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**Opalg** It certainly doesn't. What you want to do is to maximise the Σ-norm of BX subject to the Σ-norm of X being at most 1. If $\displaystyle X = \begin{bmatrix}w&x\\y&z\end{bmatrix}$ then $\displaystyle BX = \begin{bmatrix}1&2\\3&4\end{bmatrix} \begin{bmatrix}w&x\\y&z\end{bmatrix} = \begin{bmatrix}w+2y&x+2z\\3w+4y&3x+4z\end{bmatrix}$. So you want to maximise $\displaystyle |w+2y| + |x+2z| + |3w+4y| + |3x+4z|$ subject to $\displaystyle |w|+|x|+|y|+|z|\leqslant1$. That looks quite unpleasant. But suppose we cheat a bit and assume that w, x, y and z are all positive. Then the problem becomes: maximise 4w+4x+6y+6z subject to w+x+y+z=1, and the maximum is easily seen to be 6. I would guess that this is the correct answer to the problem, but I don't offhand see how to justify that (without making the cheating assumption).