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.

Thanks for your help!

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    $\begingroup$ Have you tried factoring out the $t$? $\endgroup$ – amd Mar 5 '16 at 21:32
  • $\begingroup$ @amd If I factored out the t terms won't that effect how i calculate the eigenvalues/vectors and the method for diagonalising the matrix? Thanks for the reply! $\endgroup$ – KidMe Mar 5 '16 at 21:35
  • $\begingroup$ 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. $\endgroup$ – 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: $$\begin{align} \exp(A) &= \sum_{n \geq 0}\frac{A^n}{n!} \\ &= \sum_{k \geq 0} \frac{A^{2k}}{(2k)!}+\sum_{k \geq 0}\frac{A^{2k+1}}{(2k+1)!} \\ &= \sum_{k \geq 0} \frac{(4t^2\,{\rm Id}_2)^k}{(2k)!} +\sum_{k \geq 0} \frac{(4t^2\,{\rm Id}_2)^k t}{(2k+1)!}\begin{pmatrix}0 & 1 \\ 4 & 0\end{pmatrix} \\ &= \sum_{k \geq 0} \frac{4^kt^{2k}}{(2k)!}\,{\rm Id}_2+\sum_{k \geq 0}\frac{4^kt^{2k+1}}{(2k+1)!}\begin{pmatrix}0 & 1 \\ 4 & 0\end{pmatrix} \\ &= \begin{pmatrix} \sum_{k \geq 0} \frac{4^kt^{2k}}{(2k)!} & \sum_{k \geq 0}\frac{4^kt^{2k+1}}{(2k+1)!} \\ \sum_{k \geq 0}\frac{4^{k+1}t^{2k+1}}{(2k+1)!} & \sum_{k \geq 0} \frac{4^kt^{2k}}{(2k)!} \end{pmatrix} \end{align}$$

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  • 1
    $\begingroup$ Beat me to it :) Nice answer. $\endgroup$ – parsiad Mar 5 '16 at 21:50
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    $\begingroup$ 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! $\endgroup$ – Von Neumann Mar 5 '16 at 22:13
  • $\begingroup$ @ivo Thanks for the replies, much appreciated. I wanted to say that this method hasn't been taught and probably isn't expected for us to do. However, I have re-tried to find the exponential and I think I managed to find it. But it is not in the summation form you have written. I will post this if you wish me to do so. Thanks again for the help. $\endgroup$ – KidMe Mar 5 '16 at 22:44
  • $\begingroup$ @KidMe You can add your work in the question as an edit, and then we can discuss it more :) $\endgroup$ – Ivo Terek Mar 5 '16 at 22:59

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