Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\exp(A) := \sum_{1\le k\le n}\exp(\lambda_k) P_k$.

share|improve this question
    
Where is this definition "exp is defined in terms of the spectral decomposition" –  juaninf Apr 15 '13 at 14:58
add comment

2 Answers

up vote 2 down vote accepted

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 $P_1+P_{-1}=I$)

share|improve this answer
    
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
add comment

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 $i^{2n+1}=(-1)^ni$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.