# Prove this matrix inequality

Let $$W$$ be $$n$$ by $$n$$ real symmetric positive definite matrix, let $$V$$ be an invertible $$n$$ by $$n$$ square matrix. Then the matrix $$V^TWV$$ is also symetric and positive definite. Now let $$\|V\|$$ denote the largest singular value of $$V$$. That is, the square root of the largest eigenvalue of $$V^TV$$.

Then, show that: $$$$\min_{\|x\| = 1} x^TVWV^Tx \leq \|V\|^2 \min_{\|y\| = 1} y^TWy$$$$

Fix $$x$$ such that $$\|x\| = 1$$.
Let $$y = V^\top x / \|V^\top x\|$$. Note that $$\|y\| = 1$$ and that $$\|V^\top x\| \le \|V^\top\| = \|V\|$$. Then $$x^\top V W V^\top x = \|V^\top x\|^2 \cdot y^\top W y \le \|V\|^2 \cdot y^\top W y.$$
• Why does $\|V^{T}x\| \leq \|V^T\|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$ – jmsac Nov 23 '18 at 3:06
• Sorry the second quantity is the square root of the largest e-value of $V^TV$ – jmsac Nov 23 '18 at 3:20
• @jmsac $\|V^\top x\| = \sqrt{x^\top VV^\top x} \le \lambda_{\max}(VV^\top) = \lambda_{\max}(V^\top V)$ – angryavian Nov 23 '18 at 3:21