# Bounding the smallest eigenvalue of symmetric matrix product

Let $X = ABA^T$ where $B \in \mathbb{R}^{p \times p}$ and $B$ is positive definite matrix and $A \in \mathbb{R}^{q \times p}$ so that $X \in \mathbb{R}^{q \times q}$.

My question is concerning an upper bound on the minimum eigenvalue of $X$. When $q > p$, it is clear that ${\rm rank}(X) = p < q$, so the minimum eigenvalue of X is zero. Additionally, when ${\rm rank}(A) < \min(p, q)$, the smallest eigenvalue is again zero.

Besides these two cases, can an upper bound be set on this eigenvalue such that the bound is less than $||A||^2||B||$? Either specific cases or an inequality would be very interesting.

Take $p\ge q$ and $rank(A)=p$, i.e., $A$ has full rank. Then $AA^T$ is invertible and positive definit. Let $\lambda_1$ be the smallest eigenvalue of $B$. Let $\sigma$ be the smallest eigenvalue of $AA^T$, i.e., which is the square of the smallest singular value of $A$.
Now take $x\in \mathbb R^q$. Then $$x^TXx = x^TABA^Tx \ge \lambda_1 \|A^Tx\|^2 = \lambda_1 x^TAA^Tx \ge \lambda_1\sigma_1 \|x\|^2,$$ which provides a lower bound on the eigenvalues of $X$.
An upper bound on the smallest eigenvalue can be obtained by Rayleigh quotients: Let $\mu_1>0$ be the smallest eigenvalue of $X$. Then $\mu_1^{-1}$ is the largest eigenvalue of $X^{-1}$, and $$\frac{x^Tx}{x^TX^{-1}x} \le \mu_1^{-1} \quad \forall x\ne 0,$$ which is equivalent to $$\mu_1\le \frac{x^TX^{-1}x}{x^Tx}, \quad \mu_1\le\frac{x^Tx}{x^TXx} \quad\forall x\ne0.$$ To answer your last question, any eigenvalue of $X$ is less or equal than $\|X\|$, which is smaller than $\|A^2\|\cdot \|B\|$. Provided $\|\cdot\|$ is an operator norm.