# Bounding the largest Singular value

D is a $n \times n$ diagonal matrix whose diagonal entries lies in $(0,1]$.

B is any $n \times n$ n.n.d. matrix.

What will be the sharpest upper bound on the largest eigenvalue of:

$(D+B)^{-1}D^2(D+B)^{-1}$ ?

A naive upper bound will be $\{\min_{1\leq i \leq n}D_{ii}\}^{-1}$. Does there exist a better upper bound? As the eigenvalues of $(D+B)^{-1}D$ are real and lies in $[0,1]$, I was wondering if a better bound for $\lambda_1\{(D+B)^{-1}D^2(D+B)^{-1}\}$ can be obtained. Else, is it true that the aforementioned bound is the strongest in the sense that as we let $n \to \infty$ the largest singular value of $(D+B)^{-1}D$ is bounded? For the naive bound $\{\min D\}^{-1}$ to attain the above we need $D$ to be $≻ϵI_n$ for all $n$. Are there any other non-trivial assumptions on D under which the aforementioned largest singular value is bounded as $n \to \infty$? Note, B is any nnd matrix.

• Just use the fact that for positive semidefinite (n.n.d. in your terminology) matrices $X$, $Y$, we have $\sigma_j^{1/2}(XY) \le \sigma_j( (X+Y)/2)$ for every singular value $\sigma_j$. – Suvrit Apr 7 '15 at 1:09
• @Suvrit Applying the aforementioned inequality I am still getting $\{\min_{i=1,\ldots,n} D_{ii}\}^{-1}$ for the upper bound as B can be any nnd matrix. Do you think a better bound can be derived? Thank you very much for your help. – Gourab Mukherjee Apr 7 '15 at 5:09