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