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\fancyhead[L]{\Large \textbf{Math 207 Fall 2014}}
\fancyhead[R]{\large \textbf{Practice Exam - Answers and notes}}


\begin{document}
\pagestyle{fancy}
\fancyfoot[R,L,C]{}


\bp Let $V$ be the vector space of all $n\times n$ complex matrices.
Prove that $\<A,B\>=\tr(AB^*)$ defines an inner product on $V$. 
\ba We check that the axioms of an inner product are satisfied.
\begin{itemize}
\item
\emph{Linearity in the first entry}.
If $A, A'$, and $B$ are $n\times n$ matrices, then 
\begin{multline*}
\<A+A',B\>=\tr((A+A')B^*)=\tr(AB^*+A'B^*)\\=\tr(AB^*)+\tr(A'B^*)
=\<A,B\>+\<A',B\>.\end{multline*}
If $\alpha\in \C$ then $$\<\alpha A,B\>=\tr(\alpha AB^*)=\tr(\alpha
AB^*)=\alpha \tr(AB^*)=\alpha \<A,B\>.$$
\item
\emph{Hermetian symmetry}.
$$\<A,B\>=\tr(AB^*)=\overline{\tr\left(\left(AB^*\right)^*\right)}=\overline{\tr\left(\left(BA^*\right)\right)}=\overline{\<B,A\>}.$$
The first equality follows from the fact that
$\tr(M^*)=\overline{\tr(M)}$ since $\tr(M^t)=\tr(M)$ and $\tr(\overline{M})=\overline{\tr(M)}.$
\item
\emph{Nondegeneracy}.
We have to prove that if $\<A,B\>=0$ for all matrices $B$, then
$A=0$.  

Let $B_{ij}$ be the matrix whose $i$-$j$-th entry is $1$ and whose
other entries are $0$.  Then $\tr(AB_{ij}^*)$ is
the $i$-$j$-the entry of $A$.  So if $\tr(AB^*)=0$, for all $B$, then
$\tr(AB_{ij}^*)=0$ for all $B_{ij}$, hence all the entries of $A$ must
be zero.

To see that $\tr(AB_{ij}^*)$ is
the $i$-$j$-the entry of $A$, note that the matrix $B_{ij}^*$ consists of
columns of zeroes except for the $j$-th column, which is the $i$-th standard
basis vector.  Then the $i$-th column of $AB_{ij}^*$ is the $j$-th
column of $A$, and all other entries of $AB_{ij}$ are zero.   So the
trace of $AB_{ij}^*$ is a sum of terms that are almost all zero, the only possible nonzero
term being in the $i$-$i$ entry,  in the sum  $i$-$j$-entry of $A$.
\end{itemize}
\ea
\ep

\bp Let $A$ be any $n\times n$ complex matrix.  Prove that $A$ and
$A^t$ are similar.
\ba There are two steps.  First, if $J$ is the Jordan form of $A$ then $A^t$ is similar to
$J^t$.  Second, $J^t$ is similar to $J$.  So, by transitivity of
similarity, we have $A\sim J \sim J^t\sim A^t.$

\emph{First step}.  Suppose $J=P^{-1}AP$.  Then by taking transpose we
have $$J^t=(P^{-1}A P)^t=P^t A^t (P^{-1})^t=P^t A^t (P^t)^{-1}.$$  So
$A^t\sim J^t$.

\emph{Second step}.  Note that if $B$ is a $k\times k$ Jordon block,
then for 
$$Q_k=\begin{pmatrix}
0 & 0 & \cdots & 0 & 1\\
0 & 0 & \cdots & 1 & 0 \\
\vdots & \vdots & & \vdots & \vdots \\
0 & 1 & \cdots & 0 & 0\\
1 & 0 & \cdots & 0 & 0 
\end{pmatrix}$$
we have $Q_k^{-1}BQ_k=B^t$.
Then by assembling a block-diagonal matrix $Q$ with matrices $Q_k$ of
the correct sizes as the blocks, we will have $Q^{-1}JQ=J^t$.
\ea
\ep

\bp Prove that if $A^2=A$ then $A$ is diagonalizable.
\ba If $A^2=A$ then $0=A^2-A=A(A-I)$.  Therefore, the minimum
polynomial for $A$ is either $m_A(t)=t$, $m_A(t)=(t-1)$, or
$m_A(t)=t(t-1)$.  In every case, $m_A(t)$ factors into distinct linear
factors, hence $A$ is diagonalizable.  

