A and B square matrices and A is positive semidefinite. Prove that
The is easy:
but I'm having problems with the .
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?
So if all the eigenvalues are different then all the off-diagonal elements of C must be zero. If some of them are repeated then C can have some nonzero off-diagonal elements.
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?
First fact: If is a polynomial, and A is an n×n matrix, then p(A) denotes the matrix , where I is the n×n identity matrix. If B is a matrix that commutes with A, then B commutes with all the powers of A (easy proof by induction), and of course B commutes with the identity matrix. It follows that B commutes with p(A).
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.
In fact, the result fails if A is only assumed to be hermitian. For an example, look at the 2×2 matrix . This is hermitian, and its square is the identity matrix. So any 2×2 matrix B will commute with . But you should be able to find a matrix B that does not commute with A.
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.