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Prove the following directly. (Do not use the fact that relates $\left\|A\right\|_2$ to the maximum eigenvalue of $A^{T}A$.)

(a) (I've got this one already) If $D$ is the $n\times n$ diagonal matrix $D=\text{diag}[d_1,d_2,\cdots,d_n]$, then $\left\|D\right\|_2= ^{max}_{j=1:n}\|d_j\|$.

(b) If $A\in\mathbb{R}^{m\times n}$ and $U$ is an $m\times m$ orthogonal matrix, then prove $\left\|U A\right\|_2 = \left\|A\right\|_2$

(c) If $A\in\mathbb{R}^{m\times n}$ and $V$ is an $n\times n$ orthogonal matrix, then prove $\left\|A V\right\|_2 = \left\|A\right\|_2$

So I've got part a done by setting a K such that $\frac{\left\|Dx\right\|_2}{\left\|x\right\|_2} \leq K$ and then maximizing that ratio to achieve equality.

What I'm not sure on is part b, and by extension c, which I believe should be (at least mostly) trivial to find should b make sense to me. My book provides a proof using eigenvectors, as do as any proofs I can find online. So far my guesses mostly involve $\left\|U\right\|_2=1$ because of it being orthogonal, and I'd like to have some extension to Cauchy-Schwarz that works for matrices such that I can at the very least get an upper bound than use some arbitrary K to achieve equality like in a, but nothing I've tried has worked yet.

(I'm only looking for an answer for b, although any useful hints for c would be nice if it's not as simple as I'm thinking.)

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1 Answer 1

Hint: $\|UAx\|_2=\|U(Ax)\|_2=\|Ax\|_2$.

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