# about a lemma for the proof of submultiplicativity of matrix norm

in the answer of this post: I don't understand how How $$\|Ax\|\leq ||A|| ||x||$$?

• What is $x,X$? ${}{}$ – copper.hat Oct 18 '19 at 1:38
• I edited it. $x$ is a vector. – user 42493 Oct 18 '19 at 1:39
• How do you define $\|A\|$? – copper.hat Oct 18 '19 at 1:40
• $$\lVert A\rVert=\sup\limits_{\lVert x\rVert =1}\{\lVert Ax\rVert :x\in K^n\}$$ – user 42493 Oct 18 '19 at 1:40
• So, if $x \neq 0$, then $\|A {x \over \|x\|} \| \le \|A\|$. – copper.hat Oct 18 '19 at 1:43

By definition, $$\|Ax\| \leq \|A\|$$ for all $$x$$ with $$\|x\|=1$$. In particular, if $$v$$ an arbitrary vector (distinct from zero) in $$\textsf K^n$$ then we know that $$\|tv\|=1$$ where $$t=1/\|v\|$$.

So, $$\|A(tv)\| \leq \|A\|$$ and observe that $$\|A(tv)\|=\|t(Av)\|=|t|\|Av\|=t\|Av\|$$. In conclusion, we have $$t\|Av\|\leq\|A\|$$ and then $$\|Av\|\leq\frac1t\|A\|=\|v\|\|A\|=\|A\|\|v\|$$