\emph{Remark}.  In the first case, $A$ is similar to the zero matrix,
hence is the zero matrix.  In the
second case, $A$ must be similar to the identity matrix, hence is the
identity.  
In the third case, $A$ must be
similar to a diagonal matrix with $0$'s and $1$'s on the diagonal, but
this matrix might not appear to be so simple.  For example, $$A=
\left(
\begin{array}{cccc}
 -2 & \frac{1}{2} & \frac{3}{2} & -2 \\
 30 & -4 & -15 & 20 \\
 6 & -1 & -2 & 4 \\
 15 & -\frac{5}{2} & -\frac{15}{2} & 11 \\
\end{array}
\right)$$
\ea
\ep


\bp Let $V$ be the space of real polynomials of degree $\leq 3$ and
let $\sigma:V \to V$ be the differentiation operator 
$$\sigma(p(t))=\frac{dp(t)}{dt}.$$
\ben
\item What is the adjoint of $\sigma$ if we equip $V$ with the 
inner product defined by $$\<p,q\>:=\int_0^1 p(t)q(t)dt?$$
\ba It's convenient to work with an orthonormal basis.  Performing
Gram-Schmidt on the basis $\{1,t,t^2,t^3\}$ yields the orthonormal
basis 
$$B=\left\{1,2 \sqrt{3} t-\sqrt{3},6 \sqrt{5} t^2-6 \sqrt{5}
t+\sqrt{5},20 \sqrt{7} t^3-30 \sqrt{7} t^2+12 \sqrt{7}
t-\sqrt{7}\right\}$$
The matrix for $\sigma$ in this basis is 
$$\Rep_{B,B}(\sigma)=\left(
\begin{array}{cccc}
 0 & 2 \sqrt{3} & 0 & 2 \sqrt{7} \\
 0 & 0 & 2 \sqrt{15} & 0 \\
 0 & 0 & 0 & 2 \sqrt{35} \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)$$
So, the matrix for the adjoint $\sigma^*$ in the basis $B$ is given by $$\Rep_{B,B}(\sigma^*)=(\Rep_{B,B}(\sigma))^t=\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 2 \sqrt{3} & 0 & 0 & 0 \\
 0 & 2 \sqrt{15} & 0 & 0 \\
 2 \sqrt{7} & 0 & 2 \sqrt{35} & 0 \\
\end{array}
\right).$$
\ea

\begin{remark*}
You might find it interesting to express $\sigma^*$ as a matrix in the basis
$E=\{1,t,t^2,t^3\}$.  To do so, use $\Rep_{B,E}$, which is the matrix that
changes basis from $B$ to $E$: $$\Rep_{B,E}=\left(
\begin{array}{cccc}
 1 & -\sqrt{3} & \sqrt{5} & -\sqrt{7} \\
 0 & 2 \sqrt{3} & -6 \sqrt{5} & 12 \sqrt{7} \\
 0 & 0 & 6 \sqrt{5} & -30 \sqrt{7} \\
 0 & 0 & 0 & 20 \sqrt{7} \\
\end{array}
\right).
$$  Then 
\begin{align*}
\Rep_{E,E}(\sigma^*)&=\Rep_{B,E}\Rep_{B,B}\Rep_{E,B}\\
&=\left(
\begin{array}{cccc}
 1 & -\sqrt{3} & \sqrt{5} & -\sqrt{7} \\
 0 & 2 \sqrt{3} & -6 \sqrt{5} & 12 \sqrt{7} \\
 0 & 0 & 6 \sqrt{5} & -30 \sqrt{7} \\
 0 & 0 & 0 & 20 \sqrt{7} \\
\end{array}
\right) 
\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 2 \sqrt{3} & 0 & 0 & 0 \\
 0 & 2 \sqrt{15} & 0 & 0 \\
 2 \sqrt{7} & 0 & 2 \sqrt{35} & 0 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
 1 & -\sqrt{3} & \sqrt{5} & -\sqrt{7} \\
 0 & 2 \sqrt{3} & -6 \sqrt{5} & 12 \sqrt{7} \\
 0 & 0 & 6 \sqrt{5} & -30 \sqrt{7} \\
 0 & 0 & 0 & 20 \sqrt{7} \\
\end{array}
\right) ^{-1}\\
&=\left(
\begin{array}{cccc}
 -20 & -5 & -4 & -4 \\
 180 & 60 & 58 & 60 \\
 -420 & -180 & -180 & -183 \\
 280 & 140 & 140 & 140 \\
\end{array}
\right)
\end{align*}
This implies, for example, that $\sigma^*(t^2)=-4+58t-180t^2+140t^3.$
This means that for any polynomial $p(t)\in V$, we have $$\<\sigma(p(t)),
t^2\>=\<p(t), \sigma^*(t^2)\>.$$  That is, $$\int_0^1 p'(t) t^2dt =
\int_0^1 p(t) \left(-4+58t-180t^2+140t^3\right)dt \text{ for \emph{every}
  polynomial }p(t)\in V.$$
For example, for $p(t)=\pi  t^3-\sqrt{3} t^2-\frac{t}{7}+e$, this
means that
$$\int_0^1 \left(3 \pi  t^2-2 \sqrt{3} t-\frac{1}{7}\right) t^2 dt
=\int_0^1 \left( \pi  t^3-\sqrt{3} t^2-\frac{t}{7}+e
\right)\left(-4+58t-180t^2+140t^3\right)dt.$$
Both sides are the number $-\frac{1}{21}-\frac{\sqrt{3}}{2}+\frac{3 \pi }{5}$.
\end{remark*}


