One of Schur's famous results says that if $A,B$ are positive semidefinite matrices, then the Hadamard (i.e. entrywise) product $A \circ B$ is also positive semidefinite. It's also true if "semi" is dropped in both cases, of course.

My question concerns the seemingly natural generalization of this to the case that $B$ is no longer positive definite but is Hermitian and has known inertia $(i_{+},i_{-},i_{0})$. What can be said about the inertia of $A \circ B$, assuming $A$ is positive definite?

I found some work on so-called "diagonally signed" Hermitian matrices but it seems that the subject has not been touched much. Is it because there is some basic set of counterexamples that renders the whole question moot? If yes, I'd love to see it.

• ok, deleted my answer after your clarification... – Suvrit Aug 31 '12 at 15:26
• But now I have no answer... :( – Felix Goldberg Aug 31 '12 at 16:55
• $A=P^TDP$, $P\in O(n)$, $D$ diagonal $>0$. $B=Q^T\Delta Q$, $\Delta$ diagonal, $Q\in O(n)$. Now $A\circ B=C$, with Einstein convention, $$C_{ij}x_ix_j=p_{ki}d_kp_{kj}q_{li}\delta_l q_{lj}x_ix_j= d_k\delta_l (p_{ki}q_{li}x_i) (p_{kj}q_{lj}x_j)= d_k\delta_l y_{kl}^2,$$ with $y_{kl}=p_{ki}q_{li}x_i$. This identity gives a simple proof of Schur's result, but may also be useful for your question. – Bazin Sep 4 '12 at 20:29

Here's my second attempt at an answer.

As before, I claim that in general, anything can happen. That is, the inertia of $A\circ B$ can be almost anything (i.e., can be very different from the inertia of either $A$ or $B$).

Here is a line of thought.

1. Let $A$ be psd, and let $B=ee^T$ (the all ones matrix). Then, $B$ has just nonzero eigenvalue, which is $e^Te$; but inertia of $A \circ B$ is equal to inertia of $A$.
2. Let $B=-ee^T$ instead, then inertia of $A\circ B$ is equal to "negative" inertia of $A$
3. The above two examples suggest to me that by carefully manipulating $B$ we can make $A\circ B$ have any kind of inertia.
4. To substantiate this intuition, consider the following example:

\begin{equation*} A=\begin{bmatrix} 9 & 1 & -4\\\\ 1 & 2 & -6\\\\ -4&-6 & 24 \end{bmatrix},\quad c = \begin{bmatrix} 5\\\\ -4\\\ -1 \end{bmatrix}. \end{equation*}

Now by setting $B=cc^T + \alpha ee^T$, and varying $\alpha$, you can construct several examples where the inertia of $A\circ B$ ranges from all positive eigenvalues to all negative eigenvalues (this examples excludes the all zeros case), while being different from the inertia of $B$ (or with suitable modification, of $A$).

Some of the above search was guided by inequalities such as $\lambda(A\circ B) \ge (\min_i a_ii)\lambda_\min(B)$. Thus, in general, I think it is easier to derive inequalities for the eigenvalues of Hadamard products, than full characterizations. The only other idea worth mentioning here is that since the Hadamard product matrix is a principal submatrix of the Kronecker product, perhaps a more fine-grained statement can still be made.

• Actually, I might not have been clear: $A$ is assumed still positive definite. (Editing now to reflect this). Therefore, in the diagonal case you suggest, the inertia of $A \circ B$ is exactly the same as that of $B$. – Felix Goldberg Aug 31 '12 at 14:48
• Thanks for the detailed and good answer. A small follow-up question: Do you know of any eigenvalue inequality that gives upper bounds on the smallest eigenvalues of $\lambda_{\min}(A \circ B)$? Or does the Michael's example rule out everything? – Felix Goldberg Sep 4 '12 at 22:08
• I think a bound of the form $\lambda(A\circ B) \le \max_i a_{ii}\lambda_\max(B)$ probably also holds.... – Suvrit Sep 5 '12 at 8:30

What if we take A to be an identity matrix and B a symmetric matrix with zero diagonal? Then Hadamard product of A and B is zero. So, we have no information about inertia of B.

• true! excellent example! – Suvrit Sep 4 '12 at 12:14