Matrix Norm Division

Question

Suppose $A=uv^*$ where $u$ is an $m$-vector and $v$ is an $n$-vector. For any $n$- vector $x$, we can bound $||Ax||_2$ as follows: $||Ax||_2 = ||uv^*x||_2=||u||_2|v^*x|\leq||u_2||||v||_2||x||_2$.

Why does this imply that $||A||_2 \leq ||u||_2 ||v_2||$? Can you divide $||x||_2$ like that?

Answer 1

Note that $$ \|A\|_2 = \sup_{\|x\|_2 = 1} \|Ax\|_2 \leq \sup_{\|x\|_2 = 1} \|u\|_2 \|v\|_2 \|x\|_2 = \|u\|_2 \|v\|_2 $$

Answer 2

It's due to the definition of matrix norm. The definition of $\|A\|_2$ is $$\|A\|_2 = \sup\left\{ \|Ax\| : \|x\|=1 \right\} $$

Since we are restricted to $\|x\|=1$, the inequality above follows.

Answer 3

If there is an $\alpha$ such that for any nonzero $x$, $\|Ax\|_2\leq\alpha\|x\|_2$, then by the definition of the operator 2-norm, $$ \|A\|_2=\max_{x\neq 0}\frac{\|Ax\|_2}{\|x\|_2}\leq\alpha. $$ In your case, $\alpha=\|u\|_2\|v\|_2$. Note that actually for $A=uv^*$, we have the equality: $\|A\|_2=\|u\|_2\|v\|_2$. It is easy to see this by realizing for what vectors the Cauchy-Schwarz inequality gives the equality: $$ \|Ax\|_2=\|uv^*x\|_2=\underbrace{\|u\|_2\color{red}{|v^*x|}\leq\|u\|_2\color{red}{\|v\|_2\|x\|_2}}_{\text{Cauchy-Schwarz}}. $$ The equality is attained for $x$ proportional to $v$.