\item What is the adjoint of $\sigma$ if we equip $V$ with the 
inner product defined by $$\<p,q\>:=p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1)?$$
\ba Again, it's convenient to use a basis for $V$ that is orthonormal
with respect to the given inner product:
$$C=\left\{1,(t-1),\frac{1}{2}\left(t-1\right)^2, \frac{1}{3!}\left(t-1\right)^3\right\}$$
Here, we have 
$$\Rep_{C,C}(\sigma)=
\left(
\begin{array}{cccc}
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)
\Rightarrow \Rep_{C,C}(\sigma^*)=
\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
\end{array}
\right)
$$
\ea
\begin{remark*}Again, it might be interesting to express $\sigma^*$ as
  a matrix using the basis $E=\{1,t,t^2,t^3\}$.  The result is:
$$\Rep_{E,E}(\sigma^*)=\left(
\begin{array}{cccc}
 -1 & -\frac{1}{2} & -\frac{1}{3} & -\frac{1}{2} \\
 1 & 0 & 0 & 1 \\
 0 & \frac{1}{2} & 0 & -\frac{3}{2} \\
 0 & 0 & \frac{1}{3} & 1 \\
\end{array}
\right)$$
\end{remark*}
\een
\ep

\bp Let $M=\left(
\begin{array}{ccc}
 15 & -42 & 12 \\
 -3 & 10 & -6 \\
 -10 & 30 & -11 \\
\end{array}
\right)$.  Find a matrix $N$ so that $N^2=M$.
\ep
\ba Note that for $Q=\left(
\begin{array}{ccc}
 -6 & 3 & -2 \\
 -1 & 1 & 0 \\
 2 & 0 & 1 \\
\end{array}
\right)
$ we have $Q^{-1}MQ=\left(
\begin{array}{ccc}
 4 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 9 \\
\end{array}
\right)$.  
So the matrix $$N=Q \left(
\begin{array}{ccc}
 2 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 3 \\
\end{array}\right)Q^{-1}=
\left(
\begin{array}{ccc}
 3 & -6 & 0 \\
 -1 & 4 & -2 \\
 -2 & 6 & -1 \\
\end{array}
\right)$$ satisfies $N^2=M$.
\ea


\bp Suppose that $A$ is a $2\times 2$ real matrix and let
$v\in \R^2$ be any vector.  Since 
$\R^2$ is two dimensional, the three vectors $\{v,Av,A^2v\}$ must be
linearly dependent.  Prove that, in fact, $A^2v=\alpha Av+\beta v$
where $\alpha=\tr(A)$ and $\beta=-\det(A).$
\ep


