Norm of $Ax$ creates norm of $x$?

Question

Suppose that $A = \pmatrix{3&4\\5&6}.$ Is the $2$-norm of $Ax$ a norm for $x$?

Answer 1

Yes, because given a $(n\times m)$ matrix $A$ and a norm on $\Bbb R^n$, the function $\lVert Ax\rVert$ is a norm on $\Bbb R^m$ if and only if $\ker A=\{0\}$

Answer 2

Yes, the function $n(x) = \|Ax\|$ is a norm. As always, you can prove that $n$ is a norm by checking that it satisfies the subadditive, absolute homogeneity, and point-separating properties (all defined here). The fact that $n$ is point-separating requires that you notice a particular property of the matrix $A$ that your are given.

To show that $n$ is subadditive, note that for vectors $x,y$ we have $$ n(x+y) = \|A(x+y)\| = \|Ax + Ay\| \leq \|Ax\| + \|Ay\| = n(x) + n(y). $$