# Inequality for diagonal element of the inverse of a sum of matrices

I have a square matrix of the form:

$$M = X^T X + A$$

where $A$ is a diagonal matrix whose diagonal elements are positive, and $X$ is a rectangular matrix (assume that matrix dimensions in this equation are consistent). Then, $M$ is positive definite and symmetric. In particular, it is invertible.

Is the following inequality true in general?

$$(M^{-1})_{nn} < \frac{1}{a_{nn}}$$

where $a_{nn}$ denotes the diagonal entries of $A$. I've done some numerical simulations with random matrices and it seems to hold. But I have not been able to prove it in general.

## migrated from mathoverflow.netJun 4 '18 at 22:36

This question came from our site for professional mathematicians.

• Let $B=X'X$. Then, $M=B+A$, and $M^{-1} = (B+A)^{-1}$. Intuitively, since $(b+a)^{-1} \le a^{-1}$ for positive scalars, your inequality is a matrix version of this idea. As noted by Mark below, the key idea here is to observe that $M \ge A$, and then conclude that $M^{-1} \le A^{-1}$, from which it follows that $e_n^TM^{-1}e_n=(M^{-1})_{nn} \le 1/a_{nn}$. This question is, however, not really a research question, and I feel it may be more suitable for M.SE. – suvrit Jun 4 '18 at 19:34

## 1 Answer

The claim as stated is trivially false. Choose $X =$ matrix of zeros. Then $(M^{-1})_{nn} = \frac{1}{a_{nn}}$

However, $(M^{-1})_{nn} \le \frac{1}{a_{nn}}$ is true, because $M - A$ is positive semidefinite. Therefore, $A^{-1} -M^{-1}$ is positive semidefinite (see derivation below), and thus has nonnegative diagonal elements.

Here is a derivation that $M \succeq A$ implies $M^{-1} \preceq A^{-1}$, given of course, as is the case here, that $M^{-1}$ and $A^{-1}$ exist.

By Schur complement on the identity matrix, $I$, in the below, we have that $M \succeq A$ implies $$\left[\begin{array}{l}I&A^{1/2}\\A^{1/2}&M\end{array}\right] \succeq 0$$. Now applying Schur complement on $M$, we have $$I - A^{1/2}M^{-1}A^{1/2} \succeq 0$$ Applying a similarity transform with $A^{-1/2}$ to the LHS, which leaves its eigenvalues unchanged, we have $A^{-1} - M^{-1} \succeq 0$

• So if $X^T X$ is positive-definite then the strict equality holds. – becko Jun 4 '18 at 15:50
• I did not understand the step where you claim that $A^{-1} - M^{-1}$ is positive semi-definite? How do you get to that? – becko Jun 4 '18 at 15:51
• Although this proves the result, I feel like there should be a more "direct" derivation. I still don't feel like I see why it is true. – becko Jun 4 '18 at 16:46
• For future reference, the Schur complementarity theorem needed is Theorem 1.12 of the book The Schur Complement and Its Applications, by Zhang, Fuzhen (Springer, 2005). – becko Jun 5 '18 at 0:11
• Or section A5.5 of the freely available book in pdf form, "Convex Optimization" by Boyd and Vandenberghe web.stanford.edu/~boyd/cvxbook – Mark L. Stone Jun 5 '18 at 0:29