\bp Suppose $V$ is an $n$ dimensional vector space.
An operator $T:V\to V$ is called nilpotent if $T^k=0$
for some integer $k$. 
\ben
\item Prove that if $T$ is nilpotent,
  then $T^n=0$.
\ba If $T^k=0$, then the minimum polynomial for $T$ divides $T^k$, so
the minimum polynomial $m_T(t)|t^k$.  This means that
$m_T(t)=t^r$ for some $r\leq k$.  Since every linear factor of the
characteristice polynomial $\chi_T(t)$ is a factor of $m_T(t)$, and
$\chi_T(t)$ is degree $n$, we conclude $\chi_T(t)=t^n$.  The Cayley
Hamilton theorem implies $\chi_T(T)=0 \Leftrightarrow T^n=0$.
\ea
\item Prove that if $T$ is nilpotent, then there is a basis of
  vectors of $V$ so that the matrix for $T$ is upper triangular
  with zeroes on the diagonal.
\ba The remarks above about $m_T(t)=t^r$ implies that the Jordan form
for $T$ consists of Jordon blocks of sizes $\leq r$ with zero's on the
diagonal.  As we defined it, the Jordon blocks have $1$'s on the
subdiagonal, but by re-ordering the basis vectors, we will obtain the
transpose matrix, an upper triangular matrix with zeroes on the
diagonal (and zeros and ones on the superdiagonal).
\ea
\een
\ep


\bp Let $V$ be the space of polynomials of degree less than or equal
to $2$.  Let $T:V \to V$ be the homomorphism defined by 
$$p(x)\overset{T}{\mapsto}p(1)x^2+p'(0)x+p''(0)$$
\ben
\item Compute the determinant and trace of $T$.
\ba First, do part (b).  Then, it's straightforward to determine that
$\tr(T)=2$ and $\det(T)=-2$.
\ea
\item Use the ordered basis $B=\{x^2,x,1\}$ to express $T$ as a
  $3\times 3$ matrix $A=\Rep_{B,B}(T)$.
\ba
$$A=\left(
\begin{array}{ccc}
 1 & 1 & 1 \\
 0 & 1 & 0 \\
 2 & 0 & 0 \\
\end{array}
\right)$$
\ea
\item Find a basis $E$ consisting of eigenvectors for $T$.
\ba
$$\left\{t^2-t+1,-2,t^2+2 t+2\right\}$$
\ea
\item Let $Q$ be the change of basis matrix $Q=\Rep_{B,E}(\id)$.
\ba
$$
Q=\left(
\begin{array}{ccc}
 1 & -1 & 1 \\
 0 & 0 & -2 \\
 1 & 2 & 2 \\
\end{array}
\right)$$
\ea
\item Use your the ordered basis $E$ to express $T$ as a
  $3\times 3$ matrix $B=\Rep_{E,E}(\id)$.
$$\left(
\begin{array}{ccc}
 2 & 0 & 0 \\
 0 & -1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right)$$
\een
\ep

\bp Let $$C=
\left(
\begin{array}{ccccc}
 3 & 3 & 2 & 0 & 1 \\
 3 & -5 & -2 & -4 & 1 \\
 6 & 6 & 4 & 0 & 2 \\
 3 & 5 & 3 & 1 & 1 \\
\end{array}
\right)
$$
\ben
\item Find a basis for the kernel of $C$.
\item Find a basis for the image of $C$.
\item Find a basis for the space spanned by the columns of $C$.
\item Find a basis for the space spanned by the columns of $C^t$.
\een
\ba 
It's helpful to realize that the kernel of a matrix $M$ is the same as the
kernel of a matrix $M'$ obtained from $M$ by row reduction.  
Also, the span of the rows of a matrix $M$ are the same as the span of
the rows of a matrix $M'$ obtained from $M$ by row reduction.

