Calculating matrix exponential for $n \times n$ Jordan block

Question

I want to calculate exponential of the matrix which on diagonal has some $a \in \mathbb{R}$ and ones above. The $n\times n$ matrix looks like following

$$ A = \left( \begin{matrix} a & 1 & 0 & 0 & \cdots & 0 \\ 0 & a & 1 & 0 & \cdots & 0 \\ 0 & 0 & a & 1 & 0 & 0 \\ 0 & 0 & 0 & a & \ddots & 0 \\ 0 & 0 & 0 & 0 & \ddots & 1 \\ 0 & 0 & 0 & 0 & 0 & a \end{matrix} \right) $$

I tried to do it by counting determinant of matrix $A-\lambda I$ by the following algorithm :

1. Divide last row by $a-\lambda$, so the $n$-th row is just $0$ and $1$ in the $n$-th column.

2. Subtract $(n-1)$-th row by $n$-th row. Then the $1$ in the $n$-th column and $n-1$ row disappears.

3. Divide $n-1$ row by $a-\lambda$, so the $(n-1)-$ th row is just $0$ and $1$ in the $(n-1)$-th column.

And so on so on. By algorithm above we get matrix with only ones at diagonal, so the determinant of that matrix is just $(a - \lambda)^n$. So we have $n$-th fold eigenvalue equals to $\lambda$. I now I have a problem with derivation of eigenvectors of matrix $A$. Can you give me some advice? Is there any simplest way to calculate that?

Answer 1

Actually this matrix only has a one dimensional eigenspace, spanned by the first unit vector. This is an example of a 'Jordan'-block, you will find lots of theory if you google it.

Maybe a simpler way to calculate the matrix exponential is to write \begin{align} A=a\mathbb{I}+B \end{align} where $\mathbb{I}$ is the unit matrix and $B$ has only one's above the diagonal. Since $\mathbb{I}$ and $B$ commute, you get \begin{align} \exp(A)=\exp(a\mathbb{I})\cdot \exp(B) \end{align} Calculating the exponential of $a\mathbb{I}$ is straightforward and calculating the exponential of $B$ is also not too difficult, since $B$ is nilpotent, i.e $B^k=0$ for some appropriate $k$.