Find matrix exponential of a Jordan matrix

Question

$$M = \begin{bmatrix} 1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & -1 & 1\\ 0 & 0 & 0 & 0 & -1\\ \end{bmatrix}$$

I'm trying to find the exponentiation of this matrix, $e^M$. Wolfram tells me the answer is \begin{bmatrix} e & e & e/2 & 0 & 0\\ 0 & e & e & 0 & 0\\ 0 & 0 & e & 0 & 0\\ 0 & 0 & 0 & e^{-1} & e^{-1}\\ 0 & 0 & 0 & 0 & e^{-1}\\ \end{bmatrix}

But I don't follow how to get here. I tried finding the eigenvectors of $M$ and got $(1,0,0,0,0)$ and $(0,0,0,1,0)$ but I don't know where to go from here.

Answer 1

You should try calculating various powers of $M$ until you can find a nice formula for $M^n$. Then plug your results into the series for $e^M$.

Answer 2

Observe that the matrix is block diagonal,

$$M=\begin{bmatrix}U&0\\0&L\end{bmatrix},$$ where $U$ and $L$ are upper triangular, unit diagonal and persymmetric.

It is an easy matter to check that

$$U^n=\begin{bmatrix}1&n&\dfrac{n(n-1)}2\\0&1&n\\0&0&1\end{bmatrix}$$

and

$$L^n=(-1)^n\begin{bmatrix}1&-n\\0&1\end{bmatrix}.$$

Hence using the Taylor development, we do obtain $e,e,\dfrac e2$ for the distinct elements of $U^n$, and $e^{-1},e^{-1}$ for those of $L$.

$$\sum_{n=0}^\infty\frac1{n!}=\sum_{n=0}^\infty\frac n{n!}=2\sum_{n=0}^\infty\frac{n(n-1)}{2n!}=e,$$

$$\sum_{n=0}^\infty\frac{(-1)^n}{n!}=-\sum_{n=0}^\infty\frac{(-1)^nn}{n!}=e^{-1}.$$