Let $A$ be a symmetric positive definite matrix, is it true that $$ \| BAB^T \|_2 \geq C \| B\|^2_2\| A\|_2 $$ for some constant $C$? Assuming all matrices are real and the constant may depend on the size $n$.
Here $\Vert \cdot \Vert_2$ is the induced/operator 2-norm defined as
$\| A \|_2 = \sup \limits _{x \ne 0} \frac{\| A x\| _2}{\|x\|_2}$
Yes, you can find a constant that is dependent on $A$ alone.
Based on your comment you can write $$ \begin{aligned} \|BAB^T\|_2 &= \|(BA^{1/2}) (B A^{1/2})^T\|_2 \\ &= \|BA^{1/2}\|^2_2 \\ &\geq \|B\|_2^2 \sigma_{\min}(A^{1/2})^2 \\ & = \|B\|_2^2 \sigma_{\min}(A)\\ \end{aligned} $$ where on the second line we used the fact that $\|XX^*\|_2 = \|X\|_2^2$ (transposition gives the adjoint operator if we view the vector space as a real inner product space and note that the $2$-norm of the vector can be defined via this inner product). On the third line we used the inequality found in this question. Noting that $\|A\|_2 = \sigma_{\max}(A)$ we can take $C = \frac{\sigma_{\min}(A)}{\sigma_{\max}(A)}$ which is non-trivial as $A$ is positive definite.
