# matrix operator 2 norm inequality involving product of three matrices

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

$$\| A \|_2 = \sup \limits _{x \ne 0} \frac{\| A x\| _2}{\|x\|_2}$$

• Is $B$ just any matrix? Commented Dec 15, 2020 at 14:26
• Yes, B is any matrix
– Pew
Commented Dec 15, 2020 at 14:31
• In general you should have $$\|AB \| \leq \|A\| \|B\|$$ and $\|A^T\| = \|A \|$. If $A$ is positive definite then $\|A\|_{2} = \lambda_{\text{max}}(A)$. I'm not sure about the constant. You can have $\|A\|_2 \leq \sqrt{r} \|A\|_2$ where $r$ is the rank of the matrix. It's full rank so it's $n$ Commented Dec 15, 2020 at 14:36
• In fact, I need some inequality of the other direction, for example, since $A$ is positive definite, I can let $D = \sqrt{A}$, and $BAB^T = BD (BD)^T$, then I have $\| BD (BD)^T\|_2= \| BD\|^2_2$. Now maybe there is a constant that $\|BD \|_2 \geq C\| B\|_2\|D\|_2$
– Pew
Commented Dec 15, 2020 at 15:07

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.