Show if $||\cdot||$ is a norm on $\mathbb{R}^m$ and $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is linear and one to one the following is a norm.

Question

Show if $||\cdot||$ is a norm on $\mathbb{R}^m$ and $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is linear and one to one then $||\cdot||_*: \mathbb{R}^n \rightarrow \mathbb{R}$ given by $||x||_* = ||Tx||$ is a norm on $\mathbb{R}^n$

Answer 1

Verify the three axioms of a vector space norm. You need one-to-one for nonnegativity. That is the only slightly tricky part.

Answer 2

The nonnegativity and the homogeneity and the triangle inequality of $||.||_*$are straightforward using the properties of the $||.||$ norm and the fact that $T$ is linear.

Now we prove that $$||x||_*=0\iff x=0$$ Indeed we have $$||x||_*=0\iff||T x||=0\iff Tx=0\iff x=0\;\text{since $T$ is injective}$$

Answer 3

This sounds like a homework problem, and since you haven't given any indication of how much work you've put into this I'm just going to outline the method. To show that the function $\|x\|_*=\|Tx\|$ is a norm you need to check that the following three properties are true:

1. $\|Tx\|\geq0$ for any $x$, and $\|Tx\|=0$ only if $\|x\|=0$.
2. $\|T(x+y)\|\leq\|Tx\|+\|Ty\|$.
3. $\|T(ax)\|=|a|\|Tx\|$, where $a$ is a scalar.

You'll need to use the fact that $\|\cdot\|$ is a norm for all of these and the fact that $T$ is one-to-one for part 1.