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Suppose that we have a real symmetric matrix $A = VDV^T$ with known eigenvalue decomposition ($V$ is an orthogonal matrix). How to show that $\|Ax\|_2 = \|Dx\|_2$?

It is straightorward to show that $\|Ax\|_2 = \|VDV^Tx\|_2 = \|DV^Tx\|_2$. But how to dispose of $V^T$ from the norm?

Community
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Andrei Kulunchakov
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1 Answers1

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It is not the case that $\|DV^Tx\| = \|Dx\|$ for a vector $x$, an orthogonal $V$, and a diagonal $D$.

It is the case, however, that $\|DV^T\| = \|D\|$. Note that $V^T$ is orthogonal. We may therefore say that $$ \|DV^T\| = \max_{\|x\| = 1} \|DV^Tx\| = \max_{\|x\| = 1} \|D(V^Tx)\| = \max_{\|y\| = 1} \|Dy\| = \|D\| $$

Ben Grossmann
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  • I don't understand what use you have for an example, but sure. Take $$ D = \pmatrix{1\&2}, \quad V = \pmatrix{0&1\1&0}, \quad x = \pmatrix{1\0} $$ – Ben Grossmann Mar 05 '17 at 00:08