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, $$ \|WAW^{-1}\|_1 = \max_j \sum_{i=1}^{N} \frac{w_i}{w_j} A_{i,j} = \max_j \frac{1}{w_j} \sum_{i=1}^{N} w_i A_{i,j} $$