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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: \begin{equation} \min_{\|x\| = 1} x^TVWV^Tx \leq \|V\|^2 \min_{\|y\| = 1} y^TWy \end{equation}

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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 $$x^\top V W V^\top x = \|V^\top x\|^2 \cdot y^\top W y \le \|V\|^2 \cdot y^\top W y.$$

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  • $\begingroup$ Why does $\|V^{T}x\| \leq \|V^T\|$ Hold? The second is the largest e value of $V^TV$ not of $V^T$ $\endgroup$ – jmsac Nov 23 '18 at 3:06
  • $\begingroup$ Sorry the second quantity is the square root of the largest e-value of $V^TV$ $\endgroup$ – jmsac Nov 23 '18 at 3:20
  • $\begingroup$ @jmsac $\|V^\top x\| = \sqrt{x^\top VV^\top x} \le \lambda_{\max}(VV^\top) = \lambda_{\max}(V^\top V)$ $\endgroup$ – angryavian Nov 23 '18 at 3:21
  • $\begingroup$ Ahh ok thank you! $\endgroup$ – jmsac Nov 23 '18 at 3:23

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