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\title{Linear Algebra Exam}
\target{Math 207: Fall 2014}

\begin{document}


\begin{exam}{\today}


\bp Find a $3\times 3$ matrix $A$ with nonzero integer entries so that
$A^3=A.$
\ba If $A^3=A$ then $A^3-A=0$ which implies $A$ satisfies the
polynomial $A(A-I)(A+I)$.  There are several simple diagonal matrices
that satisfy that equation, for example $D=\left(
\begin{array}{ccc}
 0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{array}
\right)$. Any matrix similar to $D$ will satisfy $A^3=A$ also.  So, to
find the matrix $A$, we conjugate $D$ by an invertible matrix $P$.  To
be sure that the result has integer entries, we choose $P$ with
integer entries and determinant $1$ so that $P^{-1}$ will have integer
entries also.  Here's a good choice, I found by performing row
operations on the identity, but only the row operations $row_i
\leadsto row_i + row_j$ which don't change the determinant.
$$P=\left(
\begin{array}{ccc}
 1 & 1 & 0 \\
 0 & 1 & 1 \\
 1 & 1 & 1 \\
\end{array}
\right) \text{ and its inverse }P^{-1}=\left(
\begin{array}{ccc}
 0 & -1 & 1 \\
 1 & 1 & -1 \\
 -1 & 0 & 1 \\
\end{array}
\right)$$
Then, we find $$A=P^{-1}DP=\left(
\begin{array}{ccc}
 -1 & -2 & -2 \\
 1 & 2 & 2 \\
 -1 & -1 & -1 \\
\end{array}
\right).$$
\ea
\ep


\bp Let $V$ be the vector space of five-times differentiable
real-valued functions
and let $W$ be the subspace of polynomials of degree
less than or equal to five.  The following formulae define symmetric
bilinear functions on $V$, which restrict to inner products on $W$.

\begin{align*}
 \<f,g\>_1&=\int_0^{10} f(t)g(t)dt \\
 \<f,g\>_2&=f(0)g(0)+f(2)g(2)+f(4)g(4)+f(6)g(6)+f(8)g(8)+f(10)g(10)\\
\<f,g\>_3&=f(1)g(1)+f(5)g(5)+f(9)g(9)+f'(1)g'(1)+f'(5)g'(5)+f'(9)g'(9)\\
\<f,g\>_4&=f(5)g(5)+f''(5)g''(5)+f''(5)g''(5)+f'''(5)g'''(5)+f''''(5)g''''(5)+f'''''(5)g'''''(5)\\
\end{align*}


The following picture shows the function $f(t)=\sin(t)$ together with
several degree five polynomials obtained by projecting $f$ onto $W$
using the given inner products.  Which curves go with which inner
products?  Justify your answer, but don't do any computations!

\begin{center}
\includegraphics[scale=0.5]{picture.pdf}
\end{center}
\ba The solid one is the graph of $f$, the dotted one corresponds to
$\<\,,\,\>_1$, dashed corresponds to $\<\,,\,\>_2$, long-dashed
corresponds to $\<\,,\,\>_3$, and dot-dashed corresponds to
$\<\,,\,\>_4$.
\ea
\ep


\bp Let $$M=\left(
\begin{array}{cccc}
 4 & -5 & -2 & 9 \\
 -5 & 7 & 5 & -10 \\
 3 & -5 & -3 & 7 \\
 -4 & 5 & 3 & -8 \\
\end{array}
\right).$$
Prove that $M$ is not diagonalizable over the real numbers but $M$ is
diagonalizable over the complex numbers.

\emph{Hint:} Compute $M^2$.
\ba
A quick check shows that $M^2=-I\Leftrightarrow M^2+I=0$.
Therefore, the minimum polynomial $m(t)$ of $M$ divides $t^2+1$, and
hence must equal $t^2+1$.  Since $m(t)$ doesn't factor into linear
factors over $\R$, $M$ is not diagonalizable over $\R$.

