2 Linear algebra questions

1)Let A be an n x n symmetric matrix. Prove eigenvalues are real.

I can do it in the complex case by proving it is equal to it's conjugate but it doesn't work in the real case.

2) if A and B are similar, that is if AP=PB for invertible P, is there an easy way to find powers of A?

Thanks in advance.

Re: 2 Linear algebra questions

Re: 2 Linear algebra questions

1) If is a real matrix then is an hermitian matrix if and only if is a symmetric matrix. You should know all the eigenvalues of an hermitian matrix are real (do you know why?). So if is symmetric and real then is an hermitian matrix thus the eigenvalues of are real.

Re: 2 Linear algebra questions

Re: 2 Linear algebra questions

Yes sorry, A is real matrix. So if A is symmetric, it is hermitian which means T(v)=Av is self adjoint w.r.t to standard inner product on . Hence,letting <.,.> denote the standard inner product and letting eigenvalue t have eigenvector v, we have,

t<v,v> =<tv,v>=<T(v),v>=<v,T(v)>=<v,tv>=t*<v,v>. v is non zero by definition so <v,v> is non zero by positive definite. Therefore t=t* and we are done.

Is this the correct reasoning?

Re: 2 Linear algebra questions

Quote:

Originally Posted by

**boromir** Yes sorry, A is real matrix. So if A is symmetric, it is hermitian which means T(v)=Av is self adjoint w.r.t to standard inner product on

. Hence,letting <.,.> denote the standard inner product and letting eigenvalue t have eigenvector v, we have,

t<v,v> =<tv,v>=<T(v),v>=<v,T(v)>=<v,tv>=t*<v,v>. v is non zero by definition so <v,v> is non zero by positive definite. Therefore t=t* and we are done.

Is this the correct reasoning?

I think that is correct. I have a slightly different proof:

Let be an hermitian matrix and supposte is an eigenvector of and suppose is the corresponding eigenvalue. Then we have and thus

Therefore and thus is real.