Let $\|.\|$ be a norm on $\mathbb{R}^n$, and let $A$ be an $n\times n$ matrix. Put $\|x\|'=\|Ax\|$. What are the precise conditions on $A$ to ensure

Question

Let $\|.\|$ be a norm on $\mathbb{R}^n$, and let $A$ be an $n\times n$ matrix. Put $\|x\|'=\|Ax\|$. What are the precise conditions on $A$ to ensure that $\|.\|'$ is also a norm?

I have to check the following:

I think that $(2)$ and $(3)$ can be fulfilled for any matrix because $\|\lambda x\|'=\|\lambda(Ax)\|=|\lambda|\|Ax\|=|\lambda|\|x\|'$ and $\|x+y\|'=\|A(x+y)\|=\|Ax+Ay\|\leq \|Ax\|+\|Ay\|=\|x\|'+\|y\|'$, I think the problem is in $(1)$ and I do not know what conditions $A$ must meet, could someone help me please? With being a diagonal matrix is ​​it fulfilled? Thank you very much.

Answer 1

You need to know that $\lVert x \rVert' = \lVert Ax \rVert > 0$ for any $x \neq 0$. We know that $\lVert y \rVert > 0$ for any $y \neq 0$ because $\lVert \cdot \rVert$ is a norm, while $\lVert 0 \rVert = 0$. So you need that $Ax \neq 0$ for any $x \neq 0$. For $n \times n$ real matrices, this is the same as demanding that $A$ is invertible.

Answer 2

$A$ has to be an invertible matrix.

Answer 3

Indeed, if $||x||' = ||Ax||$ is to be a norm, then you have already seen that it satisfies the triangle inequality, and scaling.

The first condition states that if $x = 0$ then $||x'|| = 0$. This is satisfied, since if $x = 0$ then $Ax = 0$ since $A$ is linear, and therefore $||x||' = ||Ax|| = 0$.

The second states that if $||x||' = 0$ then $x = 0$. Now, if $||x||' = ||Ax|| = 0$, then $Ax = 0$, and this implies that $x = 0$ if and only if $A$ is injective, which by rank nullity theorem happens if and only if it is invertible.

In other words, $||Ax||$ is a norm if and only if $A$ is an invertible matrix. There are many equivalent conditions : it can be injective/surjective, or have non-zero determinant, or have linearly independent columns etc.

The converse of the question is more interesting : Given two norms $|| \cdot||$ and $||\cdot||'$, does there always exist a matrix $A$ so that $||x||' = ||Ax||$ for all $x$? You may want to think about this.