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

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

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    $\begingroup$ 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. $\endgroup$
    – dan_fulea
    Feb 12, 2020 at 11:22
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    $\begingroup$ 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? $\endgroup$
    – dan_fulea
    Feb 12, 2020 at 11:26
  • $\begingroup$ Does it have to hold for all $x$, or just for one specific $x$? In the latter case, just take $x=0$. $\endgroup$ Feb 12, 2020 at 11:36

1 Answer 1

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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 $A$.

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