# Plugging a matrix multiplied by an imaginary number in the exponential function

I want to prove $\exp(iAx) = I\cos x + iA\sin x$, where $I$ is the identity matrix $\in M_n(\mathbb{C})$, $A\in M_n(\mathbb{C})$ s.t. $A^2 = I$ and $A$ is normal, $x \in \mathbb{R}$, and $\exp(iAx)$ is defined in terms of the spectral decomposition of $A$. From $A^2 = I$ I know the moduli of the eigenvalues of $A$ are equal to 1, but I don't know how to go further. How do you do that?

EDIT: I clarified the implicit precondition on $A$ which is set in bold (but the proposition seems to hold without this). Also by "$\exp$ is defined in terms of the spectral decomposition", I mean this: let $\lambda_1,\dots,\lambda_n$ be eigenvalues of $A$ and $A=:\sum_{1\le k\le n}\lambda_k P_k$, where $P_k$ is the projection into the eigenspace that belongs to $\lambda_k$, and then define $\exp(A) := \sum_{1\le k\le n}\exp(\lambda_k) P_k$.

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Where is this definition "exp is defined in terms of the spectral decomposition" –  juaninf Apr 15 '13 at 14:58

I decided to write another answer instead of editing my first, since the first one is correct in another definition of $\exp$. Now, using your def:
Since $A^2=I$ we know that for each eigenvalue $\lambda$, we have $\lambda^2=1$. Hence $\lambda=\pm 1$. So, your representation using projections could be rewritten as: $A=P_1-P_{-1}$ (the projections on eigenspaces corresponding to $-1$ and $1$ resp.).
Since the eigenvalues of $iAx$ are $i\lambda x$, they are $\pm ix$. Hence $iAx=ixP_1-ixP_{-1}$ is the decomposition of $iAx$. So, by your def, we have $\exp(iAx)=\exp(ix)P_1+exp(-ix)P_{-1}$.
Using Euler formula: $\exp(ix)=\cos x+i\sin x$, we get: \begin{align*} \exp(iAx)&=(\cos x+i\sin x)P_1+(\cos x-i\sin x)P_{-1}=\cos x(P_1+P_{-1})+i\sin x(P_1-P_{-1})=\\ &=I\cos x+iA\sin x \end{align*} (recall that $P_1+P_{-1}=I$)

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I'm a bit confused, but are the eigenvalues of $A$ necessarily real? I thought since $A \in M_n(\mathbb{C})$ is normal it has $n$ distinct eigenvalues. –  Pteromys Feb 11 '12 at 2:19
@Pteromys: It is diagonalizable, but doesn't necessarily have all distinct eigenvalue. Since $A^2=I$, if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $I$. So $\lambda^2=1$, but then $(\lambda+1)(\lambda-1)=0$, so $\lambda=\pm1$ –  Dennis Gulko Feb 11 '12 at 7:54
(I've been confusing characteristic polynomials with minimal polynomials.) If $A$ has two distinct eigenvalues, they necessarily are 1 and -1, but how do you know that? –  Pteromys Feb 11 '12 at 8:57
@Pteromys: How do I know that there are two distinct eigenvalues? I don't, and it doesn't matter, since if it only has 1 eigenvalue, then one of the projections is 0, and the same equality still holds. You can also loomat those cases as special: if $A$ has just one eigenvalue, then $A=I$ or $A=-I$. –  Dennis Gulko Feb 11 '12 at 11:02

The usual definition for $\exp$ on matrices is: $\exp(B)=\sum_{n=0}^\infty \frac{1}{n!}B^n$ for any $B\in M_{n\times n}(\mathbb{C})$. Similarly, $\cos B=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}B^{2n}$ and $\sin B=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}B^{2n+1}$. If $B$ is diagonalizable, then $B=PDP^{-1}$, $D$ is diagonal. Then $\exp(B)=P\exp(D)P^{-1}$ and $\exp(D)$ is just the $\exp$ of the elements on the diagonal of $D$. The same holds for $\cos$ and $\sin$.
Now, we have: \begin{align*} \exp(iAx) &= \sum_{n=0}^\infty \frac{1}{n!}(iAx)^n\overset{(*)}{=}\sum_{n=0}^\infty \frac{1}{(2n)!}(iAx)^{2n}+\sum_{n=0}^\infty \frac{1}{(2n+1)!}(iAx)^{2n+1} \overset{(**)}{=}\\ &=\sum_{n=0}^\infty \frac{1}{(2n)!}(-1)^nIx^{2n}+\sum_{n=0}^\infty \frac{1}{(2n+1)!}(-1)^niIAx^{2n+1}=\\ &=I\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}+iA\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}=\\ &=I\cos x+iA\sin x \end{align*} Where: $(*)$ is done by summing the odd and even powers separately.
$(**)$ is done by calculating $A^{2n}=(A^2)^n=I^n=I$ and hence $A^{2n+1}=A^{2n}A=A$. Also, $i^{2n}=(-1)^n$ and $i^{2n+1}=(-1)^ni$

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I edited my post so that the intended definition of exp would be clear. –  Pteromys Feb 10 '12 at 14:28
@Pteromys: Ok, I posted an answer that works with your definition. –  Dennis Gulko Feb 10 '12 at 15:07