How can I go about proving the triangle inequality holds for the Frobenius norm?
I worked through $\|A+B\|_F \le \|A\|_F + \|B\|_F$ and was not able to make it work =/.
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You can use the fact that it's a norm from inner product : $$\langle A,B \rangle = \text{Trace}(A^TB)$$ Then: $$\|A+B\|^2=\|A\|^2+\|B\|^2+2\langle A,B\rangle$$ using Cauchy-Schwarz inequalitie we have : $$\langle A,B\rangle \leq \|A\| \|B\|$$ and this gives: $$\|A+B\|^2 \leq \|A\|^2+\|B\|^2+2 \|A\| \|B\|= (\|A\| + \|B\|)^2$$
Mohamed
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It might be problematic using Cauchy-Schwartz, as usually you need to prove triangle-inequality to prove a function is indeed a norm, but CS requires you to know it's a norm (or an inner product for that case). – theFrok Nov 23 '21 at 11:14
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It suffices to show $$\sqrt{\sum_{i=1}^nx_i^2}+\sqrt{\sum_{i=1}^ny_i^2}\geqslant\sqrt{\sum_{i=1}^n(x_i+y_i)^2}.$$Sqaure it, it's equivalent to show$$\sqrt{\sum_{i=1}^nx_i^2}\sqrt{\sum_{i=1}^ny_i^2}\geqslant\sum_{i=1}^nx_iy_i,$$ which is the Cauchy-Schwartz Inequality.
It also works in $\mathbb{C}$ with slight modification.
