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It's positive-valued and absolutely homogeneous, but I can't prove it satisfies the triangle inequality. I tried many "counter examples", but all of them satisify the inequality.

If we take T and K matrices, we get to this expression $\sqrt{||T^TT + K^TT + T^TK + K^TK||_2}$. And I don't know how to carry on from here

sadcat_1
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2 Answers2

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In general, if you have $$|a+b|\le |a|+|b|,$$ then you also have $$\sqrt{|a+b|}\le\sqrt{|a|+|b|}\le\sqrt{|a|}+\sqrt{|b|}.$$ This means that for any norm $\|\cdot\|,$ its square root is also a norm, regardless of whether you are on a matrix space etc.

EDIT: Actually, what I wrote above is non-sense as shown in the comments. The right way to proceed is to either know/derive the following fact: $$\|A^TA\|_2 = \|AA^T\|_2 = \sigma(AA^T) = \sigma(A)^2=\|A\|_2^2$$ and then it's done. I am assuming you are using the usual induced $2$-norm.

dezdichado
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    The square root of a norm is in general not a norm as we have $$\sqrt{\Vert \lambda v \Vert} = \sqrt{\vert \lambda\vert} \sqrt{\Vert v\Vert}.$$ – Severin Schraven Jan 14 '22 at 01:42
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    Your proof is flawed, because you do not prove $A\mapsto|A^TA|_2$ is a norm. – MH.Lee Jan 14 '22 at 01:53
  • Maybe I fail to understand what is there to be proven, but from what I understand, I need to show this inequality: $\sqrt{||(K+T)^T(K+T)||_2}\leq \sqrt{||K^TK||_2} + \sqrt{||T^TT||_2}$. So how does this helps? When you open up the brackets, as seen in the OP, you get those pesky $K^TK$ and $T^TT$ – sadcat_1 Jan 14 '22 at 08:55
  • @Suspicious_1 yes and you show that inequality by getting rid of that pesky square roots by following what I wrote. – dezdichado Jan 14 '22 at 16:42
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Since $\|A^TA\|_2 = \|A\|_2^2$ the result follows.

copper.hat
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