A and B square matrices and A is positive semidefinite. Prove that

The is easy:

but I'm having problems with the .

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- Feb 12th 2010, 08:18 AMNeTTuRPositive semidefinite matrix
A and B square matrices and A is positive semidefinite. Prove that

The is easy:

but I'm having problems with the . - Feb 13th 2010, 01:54 AMOpalg
- Feb 14th 2010, 12:41 PMNeTTuR
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? - Feb 15th 2010, 03:18 AMOpalg
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. - Feb 15th 2010, 11:09 AMNeTTuR
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? - Feb 15th 2010, 12:43 PMOpalg
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. - Feb 15th 2010, 01:54 PMNeTTuR
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.