I am trying to find the closed form solution for $e^{A}$, where $$A = \begin{pmatrix} \lambda & \mu_1 \\ \mu_2 & \lambda\end{pmatrix}. \quad \mu_1\mu_2 < 0$$
Is there a way to approach this problem without having to manually calculate powers of $A$? I think the key lies in having to decompose this matrix, perhaps into SVD, but I am not sure how to go about that. Any advice would be greatly appreciated. Thank you.
If$$B=\begin{bmatrix}0&\mu_1\\\mu_2&0\end{bmatrix},$$then $A=\lambda\operatorname{Id}_2+B$ and, since $\lambda\operatorname{Id}_2$ and $B$ commute,$$\exp(A)=\exp(\lambda\operatorname{Id}_2)\exp(B)=\begin{bmatrix}e^\lambda&0\\0&e^\lambda\end{bmatrix}\exp(B).$$On the other hand,\begin{align}B&=\begin{bmatrix}0&\mu_1\\\mu_2&0\end{bmatrix}\\B^2&=\begin{bmatrix}\mu_1\mu_2&0\\0&\mu_1\mu_2\end{bmatrix}\\B^3&=\begin{bmatrix}0&\mu_1^{\,2}\mu_2\\\mu_1\mu_2^{\,2}&0\end{bmatrix}\\B^4&=\begin{bmatrix}\mu_1^{\,2}\mu_2^{\,2}&0\\0&\mu_1^{\,2}\mu_2^{\,2}\end{bmatrix}\\B^5&=\begin{bmatrix}0&\mu_1^{\,3}\mu_2^{\,2}\\\mu_1^{\,2}\mu_2^{\,3}&0\end{bmatrix}\end{align}and so on … Therefore, if $\exp(B)=\left[\begin{smallmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{smallmatrix}\right]$, then\begin{align}b_{11}=b_{22}&=1-\frac1{2!}\sqrt{-\mu_1\mu_2}^2+\frac1{4!}\sqrt{-\mu_1\mu_2}^4+\cdots\\&=\cos\left(\sqrt{-\mu_1\mu_2}\right),\end{align}and a similar computation shows that$$b_{12}=\sqrt{-\frac{\mu_1}{\mu_2}}\sin\left(\sqrt{-\mu_1\mu_2}\right)\quad\text{and that}\quad b_{12}=\sqrt{-\frac{\mu_2}{\mu_1}}\sin\left(\sqrt{-\mu_1\mu_2}\right).$$Therefore$$\exp(A)=\begin{bmatrix}e^\lambda\cos\left(\sqrt{-\mu_1\mu_2}\right)&e^\lambda\sqrt{-\frac{\mu_1}{\mu_2}}\sin\left(\sqrt{-\mu_1\mu_2}\right)\\e^\lambda\sqrt{-\frac{\mu_2}{\mu_1}}\sin\left(\sqrt{-\mu_1\mu_2}\right)&e^\lambda\cos\left(\sqrt{-\mu_1\mu_2}\right)\end{bmatrix}.$$