin{bmatrix}0\\2\\0\\0\end{bmatrix},
\begin{bmatrix}0\\0\\3\\0\end{bmatrix}
  \right)
  }$
  for this basis.

Part (b): From the RRE form obtained in
Part (a), we find that the only free variable is $x_1$,
and the general solution to the homogeneous system in vector form is
\[
{\tiny
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} = 
\begin{bmatrix}x_1\\0\\0\\0\end{bmatrix} = 
  x_1\begin{bmatrix}1\\0\\0\\0\end{bmatrix}.
  }
\]
According to our algorithm, ${\mathcal K} =
{\tiny
  \left(\begin{bmatrix}1\\0\\0\\0\end{bmatrix}\right)
    }$
  is a basis for the kernel.

  Part (c): The rank of $T$ is the dimension of the image of $T$,
  which is the size of $\mathcal I$, namely it is $3$.
  \color{black}
  \medskip

\begin{enumerate}
\item[3.]
If $A$ is an $n\times n$ matrix, then any matrix
of the form $B = C^{-1}AC$ is called a conjugate of $A$.
Explain why the following statement is true:
If $T\colon \mathbb R^n\to \mathbb R^n$
is a linear transformation whose matrix relative
to the standard basis $\mathcal E$ is
$A = {}_{\mathcal E}[T]_{\mathcal E}$,
then the matrix ${}_{\mathcal B}[T]_{\mathcal B}$
for $T$ relative to some other basis $\mathcal B$
is a conjugate of the matrix $A$.
\end{enumerate}
\medskip

\color{blue}
{\bf Solution.}
Let $A = {}_{\mathcal E}[T]_{\mathcal E}$,
$B = {}_{\mathcal B}[T]_{\mathcal B}$, 
$C = {}_{\mathcal E}[\textrm{id}]_{\mathcal B}$, and
$D = {}_{\mathcal B}[\textrm{id}]_{\mathcal E}$.
It follows from HW5(3) that $CD=
{}_{\mathcal E}[\textrm{id}]_{\mathcal B}\cdot {}_{\mathcal B}[\textrm{id}]_{\mathcal E}
= {}_{\mathcal E}[\textrm{id}]_{\mathcal E} = I$ and
$DC=
{}_{\mathcal B}[\textrm{id}]_{\mathcal E}\cdot {}_{\mathcal E}[\textrm{id}]_{\mathcal B}
= {}_{\mathcal B}[\textrm{id}]_{\mathcal B} = I$, so $C^{-1}=D={}_{\mathcal B}[\textrm{id}]_{\mathcal E}$.
Hence $C^{-1}AC= {}_{\mathcal B}[\textrm{id}]_{\mathcal E}\cdot 
{}_{\mathcal E}[T]_{\mathcal E}\cdot {}_{\mathcal E}[\textrm{id}]_{\mathcal B}
={}_{\mathcal B}[\textrm{id}\circ T\circ \textrm{id}]_{\mathcal B}
={}_{\mathcal B}[T]_{\mathcal B}= B.$
\color{black}

\end{document}

\color{blue}
{\bf Solution 2.} (Use knowledge from Calculus.)

Part (a): We know from Calculus that the derivative a polynomial
of degree at most $3$ must be a polynomial
of degree at most $2$. Conversely, if $p(x)=ax^2+bx+c$
is any polynomial of degree at most $2$, then $p(x)$
is the derivative of some polynomial of degree
at most $3$, for example $p(x) = D(q(x))$ where
$q(x) = \int_0^x p(t)\;dt = (a/3)x^3+(b/2)x^2+cx$.
Hence the image of $T$ is exactly the space of polynomials
of degree at most $2$, which we know from HW5(1) has basis
${\mathcal I}=(1,x,x^2)$.

Part (b): We know from Calculus that if $p(x)$ is a polynomial
and $D(p(x))$ is the zero polynomial, then $p$ is constant.
Thus, the kernel of the derivative transofrmation is the space
of polynomials of degree zero. We know from HW5(1) that a basis
for this space is ${\mathcal K}=(1)$.

Part (c): the rank of $T$ is the dimension of the image,
which is the size of ${\mathcal I}$, which is $3$.
\color{black}

\medskip
