Compute matrix norm induced by weight l1 vector norm For a strictly positive collection of weights $\{w_{i}\}$, consider the weighted $l_{1}$ vector norm:
$$
||x||_{W} = \sum_{i}^{N} w_{i}|x_{i}|
$$
What is (or more accurately, how would you compute) the matrix norm induced by this vector norm? That is, what is $||A||_{W}$? (You may assume $A$ is square so that the weights are well-defined.)
Note that, when $w_{i} = 1$ for all $i$, you get the standard result
$$||A||_{1} = \max_{j} \sum_{i=1}^{N} |a_{ij}|.$$
For this reason, I feel like the correct answer should be
$$||A||_{W} = \max_{j} \sum_{i=1}^{N} w_{i}|a_{ij}|.$$
But I cannot arrive at this result using any of the standard tricks.

Let $W$ be the diagonal matrix of weights. Notice then that,
$$
\|x\|_W 
= \sum_{i=1}^{N} w_i |x_i|
= \sum_{i=1}^{N} |w_ix_i|
= \|Wx\|_1
$$
By definition,
$$
\|A\|_W = \sup_{\|x\|_W = 1} \|Ax\|_W
$$
Thus, letting $y=Wx$ (so that $x = W^{-1}y$),
$$
\|A\|_W 
= \sup_{\|Wx\|_1 = 1} \|WAx\|
= \sup_{\|y\|_1 = 1} \|WAW^{-1}y\|_1
= \| WAW^{-1} \|_1
$$
Now,
$$
[WAW^{-1}]_{i,j} =  (w_i/w_j) A_{i,j}
$$
so,
$$
\|WAW^{-1}\|_1 
= \max_j \sum_{i=1}^{N} \frac{w_i}{w_j} A_{i,j}
= \max_j \frac{1}{w_j} \sum_{i=1}^{N} w_i A_{i,j}
$$
