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.
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 $A^{-1} - M^{-1} \succeq 0$