# Tight bounds on maximum and minimum eigenvalues of product of a matrix with a diagonal matrix: of the form $\ A^T D A$

Lets say, $A\in \mathbb{R}^{m\times n}$, and $D\in \mathbb{R}^{m\times m}$, where $D$ is a diagonal matrix with positive diagonal elements, and all the elements are $\le 1$. For simplicity, assume that $A^TA$ is positive definite.

It is easy to see that $$\lambda_{\max}(A^TDA)\le \lambda_{\max}(A^TA)\cdot \max_{i}D_{ii},\\\lambda_{\min}(A^TDA)\ge \lambda_{\min}(A^TA)\cdot \min_{i}D_{ii}$$

However, are these bounds generally tight? For example, if $D$ is such that one of its elements is $1$, and all others are equal to some small number $\epsilon$, then does the lower bound on the least eigenvalue produce a sever underestimation of the lowest eigenvalue of $A^TDA$. Specifically,

Do there exist explicit upper bounds on $|\lambda_{\min}(A^T D A)-\lambda_{\min}(A^TA)\cdot \min_{i}D_{ii}|$, $|\lambda_{\max}(A^T D A)-\lambda_{\max}(A^TA)\cdot \max_{i}D_{ii}|$? Here $A$ is not a diagonal matrix, otherwise, the bounds would be trivially satisfied with equality. Also, to avoid triviality, $D$ is assumed not to be an identity matrix.

Any references on this subject? Thanks in advance.

• If $A=I$, your inequalities are equalities. – Robert Israel Dec 4 '16 at 21:14
• Yes, I know. But I really want to know about the cases where $A$ is not diagonal, and probably not even square, so that we do not have trivial examples like this. I will include that in my question. Thanks. – Samrat Mukhopadhyay Dec 5 '16 at 8:06
• If it's very close to $I$, the inequalities will be very close to equalities. There's no hope of improving the estimates without something that specifies how far $A$ is from matrices that make the inequality an equality. – Robert Israel Dec 7 '16 at 5:52
• Oh. Actually the matrices $A$ that I am interested in have more rows than columns, have full column rank and their eigenvalues lie in a range $[1-\delta,1+\delta]$, where $\delta\in (0,1)$. Does that make the situation any better? Actually I am not been able to grasp what will happen if $D$ has entries, for example, all $0$,except only one diagonal element $1$. Is it possible then that the difference between the bound and the actually eigenvalues are very small? – Samrat Mukhopadhyay Dec 7 '16 at 7:23
• @Robert, see Rayleigh quotient for a derivation. – Samrat Mukhopadhyay Jan 14 '18 at 6:53