Exponential of a Jordan Block using Cayley-Hamilton Theorem

Question

For the sake of simplicity, I will only consider $2\times2$ matrices. The Cayley-Hamilton theorem allows us to conclude that

$$e^{At} = \alpha_0I + \alpha_1 A$$

where $\alpha_0$ and $\alpha_1$ can be obtained from the following system of equations

$$e^{\lambda_1 t} = \alpha_0 + \alpha_1 \lambda_1 $$

$$e^{\lambda_2 t} = \alpha_0 + \alpha_1 \lambda_2 $$

$\lambda_1$ and $\lambda_2$ being the eigenvalues of $A$.

Now consider $A=\left( \begin{matrix} 0&1 \\ 0&0 \end{matrix}\right)$. It is straightforward that $e^{At}=\left( \begin{matrix} 1&t \\ 0&1 \end{matrix}\right)$. This result can be obtained by using the decomposition

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

And then using $e^{A+B} = e^Ae^B$, where $e^B$ is obtained using the series definition of the exponential.

Suppose that we wanted to use the Cayley-Hamilton theorem to reach the same result. My intuition is that the previous system of equations is no longer valid, since it arises from the exponential of a diagonal matrix, and this matrix is not diagonalizable. If we ignore this and proceed as usual, the previous equations become the same (since the eigenvalues are equal), from which we conclude that $\alpha_0 = 1$.

While searching for solutions to this, I came across this, where it mentions how to deal with repeated eigenvalues by using the derivatives of the first expression I wrote in this question. However all my attempts at using that information to compute $\alpha_1$ have failed.

Is there something that I am missing, or is it simply not possible to compute the exponential of a Jordan block using the CH theorem?

Answer 1

Computing the exponential from the Jordan normal form is actually quite simple. Write the Jordan block as $$J = \lambda I + N,$$ where $\lambda$ is the eigenvalue corresponding to the block and $N$ nilpotent. As $\lambda I$ and $N$ commute, it suffices to compute the exponentials separately. $\exp(\lambda I)$ is trivial and $N^d = 0$ (by Cayley-Hamilton if you want). Hence you only need to consider the first $d$ terms in the expansion of the exponential.