Over the complex numbers, the minimum polynomial of $M$ factors
 $m(t)=t^2+1=(t-i)(t+i)$, and we
find that $M$ is diagonalizable over $\C$.  In fact, $M$ is similar to
a diagonal matrix with either $i$ or $-i$ on the diagonal.
\ea


\ep

\bp Compute $e^B$ where $B=\left(
\begin{array}{cc}
 -2\pi  & -\pi \\
 5\pi & 2\pi \\
\end{array}
\right).$
\ba It's easier to work with a simpler matrix similar to $B$.  
We find $\chi_B(t)=t^2+\pi^2=(t-i\pi)(t+i\pi)$
so we have eigenvalues $\pm i \pi$.  So, if we arrange corresponding
eigenvectors as columns in a matrix $P$, we have $P^{-1}BP=D$ where $$D=\left(
\begin{array}{cc}
 i\pi  & 0 \\
 0 & -i\pi \\
\end{array}
\right)$$
Then 
\begin{align*}
e^B&=e^{PDP^{-1}} \\
&=Pe^{D}P^{-1} \\
&=P
\left(
\begin{array}{cc}
 e^{i \pi} & 0 \\
 0 & e^{-i\pi} \\
\end{array}
\right) P^{-1}\\
&=P\left(
\begin{array}{cc}
 -1 & 0 \\
 0 & -1 \\
\end{array}
\right)P^{-1}\\
&=\left(
\begin{array}{cc}
 -1 & 0 \\
 0 & -1 \\
\end{array}
\right).
\end{align*}
The second to last equality follows from the fact that $e^{i x} =
\cos(x)+i \sin(x)\Rightarrow e^{i \pi}=-1 \text{ and }e^{-i \pi}=-1.$
\ea
\ep

\bp Let $V$ be the vector space of polynomials of degree less than or
equal to two and define an inner product on $V$ by 
$$\<p,q\>=p(-1)q(-1)+p(0)q(0)+p(1)q(1).$$
A convenient orthonormal basis for $V$ is 
$$B=\left\{\frac{1}{2} (t-1) t,1-t^2,\frac{1}{2} t (t+1)\right\}.$$
Consider $T:V \to V$ defined by $T(p)=p'(t)$.  Let $T^*$ denote the
adjoint of $T$.
\bigskip

Find $T^*(t)$.
\ba First, note that $\<p,e_1\>=p(-1)$, $\<p,e_2\>=p(0)$, and
$\<p,e_3\>=p(1)$ for the given basis $B=\{e_1, e_2, e_3\}$.  To
determine the matrix for the operator $T$, we differentiate each $e_i$
and express the result in the basis $B$ easily by evaluating
$e_i'(-1)$, $e_i'(0)$, and $e_i'(1)$.
The result is the following matrix:
$$\Rep_{B,B}(T)=\left(
\begin{array}{ccc}
 -\frac{3}{2} & 2 & -\frac{1}{2} \\
 -\frac{1}{2} & 0 & \frac{1}{2} \\
 \frac{1}{2} & -2 & \frac{3}{2} \\
\end{array}
\right)$$
Then, the matrix for $T^*$ is simply the transpose:
$$\Rep_{B,B}(T^*)=\left(
\begin{array}{ccc}
 -\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} \\
 2 & 0 & -2 \\
 -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \\
\end{array}
\right).$$
In the basis $B$, the polynomial $t$ is the vector
$\Rep_B(t)=\left( \begin{array}{c} -1\\0\\1\end{array}\right)$.  So,
  $T^*(t)$ is represented by the vector
$$\left(
\begin{array}{ccc}
 -\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} \\
 2 & 0 & -2 \\
 -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \\
\end{array}
\right)\left( \begin{array}{c} -1\\0\\1\end{array}\right)=
=\left( \begin{array}{c} 2\\-4\\2\end{array}\right).$$
So, $T^*(t)=2e_1-4e_2+2e_3=6t^2-4$.
\ea
\ep


\end{exam}
\end{document}


