Proving the identity $e^{J_nA} = cos(A) + J_nsin(A)$ where $J_n^2=-I_n$

Question

My professor has told me that the following identity is true: $$e^{J_nA} = \cos(A) + J_n\sin(A)$$ where $e^A$, $\sin(A)$, and $\cos(A)$ are all matrix functions defined by their Taylor series, $A$ is a $n\times n$ matrix, and $$J_n^2=-I_n$$ I know one possible value of $J_2$ here: $$J_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ As a sidenote, n must be even if we want $J_n$ to have only real entries. I am having trouble deriving the identity. I know that the proof for $e^{it}=\cos(t)+i\sin(t)$ involved separating the Taylor series for sin and cos depending on the behavior of $i^n$. I tried a similar method for the matrix exponential and got here. $$e^{J_nA}=I + \frac{J_nA}{1} + \frac{J_nAJ_nA}{2!} + \frac{J_nAJ_nAJ_nA}{3!}+\dotsb $$ I would like to write it as: $$e^{J_nA}=I + \frac{J_nA}{1} + \frac{-A^2}{2!} + \frac{-J_nA^3}{1}...$$ The identity only works if $J_n$ and $A$ commute, though I can't see how this can be true. It could work if $J_n$ and $A$ are simultaneously diagonalizable, but that would impose restrictions on what matrix $A$ is. Is there a way to prove the identity? Is the identity just not true? I know there are multiple possible values for $J_n$. Would I need to impose restrictions on what the value of $J_n$ is to get the identity to work? Thank you

Answer 1

You're absolutely right: this identity simply doesn't work, simply because $A$ and $J_n$ need not commute. If you wanted the identity to work, then imposing $A$ and $J_n$ commute would be, in my opinion, the most natural condition to impose.

As a counterexample, consider the $J_n$ in your question, and $$A = \begin{pmatrix} -1 & 1 \\ 0 & 1\end{pmatrix}.$$ Note that $A^2 = I$, so \begin{align*} \cos(A) &= I - \frac{I}{2!} + \frac{I}{4!} - \ldots = \cos(1)I \\ \sin(A) &= A - \frac{A}{3!} + \frac{A}{5!} - \ldots = \sin(1)A \end{align*} On the other hand, $$J_nA = \begin{pmatrix} 0 & -1 \\ -1 & 1\end{pmatrix}.$$ The eigenvalues of this matrix are $\frac{1 \pm \sqrt{5}}{2}$, which means that $e^{J_n A}$ has eigenvalues $\exp\left(\frac{1 \pm \sqrt{5}}{2}\right)$. However, the eigenvalues of $$\cos(A) + J_n\sin(A) = \cos(1)I + \sin(1)J_nA,$$ are $\cos(1) + \sin(1)\frac{1 \pm \sqrt{5}}{2}$, which are distinct from $\exp\left(\frac{1 \pm \sqrt{5}}{2}\right)$.

... OK, I haven't actually proven that, or even estimated these quantities with a calculator, but I'm sure that they're distinct!

Answer 2

If we take $ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $ then $$ JA = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} $$ and $$ (JA)^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ while $$ J^2A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} $$ so in this case we will get $$ e^{JA} = \cosh(A) + J_n\sinh(A). $$