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The original problem I'm looking at is: given a bound on the operator norm of $\Lambda A \Lambda,$ where $\Lambda, A$ are positive definite matrices and $\Lambda$ is diagonal, what is the tightest bound on the operator norm of $A \Lambda^2.$

My starting point is the fact that these two matrices have the same eigenvalues, so the operator norm of $\Lambda A \Lambda$ upper bounds the spectral radius of $A \Lambda^2.$

For normal matrices, the numerical radius is the same as the spectral radius and the operator norm. While $A \Lambda^2$ is not normal, one might hope that it is nice enough that there is still some nontrivial connection between its numerical and spectral radii ( a bound on the former is a bound on the operator norm, up to a constant). Is this the case, or am I barking up the wrong tree?

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Oog, that's hard to read. Try $DAD$ instead? –  Qiaochu Yuan Apr 24 '13 at 2:09

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This was too long for a comment--I'm not sure if this helps you or not, but it seems like there can't be a bound independent of the sizes of the matrices. Work in size $n\times n$ and consider $$ \Lambda = diag(1,0,0\dots 0) $$ (This isn't positive definite, of course, but it's the limiting case of $\Lambda = diag(1,\epsilon,\dots \epsilon)$.) For any $A$, we have that the norm of $\Lambda A\Lambda$ is equal to the $(1,1)$ entry of $A$, while the norm of $A\Lambda^2$ is equal to the norm of the first column of $A$; so if $A$ is the matrix of all $1$'s then $\|A\Lambda^2\|$ exceeds $\|\Lambda A\Lambda\|$ by a factor of $\sqrt{n}$.

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