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.

  • $\begingroup$ $Q=D^{-1/2}RD^{-1/2}$, diagonal elements of $Q$ are equal to 1. $\endgroup$ – Fedor Petrov Mar 7 '19 at 20:01
  • $\begingroup$ There is no theory, I simply define $Q$ this way. $\endgroup$ – Fedor Petrov Mar 7 '19 at 20:10
  • $\begingroup$ We know that $D^{1/2}$ exists, and so does $D^{-1/2}$, and we define $Q$ as $Q:=D^{-1/2}RD^{-1/2}$. $\endgroup$ – Fedor Petrov Mar 7 '19 at 20:12
  • $\begingroup$ I got it. Thank you. $\endgroup$ – Mamal Mar 7 '19 at 20:18

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