# Exponential of a matrix with t terms

I am trying to find the exponential function of the following matrix: $$\begin{pmatrix}0&t\\4t&0\\ \end{pmatrix}$$

I tried finding the eigenvalues, and got two distinct eigenvalues, but when trying to find the eigenvectors it gets more complicated and I'm not sure if I'm doing it the right way.

However, looking at other examples on the internet, the matrices I find have no $t$ entries, only constant numbers, and the method appears to work fine with those matrices.

Thus, I was wondering if I need to use other methods than those matrices that don't include the $t$ terms.

• Have you tried factoring out the $t$? – amd Mar 5 '16 at 21:32
• The eigenvalues will just be $t$ times the eigenvalues of the matrix without the $t$’s. See Ivo Terek’s answer below for a straightforward approach to solving this. – amd Mar 5 '16 at 21:46
I think that you're overcomplicating it. If $A$ is that matrix, then it is quick to check that $A^2 = 4t^2\,{\rm Id}_2$. Then: \begin{align} \exp(A) &= \sum_{n \geq 0}\frac{A^n}{n!} \\ &= \sum_{k \geq 0} \frac{A^{2k}}{(2k)!}+\sum_{k \geq 0}\frac{A^{2k+1}}{(2k+1)!} \\ &= \sum_{k \geq 0} \frac{(4t^2\,{\rm Id}_2)^k}{(2k)!} +\sum_{k \geq 0} \frac{(4t^2\,{\rm Id}_2)^k t}{(2k+1)!}\begin{pmatrix}0 & 1 \\ 4 & 0\end{pmatrix} \\ &= \sum_{k \geq 0} \frac{4^kt^{2k}}{(2k)!}\,{\rm Id}_2+\sum_{k \geq 0}\frac{4^kt^{2k+1}}{(2k+1)!}\begin{pmatrix}0 & 1 \\ 4 & 0\end{pmatrix} \\ &= \begin{pmatrix} \sum_{k \geq 0} \frac{4^kt^{2k}}{(2k)!} & \sum_{k \geq 0}\frac{4^kt^{2k+1}}{(2k+1)!} \\ \sum_{k \geq 0}\frac{4^{k+1}t^{2k+1}}{(2k+1)!} & \sum_{k \geq 0} \frac{4^kt^{2k}}{(2k)!} \end{pmatrix} \end{align}
• Nice answer, but for the sake of completeness, you should warn the OP that this procedure can be done if and only if you can determine (with recurrence) what $A^n$ and $A^{n+1}$ are, otherwise it won't work! Here, however, it works! – Von Neumann Mar 5 '16 at 22:13