Does there exist a matrix $A$ such that $\|Ax\|_{\infty} = \|x\|_1$?

Question

$A \in \mathbb R^{2\times2}, x \in \mathbb R^{2\times1}$

Does there exist matrix $A$: $\|Ax\|_{\infty} = \|x\|_1$, how to prove it for any $x$, or just find one example matrix $A$?

Answer 1

Let $$x=\pmatrix{x_1\\ x_2}$$

We have three possible cases: $$ x_1x_2=0 \\ x_1x_2 <0 \\ x_1x_2>0 $$

If $x_1x_2=0$, then $|x_1|+|x_2|=|x_1+x_2|=|x_1-x_2|$.

If $x_1x_2 <0$ then $|x_1|+|x_2|=|x_1-x_2|$ and $|x_1-x_2| \geq |x_1+x_2|$.

If $x_1x_2 >0$ then $|x_1|+|x_2|=|x_1+x_2|$ and $|x_1+x_2| \geq |x_1-x_2|$.

Therefore the matrix $$ A=\pmatrix{1 & 1 \\ 1 & -1} $$ which sends $$x=\pmatrix{x_1\\ x_2}$$ to $$\pmatrix{x_1+x_2\\x_1- x_2}$$ satisfies $\|Ax\|_\infty=\|x\|_1$ for all $x \in \mathbb{R}^2$.

In fact there are $8$ different matrices that satisfies the relation, just switch columns, rows or signs of $A$.