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

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

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

• Is $A$ fixed? Is $x$ fixed? Please use that "for all" and "exists" (in the right order) to give the question the right shape. And show the own attempts to solve the issue. Feb 12, 2020 at 11:22
• Hint: Look at the sets $\{\ x\in\Bbb R ^2\ : \ \|x\|_1=1\ \}$ and $\{\ y\in\Bbb R ^2\ : \ \|y\|_\infty=1\ \}$. Is there any linear transform bringing the one into the other one? Feb 12, 2020 at 11:26
• Does it have to hold for all $x$, or just for one specific $x$? In the latter case, just take $x=0$. Feb 12, 2020 at 11:36

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 $$A$$.