Prove: If A is self-adjoint and if A^k=1 for some strictly positive integer k, then A^2=1
Please, help

Question

Prove: If A is self-adjoint and if A^k=1 for some strictly positive integer k, then A^2=1
Please, help

loser66 · Answer

I appreciate any tip, pleaaaaaaaaaase

usukidoll · Answer

:(

loser66 · Answer

Please

usukidoll · Answer

depends on the subject. what is this?

loser66 · Answer

theoretical linear algebra

usukidoll · Answer

don't know sorry D:

loser66 · Answer

No worry, I will ask my prof for help although the chance to get help from him is low. hihihihi

usukidoll · Answer

my question is worse :(

loser66 · Answer

what course? I know your level is higher than me, but still want to take a look

usukidoll · Answer

intro to advanced mathematics ... I'll bump my question

loser66 · Answer

Woah.... above my head, hihihihi

loser66 · Answer

@KingGeorge  any tip?

kinggeorge · Answer

What are some of the properties of self-adjoint matrices that you're allowed to use?

loser66 · Answer

sorry, I confuse , that is A* =A

loser66 · Answer

e. values are real

kinggeorge · Answer

Hm. I'm not completely sure. I'll have to think about it some.

loser66 · Answer

Thanks in advance. I need it for my final review. If you have something, please post it.

kinggeorge · Answer

Well what if we let $\vec{v}$ be an eigenvector of $A$ with eigenvalue $\lambda$. Then$$A^k\vec{v}=\lambda^k\vec{v}$$and$$A^k\vec{v}=I\vec{v}=\vec{v}$$So $\lambda^k=1$ which implies that $\lambda=\pm1$. I'm not sure how to continue from here, but http://en.wikipedia.org/wiki/Unitary_matrix claims that what we have is equivalent to $A$ being unitary, and therefore $$AA^*=I\implies A^2=I$$