# Calculating matrix exponential for $n \times n$ Jordan block [duplicate]

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?

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 $$k$$.
• Okey, so calculating $aI$ is really simple. But B is a nilpotent matrix, but using your notation the $k$ that $B^k=0$ equals to $n$. $B^2$ is a matrix $B$ "moved by one unit" to right side, $B^3$ is moved by two etc. But it's nightmare to write this in formal way $n-th$ times. There is no other way that strictly calculation ? – John Jun 7 at 13:37
• I would simply prove by induction how the $k$-th power of $B$ looks like. I don't think that this calculation is a 'nightmare'. – Jake28 Jun 7 at 14:11