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Suppose $D$ is a $n$ by $n$ diagonal matrix with positive entries on the diagonal, and $A$ is a $m$ by $n$ matrix, is there anything we can say about the operator norm (or value of maximum entry) of the matrix $DA^T(ADA^T)^{-1}A$ in terms of the operator norms of $A$ and $D$? Intuitively $DA^T(ADA^T)^{-1}A$ should be about as 'large' as a constant (so not dependent on $n$ or $m$) because there are the same number of $D$ and $A$ in the 'denominator' as there are in the 'numerator', but I'm not sure if this is true or how to formally show it, since $A$ is not a square matrix?

One line of thought is $\|DA^T(ADA^T)^{-1}A\| \leq \|D\|\|A^T\|\|A\| \|(ADA^T)^{-1}\|$ but I'm not sure how to proceed from here as I only know how to bound $\|(ADA^T)^{-1}\|$ from below i.e. $\|(ADA^T)^{-1}\| \geq \frac{1}{\|ADA^T\|} \geq \frac{1}{\|A\|\|D\|\|A^T\|}$.

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Here, $\|\cdot\|$ will refer to the operator norm induced by the standard $2$-norm. That is, $\|\cdot\|$ is the spectral norm, which is equal to the largest singular value.

Let $M = DA^T(ADA^T)^{-1}A$. Note that $$ \begin{align} M^2 &= [DA^T(ADA^T)^{-1}A]^2 \\ & = DA^T(ADA^T)^{-1}(ADA^T)(ADA^T)^{-1}A \\ & = DA^T(ADA^T)^{-1}A = M. \end{align} $$ Thus, the eigenvalues of $M$ are all equal to $0$ or $1$. Moreover, $M$ is similar to the symmetric matrix $$ S = D^{-1/2}MD^{1/2} = D^{1/2}A^T(ADA^T)^{-1}AD^{1/2}. $$ Because $S$ is symmetric with eigenvalues $0,1$, we must have $\|S\| = 1$ or $S = 0$. In the case that $\|S\| = 1$, we can conclude that $$ \|M\| = \|D^{1/2}SD^{-1/2}\| \leq \|D^{1/2}\|\cdot\|D^{-1/2}\| \cdot \|S\| = \kappa(D^{1/2}), $$ where $\kappa(D^{1/2})$ denotes the condition number of $D^{1/2}$. Because $D$ is diagonal, we have $$ \kappa(D^{1/2}) = \sqrt{\frac{\lambda_{\max}(D)}{\lambda_{\min}(D)}}, $$ where $\lambda_{\max}(D)$ and $\lambda_{\min}(D)$ are simply the largest and smallest diagonal entries of $D$ (respectively).

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