<s>[INST] What is $e^{A}$ where A is an anti-diagonal matrix I am trying to get a closed form for the matrix produced by the following operation: $$e^A$$ where $A$ is an anti diagonal matrix, say, of size $2\times 2$:
$$A=\begin{pmatrix}
0 &b \\ 
c &0 
\end{pmatrix}$$
Using Mathematica MatrixExp I got
$$ \left(
\begin{array}{cc}
 \frac{1}{2} e^{-\sqrt{b} \sqrt{c}}+\frac{1}{2} e^{\sqrt{b} \sqrt{c}} & \frac{\sqrt{b} e^{\sqrt{b} \sqrt{c}}}{2 \sqrt{c}}-\frac{\sqrt{b} e^{-\sqrt{b} \sqrt{c}}}{2 \sqrt{c}} \\
 \frac{\sqrt{c} e^{\sqrt{b} \sqrt{c}}}{2 \sqrt{b}}-\frac{\sqrt{c} e^{-\sqrt{b} \sqrt{c}}}{2 \sqrt{b}} & \frac{1}{2} e^{-\sqrt{b} \sqrt{c}}+\frac{1}{2} e^{\sqrt{b} \sqrt{c}} \\
\end{array}
\right)
$$
But here one can find a formula for the computation on a general $2\times 2$ matrix, at the bottom of the page. Using that formula I got a different result.
What is the correct answer?
 [/INST] $$A=\begin{pmatrix}
0 &b \\ 
c &0 
\end{pmatrix}$$
So
$$A^2 = bc I$$
Then
$$e^A = \sum_{k=0}^\infty\frac{A^k}{k!} = \sum_{k=0}^\infty\frac{A^{2k}}{(2k)!} + \frac{A^{2k + 1}}{(2k+1)!} \\
= \sum_{k=0}^\infty\frac{(bc)^k}{(2k)!} I + \frac{(bc)^k }{(2k+1)!} A $$
And you can then sum this component-wise.
</s>