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Let $X = A^{\top} D A$ where $A$ is a rectangular matrix such that $A^{\top}A$ is positive definite (can assume strictly positive definite) and $D$ is a diagonal matrix (feel free to assume all elements of $D$ are positive and less than $1$).

I am interested in lower bounds on $\lambda_{\min}\left(X\right)$ that depend on any norm of the diagonal of $D$ (in the problem I am working on, the norm of the diagonal of $D$ will approach $\infty$ and I want to show the minimum eigenvalue diverges).

Without further restriction this is not possible because letting $A = I$, the identity matrix,

$\lambda_{\min}\left(X\right) = \min D_{ii}$.

Which is not a norm of $D$.

Let $n$ be the number of rows of $A$ and $m$ be the number of columns (assume $n > m$) and let $a_{i}$ be the $i$'th row of $A$. My basic thought about why there might be a solution is that:

$X = \sum_{i=1}^{n} a_{i} a_{i}^{\top} d_{i}$.

Intuitively: if we required that $a_{i} a_{i}^{\top}$ had to be a "full" matrix with only non-zero elements than it seems that any $d_{i}$ becoming large would make each element of the $X$ matrix large and so the smallest eigenvalue would be large.

Bottom line: I am looking for a lower bound on the minimum eigenvalue such that the bound becomes large as the norm of the diagonal of $D$ becomes large. Is there any such lower bound for some conditions on $A$?

Additionally: or if the lower bound can depend only $||A^{\top} D||$, that would answer my question as well.

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There cannot be a better lower bound than $\lambda_\min(X) = \min D_{ii}$: If there is a vector $x\ne0$ such that $DAx=\lambda_\min(D)Ax$, then the estimate is sharp.

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