How to calculate matrix exponential of a $2\times 2$ matrix with repeated e values

Question

Specifically, I am trying to calculate the matrix exponential, $e^{At}$, where A = $\begin{bmatrix}-1 & 1\\-9 & 5\end{bmatrix}$. I calculated the the E values to be 2 with a multiplicity of 2 but with only one e vector. Since I cannot diagonalize I thought of splitting A into $\begin{bmatrix}-1 & 0\\0 & 5\end{bmatrix}$ + $\begin{bmatrix}0 & 1\\-9 & 0\end{bmatrix}$ and then $e^{At}$ = $\begin{bmatrix}e^{-t} & 0\\0 & e^{5t}\end{bmatrix}$ + $\begin{bmatrix}cos(t) & sin(t)\\-9sin(t) & cos(t)\end{bmatrix}$ = $\begin{bmatrix}e^{-t}+cos(t) & sin(t)\\-9sin(t) & e^{5t}+cos(t)\end{bmatrix}$. Is this correct?

Answer 1

(The variable $t$ went lost in this answer. JMoravitz's answer takes care of everything.)

Your matrix $A$ is similar to its Jordan normal form $$B:=2I+N, \qquad N=\left[\matrix{0&1\cr 0&0\cr}\right]\ .$$ As $I$ and $N$ commute, and $N^2=0$, one has $$\exp(B)=e^2 I\cdot \exp(N)=e^2(I+N)=\left[\matrix{e^2&e^2\cr 0&e^2\cr}\right]\ .$$ So what you need is the matrix $P$ converting $A$ to its normal form $B$.

Answer 2

We find the Jordan form of the matrix: $A=PJP^{-1}$

$\begin{bmatrix} -1&1\\-9&5\end{bmatrix} = \begin{bmatrix}1&-\frac{1}{3}\\3&0\end{bmatrix}\begin{bmatrix}2&1\\0&2\end{bmatrix}\begin{bmatrix}0&\frac{1}{3}\\-3&1\end{bmatrix}$

So, $e^{At} = \sum\limits_{k=0}^\infty \frac{1}{k!}(At)^k$

$=\sum\limits_{k=0}^\infty \frac{1}{k!}(P\left[\begin{smallmatrix}2&1\\0&2\end{smallmatrix}\right]P^{-1}t)^k = \sum\limits_{k=0}^\infty \frac{1}{k!}P(\left[\begin{smallmatrix}2&1\\0&2\end{smallmatrix}\right]t)^kP^{-1} = P\left(\sum\limits_{k=0}^\infty \frac{1}{k!}(\left[\begin{smallmatrix}2&1\\0&2\end{smallmatrix}\right]t)^k\right)P^{-1}$

$=P\left(e^{\left[\begin{smallmatrix}2&1\\0&2\end{smallmatrix}\right]t}\right)P^{-1} = P\left(e^{\left[\begin{smallmatrix}2&0\\0&2\end{smallmatrix}\right]t+\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]t}\right)P^{-1}$

$=P\left(\sum\limits_{k=0}^\infty \frac{1}{k!}\left[\begin{smallmatrix}2t&0\\0&2t\end{smallmatrix}\right]^k\right)\left(\sum\limits_{n=0}^\infty\frac{1}{n!}\left[\begin{smallmatrix}0&1t\\0&0\end{smallmatrix}\right]^n\right)P^{-1}$

$=P\begin{bmatrix}e^{2t}&0\\0&e^{2t}\end{bmatrix}\left(\begin{bmatrix}1&0\\0&1\end{bmatrix}+\begin{bmatrix}0&1t\\0&0\end{bmatrix}+0+0+\cdots\right)P^{-1}~~~~~(\dagger)$

$=P\begin{bmatrix}e^{2t}&te^{2t}\\0&e^{2t}\end{bmatrix}P^{-1}$

$=\begin{bmatrix}e^{2t}-3te^{2t}&te^{2t}\\-9te^{2t}&e^{2t}+3te^{2t}\end{bmatrix}$

Once familiar with the method, one usually skips straight to $(\dagger)$ after finding the Jordan form.

WolframAlpha returns the same result.