I was thinking something like this:
, where D diagonal matrix with eigenvalues of A and U unitary.
This gives and is an eigenvalue to A. If I take the square root of these, then will commute.
So for a Hermitian matrix, there will be a positive semidefinite matrix that commutes with and whose square is . Now because and therefore also is positive semidefinite, there exist an unique positive semidefinite matrix whose square is and this matrix is of course A.
Does this look right?
Notice that if p(x) is a polynomial then is a diagonal matrix whose diagonal elements are . If p(x) is as in my previous comment then it follows that . Since , it follows that . I think you'll find that is a much simpler way to show that AB = BA, rather than trying to work with the matrix C.
Thank you so much for your time. I really appreciate it.
I do not however fully understand the your proof yet. I haven't seen the theorem your using to be able to say that or that B obviously should commute with the polynomial expression. It's important because there is a follow up question saying: "What can you say if A is only assumed to be Hermitian?". Is the extra assumption that A is positive semidefinite unnecessarily strong?
Second fact: If D is a diagonal matrix, with diagonal entries , then each power of D is also diagonal, with entries (again, an easy proof by induction). It follows that p(D) is diagonal, with entries .
I hope those facts will help to make sense of the claims that I made in previous comments.
But where in the proof does the assumption that A is positive semidefinite come in?
Never mind. You should never do math when you are tired. Of course the polynomial is only guaranteed to exist if all the eigenvalues are non-negative. Otherwise the polynomial might have to take multiple values for the same . The eigenvalues in your example are , which gives and .
Thank you so much for your help.