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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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  • $\begingroup$ It is not true. $I$ and $-I$ have norm one, but $\frac{1}{2}I + \frac{1}{2}(-I) = 0$ which has norm zero. $\endgroup$ – copper.hat Sep 27 '13 at 21:46
  • $\begingroup$ Did you mean norm $\le 1$ by any chance? $\endgroup$ – copper.hat Sep 27 '13 at 21:47
  • $\begingroup$ Yes. Apologies, I meant spectral norm atmost 1. I'll edit the question. $\endgroup$ – John Sep 27 '13 at 23:44
  • $\begingroup$ The unit spectral norm ball is a spectrahedron. Take a look at this. $\endgroup$ – Rodrigo de Azevedo Jul 22 '18 at 9:16
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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 $\|\mu A + (1-\mu)B \| \le 1$.

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