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?
 A: 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: $$A^{2k}=\begin{pmatrix}b^kc^k&0\\0&b^kc^k\end{pmatrix},A^{2k+1}=\begin{pmatrix}0&b^{k+1}c^k\\b^kc^{k+1}&0\end{pmatrix}.$$
A: $$A=\begin{pmatrix}
0 &b \\ 
c &0 
\end{pmatrix}$$
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.
A: Assuming that by "got a different result" you mean that you got different formulas, the solutions are reformulations of each other. It is easy to verify this numerically (for instance with Julia which has a built-in matrix exponential)
b=5; c=pi*1.1;
EE0=exp([0 b; c 0]);

which gives
2×2 Array{Float64,2}:
  31.9404  38.4008
  26.5408  31.9404

Your formula:
s=sqrt(b*c);
e11=0.5*exp(-s)+0.5*exp(s)
e12=sqrt(b)*(exp(s)-exp(-s))/(2*sqrt(c))
e21=sqrt(c)*(exp(s)-exp(-s))/(2*sqrt(b))
e22=0.5*(exp(-s)+exp(s))
EE1=[e11 e12; e21 e22] 

gives
2×2 Array{Float64,2}:
 31.9404  38.4008
 26.5408  31.9404

For completeness, there is a shorter formula using hyperbole functions:
$$
e^A=\begin{bmatrix}\cosh(\sqrt{bc})&\sinh(\sqrt{bc})\frac{\sqrt{b}}{\sqrt{c}}\\ \sinh(\sqrt{bc})\frac{\sqrt{c}}{\sqrt{b}}&\cosh(\sqrt{bc})\end{bmatrix}
$$
Or in programming:
e11=cosh(s);
e12=sinh(s)*(sqrt(b)/sqrt(c))
e21=sinh(s)*(sqrt(c)/sqrt(b))
e22=cosh(s)
EE2=[e11 e12; e21 e22]

which also gives
2×2 Array{Float64,2}:
 31.9404  38.4008
 26.5408  31.9404

