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\noindent
{\bf Solutions to HW 9.}

\medskip
  
\begin{enumerate}
\item[1.]
Show that if $\lambda$ is an e-value for $A$ and $k$ 
is an integer,
then $\lambda^k$ is an e-value for $A^k$. 
\end{enumerate}

\medskip

\color{purple} The problem should indicate that $k$
is a \underline{positive} integer, since the statement is false
for $\lambda=0$ if $k<0$.

\medskip

\color{blue}
{\bf Solution.}

We prove this by induction on $k$. Throughout, assume
that $v$ is a $\lambda$ e-vector for $A$.

{\bf Base Case.} $k=1$

In this case, $A^kv = Av = \lambda v = \lambda^k v$.

{\bf Inductive step.} Assume the statement is true for $k=n$.
Let's prove it for $k=n+1$.

$A^kv = A^{n+1}v = A(A^nv)v = A(\lambda^n v) = \lambda^n Av
= \lambda^n \cdot \lambda v = \lambda^{n+1}v$.
\color{black}

\medskip

\begin{enumerate}
\item[2.]
Suppose that $A$ is a $2\times 2$-matrix and that 
$\det(A-I) = -16$ and $\det(A-2I) = -15$. What are the e-values of $A$?
\end{enumerate}
\medskip

\color{blue}
{\bf Solution.}
If $X$ is a $2\times 2$ matrix, then $\textsf{det}(X)=\textsf{det}(-X)$.
Hence $\chi_A(\lambda)=\textsf{det}(\lambda I - A)=\textsf{det}(A-\lambda I)$,
and this polynomial is 
$\lambda^2-\textsf{tr}(A)\lambda+\textsf{det}(A)$.
We are given that 
$\textsf{det}(A-I) = -16$ 
so $\chi_A(1)=-16$. Similarly, from $\textsf{det}(A-2I) = -15$
we derive that $\chi_A(2)=-15$. Substituting these values into
$\lambda^2-\textsf{tr}(A)\lambda+\textsf{det}(A)$ yields
\[
\begin{array}{rl}
1-\textsf{tr}(A)+\textsf{det}(A)&=-16\\
4-2\textsf{tr}(A)+\textsf{det}(A)&=-15\\
\end{array}
\quad
\textrm{or}
\quad
\begin{array}{rl}
-\textsf{tr}(A)+\textsf{det}(A)&=-17\\
-2\textsf{tr}(A)+\textsf{det}(A)&=-19\\
\end{array}
\]
or
\[
\begin{bmatrix}
  -1&1\\
  -2&1
\end{bmatrix}
\begin{bmatrix}
\textsf{tr}(A)\\ \textsf{det}(A)
\end{bmatrix}=
\begin{bmatrix}
-17\\-19
\end{bmatrix}
\]
The solution to this system is
$\textsf{tr}(A)=2$ and $\textsf{det}(A)=-15$,
so $\chi_A(\lambda)=\lambda^2-2\lambda-15$.
The roots are $\lambda_1=-3$ and $\lambda_2 = 5$.
\color{black}
  \medskip

\begin{enumerate}
\item[3.]
Let $T\colon M_{2\times 2}(\mathbb R)\to M_{2\times 2}(\mathbb R)\colon
M\mapsto M^t$ be the operation of transpose.
Find the characteristic polynomial, e-values, and
e-spaces of $T$.
\end{enumerate}
\medskip

\color{blue}
{\bf Solution.}
Choose the basis ${\mathcal B} = (E_{11},E_{12},E_{21},E_{22})$
for $M_{2\times 2}(\mathbb R)$. With respect to this basis,
the matrix for $T$ is
\[
  {}_{\mathcal B}[T]_{\mathcal B}=A =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}.
\]
We compute $\chi_A(\lambda)$ by evaluating a determinant
(I used the Laplace expansion along whichever row had the most zeros):
\[
\chi_A(\lambda)=
\det(\lambda I-A) =
\det\begin{bmatrix}
\lambda-1 & 0 & 0 & 0\\
0 & \lambda & -1 & 0\\
0 & -1 & \lambda & 0\\
0 & 0 & 0 & \lambda-1\\
\end{bmatrix} = (\lambda-1)^2(\lambda^2-1)=(\lambda-1)^3(\lambda+1),
\]
so $\chi_A(\lambda)=(\lambda-1)^3(\lambda+1)$.
The e-values of $T$ (or $A$) are the roots of this polynomial,
so they are $\lambda_1 = +1$ (algebraic multiplicity = $3$)
and 
$\lambda_2 = -1$ (algebraic multiplicity = $1$).

$V_1=\textrm{null}(A-I) = \ker(T-I)
= \{M\in M_{2\times 2}(\mathbb R)\;|\;T(M) = M\}
= \{M\in M_{2\times 2}(\mathbb R)\;|\;M^t = M\}$,
which is the subspace of $M_{2\times 2}(\mathbb R)$
consisting of symmetric $2\times 2$ matrices.
A basis for this subspace is
${\mathcal S}=(E_{11},E_{12}+E_{21},E_{22})=
\left(
\begin{bmatrix}1&0\\0&0\end{bmatrix},
\begin{bmatrix}0&1\\1&0\end{bmatrix},
\begin{bmatrix}0&0\\0&1\end{bmatrix}
\right)$.
Written in the ${\mathcal B}$-basis, this is
\[
{\mathcal S} =
\left(
\begin{bmatrix}1\\0\\0\\0\end{bmatrix}_{\mathcal B},
\begin{bmatrix}0\\1\\1\\0\end{bmatrix}_{\mathcal B},
\begin{bmatrix}0\\0\\0\\1\end{bmatrix}_{\mathcal B}\right).
  \]
  One way to discover this basis is to apply the
  nullspace algorithm to
  \[
A-I = \begin{bmatrix}
0 & \hphantom{-}0 & \hphantom{-}0 & 0\\
0 & -1 & \hphantom{-}1 & 0\\
0 & \hphantom{-}1 & -1 & 0\\
0 & \hphantom{-}0 & \hphantom{-}0 & 0\\
\end{bmatrix}.
  \]

  Similarly, $V_2=\textrm{null}(A+I) = \ker(T+I) =
  \{M\in M_{2\times 2}(\mathbb R)\;|\;T(M) = M^t = -M\}$,
which is the subspace of $M_{2\times 2}(\mathbb R)$
consisting of antisymmetric $2\times 2$ matrices.
A basis for this subspace is ${\mathcal A}=(E_{12}-E_{21})$.
Written in the ${\mathcal B}$-basis, this is
\[
{\mathcal A} =
\left(
\begin{bmatrix}\hphantom{-}0\\\hphantom{-}1\\-1\\\hphantom{-}0\end{bmatrix}_{\mathcal B}\right).
\]
\color{black}

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