Here are two matrices obtained from $C$ and $C^t$ by row reduction,
which will be helpful for answering all the questions.
$$\left(
\begin{array}{ccccc}
 1 & 0 & \frac{1}{6} & -\frac{1}{2} & \frac{1}{3} \\
 0 & 1 & \frac{1}{2} & \frac{1}{2} & 0 \\
 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right)\text{ and }\left(
\begin{array}{cccc}
 1 & 0 & 2 & \frac{5}{4} \\
 0 & 1 & 0 & -\frac{1}{4} \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right).$$
\ben
\item From the reduced row echelon form
of $C$ we see that $\left(
\begin{array}{c}
 v \\
 w \\
 x \\
 y \\
 z \\
\end{array}
\right) \in \ker(C)$ if and only if $w+\frac{1}{2}x+\frac{1}{2}y=0$
and $u+\frac{1}{6}x-\frac{1}{2}y+\frac{1}{3}z=0$.  The solutions to
these two equations is a three dimensional space spanned by 
$$\left\{ 
\left(
\begin{array}{c}
 -1 \\
 0 \\
 0 \\
 0 \\
 3 \\
\end{array}
\right),
\left(
\begin{array}{c}
 1 \\
 -1 \\
 0 \\
 2 \\
 0 \\
\end{array}
\right),
\left(
\begin{array}{c}
 -1 \\
 -3 \\
 6 \\
 0 \\
 0 \\
\end{array}
\right)
\right\}
$$
\item To find the image of $C$, note that 
$$C.\left(
\begin{array}{c}
 v \\
 w \\
 x \\
 y \\
 z \\
\end{array}
\right)
=
v \left(
\begin{array}{c}
 3\\3 \\ 6 \\ 3 
\end{array}
\right)
+w \left(
\begin{array}{c}
 3\\-5 \\ 6 \\ 5 
\end{array}
\right)
+x \left(
\begin{array}{c}
 2\\-2 \\ 4 \\ 3 
\end{array}
\right)
+y \left(
\begin{array}{c}
 0\\-4 \\ 0\\ 1  
\end{array}
\right)
+z \left(
\begin{array}{c}
 1\\1 \\ 2 \\ 1 
\end{array}
\right)
$$
That is, the image of $C$ is the span of the columns of $C$, which is
the same as the span of the rows of $C^t$, which is the same as the
span of the the matrix in reduced echelon form obtained by row
reducing $C^t$.  So, the image of $C$ is spanned by 
$$\left\{\left(
\begin{array}{c}
 1 \\ 0 \\ 2 \\ \frac{5}{4}
\end{array}
\right)
\left(
\begin{array}{c}
 0 \\ 1 \\ 0 \\ -\frac{1}{4}
\end{array}
\right)\right\}
$$
\item The space spanned by the columns of $C$ is exactly the image of
  $C$.  So, $$\left\{\left(
\begin{array}{c}
 1 \\ 0 \\ 2 \\ \frac{5}{4}
\end{array}
\right)
\left(
\begin{array}{c}
 0 \\ 1 \\ 0 \\ -\frac{1}{4}
\end{array}
\right)
\right\}
$$
\item The space spanned by the columns of $C^t$ is the span of the
  rows of $C$, which is the same as the span of the rows of the matrix
  obtained by row reducing $C$.  We can read this space off as the
  span of 
$$\left\{\left(
\begin{array}{c}
 1 \\ 0 \\ \frac{1}{6} \\ -\frac{1}{2} \\ \frac{1}{3}
\end{array}\right),
\left(
\begin{array}{c}
 0 \\ 1 \\ \frac{1}{2} \\ \frac{1}{2} \\ 0 
\end{array}
\right)\right\}$$
\een

\ea
\ep


\bp Let 
$$A=
\left(
\begin{array}{ccc}
 -1 & 8 & 2 \\
 -1 & -7 & -1 \\
 2 & 8 & -1 \\
\end{array}
\right)
\text{ and }
B=\left(
\begin{array}{ccc}
 7 & 1 & 3 \\
 7 & 3 & 4 \\
 -10 & -2 & -4 \\
\end{array}
\right)
$$
\ben
\item Find the characteristic and minimum polynomials and the Jordon
  canonical forms for $A$ and $B$.
\item Find matrices $P$ and $Q$ so that $P^{-1}AP$ and $Q^{-1}BQ$ are in
Jordon form.  
\ba
$$\left(
\begin{array}{ccc}
 2 & -3 & -1 \\
 -1 & 1 & 0 \\
 2 & 0 & 1 \\
\end{array}
\right)\left(
\begin{array}{ccc}
 7 & 1 & 3 \\
 7 & 3 & 4 \\
 -10 & -2 & -4 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
 1 & 3 & 1 \\
 1 & 4 & 1 \\
 -2 & -6 & -1 \\
\end{array}
\right)=\left(
\begin{array}{ccc}
 2 & 1 & 0 \\
 0 & 2 & 1 \\
 0 & 0 & 2 \\
\end{array}
\right)$$

$$\left(
\begin{array}{ccc}
 1 & 3 & 1 \\
 1 & 4 & 1 \\
 -2 & -6 & -1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
 -1 & 8 & 2 \\
 -1 & -7 & -1 \\
 2 & 8 & -1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
 2 & -3 & -1 \\
 -1 & 1 & 0 \\
 2 & 0 & 1 \\
\end{array}
\right)
=
\left(
\begin{array}{ccc}
 -3 & 1 & 0 \\
 0 & -3 & 0 \\
 0 & 0 & -3 \\
\end{array}
\right)$$
\ea
\item Compute $e^A$ and $e^B$.
$$e^B=\left(
\begin{array}{ccc}
 1 & 3 & 1 \\
 1 & 4 & 1 \\
 -2 & -6 & -1 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
 e^2 & e^2 & \frac{e^2}{2} \\
 0 & e^2 & e^2 \\
 0 & 0 & e^2 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
 2 & -3 & -1 \\
 -1 & 1 & 0 \\
 2 & 0 & 1 \\
\end{array}
\right)
=\left(
\begin{array}{ccc}
 7 e^2 & e^2 & \frac{7 e^2}{2} \\
 8 e^2 & 2 e^2 & \frac{9 e^2}{2} \\
 -12 e^2 & -2 e^2 & -6 e^2 \\
\end{array}
\right).$$
\een
\ep

