# Bounding the operator norm of a matrix after multiplication and taking inverses.

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. $$\|(ADA^T)^{-1}\| \geq \frac{1}{\|ADA^T\|} \geq \frac{1}{\|A\|\|D\|\|A^T\|}$$.

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).