# Inverse of a matrix and the inverse of its diagonals

While researching a question, I faced with the following problem: I have to prove that for a positive definite matrix we have

$${\mathbf n}^T {\mathbf R}^{-1}{\mathbf n}\geq {\mathbf n}^T {\mathbf D} {\mathbf n}$$ for all $${\mathbf n}$$, where $${\mathbf R}$$ is a $$K$$ times $$K$$ positive definite matrix and the diagonal matrix $$\mathbf{D}$$ is defined as

$${\mathbf D} \triangleq \frac{1}{K} ({\mathbf R} \odot \mathbf{I})^{-1}$$

and $$\odot$$ is elementwise product (Hadamard product), $${\mathbf I}$$ is the identity matrix. It looks like a classic problem which provides a bound for the inverse matrix and the inverse of its diagonal elements. Is there any proof for this bound?

So far I have realized that may be the bounds for matrix norm would be useful.

Let me change the notations and reformulate the problem. You have a positive definite $$n\times n$$ ($$n$$ is your $$K$$) matrix $$R$$ with diagonal $$D$$ (your $$D$$ is $$n$$ times less than mine), and you have to prove that $$nR^{-1}-D^{-1}$$ is non-negative definite. Denote $$R=D^{1/2}QD^{1/2}$$, then $$Q=D^{-1/2}RD^{-1/2}$$ is a positive definite symmetric matrix with all diagonal elements equal to 1. And we have to prove that $$nR^{-1}-D^{-1}=D^{-1/2}(nQ^{-1}-I)D^{-1/2}$$ is non-negative definite. Note that the sum of eigenvalues of $$Q$$ equals to the trace of $$Q$$, which equals to $$n$$. Therefore all eigenvalues of $$Q$$ belong to $$(0,n)$$, and all eigenvalues of $$Q^{-1}$$ belong to $$(1/n,\infty)$$, that just means that $$nQ^{-1}-I$$ is positive definite.
• $Q=D^{-1/2}RD^{-1/2}$, diagonal elements of $Q$ are equal to 1. – Fedor Petrov Mar 7 '19 at 20:01
• There is no theory, I simply define $Q$ this way. – Fedor Petrov Mar 7 '19 at 20:10
• We know that $D^{1/2}$ exists, and so does $D^{-1/2}$, and we define $Q$ as $Q:=D^{-1/2}RD^{-1/2}$. – Fedor Petrov Mar 7 '19 at 20:12