How would I show that the convex combination of orthogonal matrixes has spectral norm $ \leq 1$? (I have some idea how to do it ... but right now I'm stuck). Also, how would I prove that the unit spectral norm ball is a convex set?
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It is not true. $I$ and $-I$ have norm one, but $\frac{1}{2}I + \frac{1}{2}(-I) = 0$ which has norm zero. – copper.hat Sep 27 '13 at 21:46
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Did you mean norm $\le 1$ by any chance? – copper.hat Sep 27 '13 at 21:47
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Yes. Apologies, I meant spectral norm atmost 1. I'll edit the question. – John Sep 27 '13 at 23:44
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The unit spectral norm ball is a spectrahedron. Take a look at this. – Rodrigo de Azevedo Jul 22 '18 at 09:16
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If $A$ is orthogonal, then $\|A\| = 1$ (because $\|Ax\| = \|x\|$ for all $x$).
Suppose $A,B$ are orthogonal and $\mu \in [0,1]$.
Then $\|\mu A + (1-\mu)B \| \le \|\mu A \| + \| (1-\mu)B \| = \mu \|A\| + (1-\mu) \|B\|= 1 + (1-\mu) = 1$.
Any norm ball is convex because the function $x \mapsto \|x\|$ is convex, and if we let $\phi(x) = \|x\|$, then we see that the unit ball is given by $\phi^{-1} (-\infty,1]$, and hence convex (since the level sets of a convex function are convex).
Alternatively, you could note that if $\|A\| \le 1$, $\|B\| \le 1$, then the same formula above (with the second last $=$ replaced by $\le$) shows that $\|\mu A + (1-\mu)B \| \le 1$.
copper.hat
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