# Any implications of this diagonal element inequality?

1. Nov 20, 2008

### winterfors

Given two definite positive definite matrices A and B of identical size with the following relationship of their diagonal elements:

$$A_{ii} \geq B_{ii}$$ (no summation)

which also holds after any unitary change of coordinates

$$\textbf{A}'=\textbf{U}^T\textbf{A}\textbf{U}$$

where $$\textbf{U}$$ complete set of orthonormal vectors.

Question: What does this imply in terms of inequalities on the eigenvalues of the matrices?

The sum of eigenvalues of A is of course equal of greater than the sum of eigenvalues B:s eigenvalues, but this is true even without allowing for change of coordinates. I'm sure you must be able to deduce something stronger when the inequality holds under any unitary coordinate change...

2. Nov 23, 2008

### acarchau

My linear algebra is very rusty so be careful of the argument below.

A-B is p.s.d .

I am assuming the matrices are real.

Now from the given condition we have for any orthogonal matrix U, the diagonal elements of $U^T (A-B) U$ are non negative.

In particular choose $U$ to be the orthogonal matrix corresponding to the spectral decomposition of $A-B$ and we get all its eigenvalues as non negative.

Last edited: Nov 23, 2008
3. Nov 23, 2008

### acarchau

I think this also implies that the eigenvalues of A dominate the eigenvalues of B in the following way: if $\lambda_i(A)$ is the $i^{th}$ largest eigenvalue of A and $\lambda_i(B)$ is similarly defined. Then $\sum_{i=1}^{k} \lambda_i(A) \geq \sum_{i=1}^{k} \lambda_i(B)$ for $k=1,\dots,n$.

This follows from the fact that for two hermitian matrices A and B of the same order we have (4.3.27 Horn and Johnson)
$\sum_{i=1}^{k} \lambda_i(A+B) \geq \sum_{i=1}^{k} \lambda_i(A) + \sum_{i=1}^{k} \lambda_i(B)$, in particular substituting the pair $A-B,B$ and noting $\lambda_i(A-B) \geq 0$ we get the result.

Last edited: Nov 23, 2008
4. Nov 23, 2008

### winterfors

This was exactly what I was looking for, many thanks!