Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

share|cite|improve this question

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.