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\begin{center}
\textbf{Mathematics 262} \\
\textbf{Homework 7 solutions}
\end{center}
\vspace{-0.05in}

Assignment:
\begin{itemize}
\item Prove the Cayley-Hamilton theorem for upper triangular matrices:
if $A$ is upper triangular with characteristic polynomial $f(x)$, then
$f(A)=0$.
\item Section 6.5: 1(b), 2, 4 (extra credit)
\end{itemize}

\textbf{Cayley-Hamilton:}
In order to show that $f(A)=0$, I will use induction on $n$.  Every $1
\times 1$ matrix $B$ satisfies its characteristic polynomial (i.e.,
$f(B)=0$), so that starts the induction.  Now assume that every $(n-1)
\times (n-1)$ upper triangular matrix satisfies its characteristic
polynomial.  Let $A$ be an $n\times n$ upper triangular matrix with
characteristic polynomial $f(x)$.  One way to show that $f(A) = 0$ is
to show that $f(A) \alpha = \vec{0}$ for every $n$-dimensional vector
$\alpha$.  Even better, it's good enough to show that $f(A) \alpha =
\vec{0}$ for every $\alpha$ in some basis.  I'll use the standard
basis: the basis with respect to which $A$ is upper triangular.  Call
the basis elements $\varepsilon_{1}$, $\varepsilon_{2}$, \dots,
$\varepsilon_{n}$.

I can also write down a formula for $f(x)$: if the diagonal entries of
$A$ are $c_{1}$, $c_{2}$, \dots, $c_{n}$, (these are not necessarily
distinct) then
\[
f(x) = (x-c_{1}) (x-c_{2}) \dots (x-c_{n}).
\]
Because $A$ is upper triangular, the subspace $W$ spanned by
$\varepsilon_{1}$, \dots, $\varepsilon_{n-1}$ is invariant under $A$;
therefore $A$ determines a linear operator $B : W \rightarrow W$.  The
matrix for $B$ is $A$ with the last row and column deleted; in
particular, $B$ is upper triangular.  The characteristic polynomial
for $B$ is $g(x) = (x-c_{1}) \dots (x-c_{n-1})$, and by induction, we
know that $g(B) = 0$.  In particular, $g(B) \varepsilon_{i} = \vec{0}$
for all $i \leq n-1$.

Because of this, $g(A)\varepsilon_{i} = \vec{0}$ for all $i \leq n-1$;
since $f(A) = (A-c_{n}I) g(A)$, this means that $f(A) \varepsilon_{i}
= \vec{0}$ for all $i \geq n-1$.  If we can show that $f(A)
\varepsilon_{n} = \vec{0}$, we will be finished.  Because $A$ is upper
triangular with $A_{nn} = c_{n}$, then $(A-c_{n}I)\varepsilon_{n}$ is
a linear combination of $\varepsilon_{1}$, \dots,
$\varepsilon_{n-1}$.  Therefore 
\begin{align*}
f(A)\varepsilon_{n} &= g(A) (A-c_{n}I) \varepsilon_{n} \\
  &= g(A) (r_{1}\varepsilon_{1} + \dots + r_{n-1}\varepsilon_{n-1}) \\
  &= r_{1} g(A) \varepsilon_{1} + \dots + r_{n-1} g(A)
\varepsilon_{n-1} \\
  & = \vec{0}.
\end{align*}

\textbf{\S 6.5, \#1(b):}
To find the invertible matrix $P$, I have to find a basis which
consists of eigenvectors for both $A$ and $B$.  $A$ has eigenvalues 0
and 2, with corresponding eigenvectors $(1, -1)$ and $(1,1)$.  You can
easily check that these are also eigenvectors for $B$, with
eigenvalues $1-a$ and $1+a$, respectively.  Now let $P =
\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}$.  Then $P^{-1} =
\frac{1}{2} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$.  One can
check that $P^{-1}AP$ and $P^{-1}BP$ are diagonal (and, of course, the
diagonal entries are the eigenvalues of the matrices).  Since
eigenvalues are not unique, and since they could have been chosen in
either order, there are many correct answers here.

\textbf{\S 6.5, \#2:}
Since the matrices are defined over the complex numbers, they are all
similar to triangular matrices.  Since they commute, they are
simultaneously triangulable.  So the question becomes, how many
linearly independent upper triangular matrices are there?  Since the
matrices are $3 \times 3$ and upper triangular, there are 6 possible
nonzero entries; therefore, there are at most 6 linearly independent
matrices in $\mathcal{F}$.  In the $n \times n$ case, there are at
most
\[
n + (n-1) + (n-2) + \dots + 2 + 1 = \binom{n+1}{2} = \frac{n(n+1)}{2}
\]
such matrices.

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