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I am reading a paper, and I am confused about the following equality:

$$\exp\bigg(\begin{bmatrix}0 & P \\ P & 0\end{bmatrix}t\bigg)=\begin{bmatrix} \cosh(Pt) & \sinh(Pt)\\ \sinh(Pt) & \cosh(Pt)\end{bmatrix}$$

where $P\in \mathbb{R}^{n\times n}$.

We know that $$\cosh(P) = \frac{1}{2}[\exp(P)+\exp(-P)],\quad \sinh(P) = \frac{1}{2}[\exp(P)-\exp(-P)]$$

and we know if $P$ is a diagonal matrix, then $\exp(P)$ can be obtained by putting exponential function to each element on the diagonal of $P$.

However, today $P$ is general, how to derive the quality?

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  • $\begingroup$ Why don't you just calculate the powers of your matrix. Then you will see it. $\endgroup$ – Friedrich Philipp Apr 3 '17 at 4:46
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Note that $$\begin{bmatrix}0&P\\P&0\end{bmatrix}^k=\begin{bmatrix}P^k&0\\0&P^k\end{bmatrix} $$ if $k$ is even, while $$\begin{bmatrix}0&P\\P&0\end{bmatrix}^k=\begin{bmatrix}0&P^k\\P^k&0\end{bmatrix} $$ is $k$ is odd. Therefore $$ \exp\Big(t\begin{bmatrix}0&P\\P&0\end{bmatrix}\Big)=\sum_{k=0}^{\infty}\frac{t^k}{k!}\begin{bmatrix}0&P\\P&0\end{bmatrix}^k$$ $$=\sum_{k=0}^{\infty}\frac{t^{2k}}{(2k)!}\begin{bmatrix}P^{2k}&0\\0&P^{2k}\end{bmatrix}+\sum_{k=0}^{\infty}\frac{t^{2k+1}}{(2k+1)!}\begin{bmatrix}0&P^{2k+1}\\P^{2k+1}&0\end{bmatrix} $$ $$ =\begin{bmatrix}\cosh(tP)&\sinh(tP)\\\sin(tP)&\cosh(tP)\end{bmatrix}$$

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