What is $e^{A}$ where A is an anti-diagonal matrix

Question

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?

Answer 1

$$A=\begin{pmatrix} 0 &b \\ c &0 \end{pmatrix}$$ 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.

Answer 2

Recall that: $$e^A=\sum_{k=0}^{+\infty}\frac{A^k}{k!}.$$ Moreover, notice that in your case, one has: $$A^2=\begin{pmatrix}bc&0\\0&bc\end{pmatrix}.$$ Therefore, you can compute $e^A$ summing on even and odd integers. Indeed, one has: $$A^{2k}=\begin{pmatrix}b^kc^k&0\\0&b^kc^k\end{pmatrix},A^{2k+1}=\begin{pmatrix}0&b^{k+1}c^k\\b^kc^{k+1}&0\end{pmatrix}.$$