0

I found this answer, but actually, I can not follow the proof inside the given answer. My question is basically the same. Why is $||AB|| \leq ||A|| ||B||$? Can someone give an easy explanation?

Does it matter what dimensions are given for $A$ and $B$, as long as both sides of the inequality can be computed? Is it true that this inequality always holds for the 2-norm? Are there any properties that $A$ and $B$ must fullfill?

Lemonbonbon
  • 171
  • No, this is the answer that I linked in my question. – Lemonbonbon Sep 29 '21 at 17:09
  • Can you explain what you don't understand about the answer? Have you looked at the book that is referenced in the answer? The proof provided in that question works for all matrix dimensions. Your question about the 2-norm is answered here (https://math.stackexchange.com/questions/1393301/frobenius-norm-of-product-of-matrix/1393667) – Pax Sep 29 '21 at 17:11
  • The linked answer has basically 3 steps. The first step somehow defines a matrix norm with $||x|| = 1$, for some reason, that I do not understand. In the next step, it follows $||Ax|| \leq ||A|| ||x||$ for some magical reason. How is this derived? It seems that the answer somehow already assumes that $||AB|| \leq ||A||||B||$ is already true, without actually proving it. – Lemonbonbon Sep 29 '21 at 17:17
  • $|Ax|/|x|=|A(x/|x|)|\le \sup_{y:|y|=1}|Ay|= |A|$. The proof then applies this inequality twice, once for $A$ and $Bx$, then then for $B$ and $x$. – Pax Sep 29 '21 at 17:18

0 Answers0