\bp Solve the two second order differential equations:
\begin{align*}
y_2''&=y_2-2y'_1\\
y_1''&=y_1+2y_2'
\end{align*}
subject to the initial conditions $$y_1(0)=0,\quad y_2(0)=0, \quad
y_1'(0)=0, \quad y_2'(0)=1.$$
By translating them into a single system of the form $Y'(t)=A Y(t)$
for a $4\times 4$ matrix $A$.
\ep
\ba
Here, we translate the system into the single, first order matrix
equation
$$Y'=AY, \quad Y(0)=B$$
where 
$$A=\begin{pmatrix} 0 & 0 & 1 & 0\\0 & 0 & 0 &1\\
  1 & 0 & 0 & 2 \\0 & 1 & -2 & 0\end{pmatrix}, \quad
B=\begin{pmatrix}0\\0\\0\\1\end{pmatrix}, \text{and }Y=\begin{pmatrix}y_1\\y_2\\y_1'\\y_2'\end{pmatrix}
.$$
We know the solution space for $Y$ is the span of the columns of
$e^{tA}$ and the solution we are interested in is given by $e^{tA}B$.
  
In order to find $e^{tA}$, we study the matrix $A$. We compute
$\chi_A(t)=t^4+2t^2+1=(t^2+1)^2=(t-i)^2(t+i)^2.$  Note that 
$m_A(t)=\chi_A(t)$ does not have distinct factors, so $A$ is not
diagonalizable.  However, we can change basis so as to put $A$ in a
form simple enough to compute the exponential.  We have $P^{-1}AP=A'$ 
where
$$A'=\begin{pmatrix} i & 1 & 0 & 0\\0 & i & 0 &0\\
  0 & 0 & -i & 1 \\0 & 0 & 0 & -i\end{pmatrix} \text{ and }
P=\begin{pmatrix}
-1&-i&-1&i\\
-i&1&i&1\\
-i&0&i&0\\
1&0&1&0
\end{pmatrix}.
$$  
Now, we compute $e^{tA}B=e^{tPA'P^{-1}}=Pe^{tA'}P^{-1}B$ and explicitely
\begin{multline*}
Pe^{tA'}P^{-1}B
 = P \begin{pmatrix}
e^{ti} &te^{it}&0&0\\
0& e^{ti} & 0 & 0\\
0&0&e^{-ti}&te^{-ti}\\
0&0&0&e^{-ti}
\end{pmatrix}
P^{-1}\begin{pmatrix}0 \\0\\0\\1\end{pmatrix}
\\
=\frac{1}{2}
\begin{pmatrix}
(e^{it}+e^{-it})-it(e^{it}-e^{-it}) & -i(e^{it}+e^{-it})-t(e^{it}+e^{-it} &
t(e^{it}+e^{-it}) & -it(e^{it}-e^{-it}) \\
i(e^{it}-e^{-it})+t(e^{it}+e^{-it}) & (e^{it}+e^{-it})-it(e^{it}-e^{-it} &
i(e^{it}-e^{-it}) & t(e^{it}+e^{-it}) \\
* & * & * & *\\
* & * & * & *
\end{pmatrix}
\begin{pmatrix}0 \\0\\0\\1\end{pmatrix}\\
=\frac{1}{2}\begin{pmatrix}
 -it(e^{it}-e^{-it}) \\
 t(e^{it}+e^{-it}) \\
 *\\
 *
\end{pmatrix}
\end{multline*}
(We ignore the last two rows since they're just the derivatives of the
first two.)

Using the basic identities,
$\cos(t)=\frac{1}{2}(e^{it}+e^{-it})$ and
$\sin(t)=\frac{1}{2i}(e^{-it}-e^{-it})$, we simplify:
\begin{gather*}
y_1(t)=t\sin(t)\\
y_2(t)=t\cos(t).
\end{gather*}


\ea



\end{document}

