I am asked to find a formula for computing $||A||_{1,2}$ defined as $max_{x\neq 0}\frac{||Ax||_2}{||x||_1}$ ,where $||.||_p$ is the general p norm on the vector.
I know that $||A||_2$ = $max_{i=1:n}\sqrt{(λ_i(A^TA))}$ where $λ_i(A^TA)$ is the i th eigenvalue of $A^TA$ and $||x||_2 \leq ||x||_1 \leq \sqrt{n}||x||_2$
what I have tried :
$||A||_{1,2}$ = $max_{x\neq 0}\frac{||Ax||_2}{||x||_1}$
$\leq max_{x\neq 0}\frac{||Ax||_2}{||x||_2}$= $max_{i=1:n}\sqrt{(λ_i(A^TA))}$
and I cannot argue any further.
I am trying to find a formula just like I found for $||A||_2$
Let $a_i$ be the $i$-th column of $A$ and $x_i$ the $i$-th component of $x.$ Then $$ \|Ax\|_2 = \left\| \sum_{i=1}^n a_i x_i \right\|_2 \leq \sum_{i=1}^n \left\| a_i x_i \right\|_2 \\ = \sum_{i=1}^n |x_i| \;\|a_i\|_2 \leq \sum_{i=1}^n \left(|x_i| \, \max_j \|a_j\|_2 \right) \\ =\left(\sum_{i=1}^n |x_i|\right)\left(\max_j \|a_j\|_2\right) = \|x\|_1 \,\max_j \|a_j\|_2 $$ Therefore $$ \frac {\|Ax\|_2}{\|x\|_1} \leq \max_j \|a_j\|_2 $$ It can easily be seen that equality is obtained if $x = e_i,$ where $e_i$ is the $i$-th element of the standard basis and $i$ is chosen such that $\|a_i\|_2 = \max_j \|a_j\|_2.$ Therefore $$ \|A\|_{1,2} = \max_j \|a_j\|_2 = \sqrt{\max\operatorname{diag}(A^TA)} $$