Proof of subordinate matrix norm equality

Question

Let $||\cdot||_v$ be a norm in $\mathbb{K^n}$ and $Q \in M_n(\mathbb{K)}$ a regular matrix. We define ${||\cdot||_v}^*: \mathbb{K^n} \rightarrow \mathbb{R}^+$ such that ${||X||_v}^*=||QX||_v$.

1. Prove that ${||\cdot||_v}^*$ is also a norm.

2. Let ${||\cdot||_M}$ and ${||\cdot||_M}^*$ their respective subordinate matrix norms in $M_n(\mathbb{K})$. Prove that $\forall A \in M_n(\mathbb{K^n})$ it is true that ${||A||_M}^*={||QAQ^{-1}||_M}$.

My definition of subordinate matrix norm is the following: If $||\cdot||_v$ is a norm then its subordinate matrix norm is ${||A||_M}= \max_{X\neq0} \frac{||AX||_v}{||X||_v}=\max_{||X||=1} {||AX||_v}=\max_{||X||\leq1} {||AX||_v}$.

The first question is pretty straight forward but I'm stuck in the second one.

Answer 1

First of all, we notice that:

• ${||A||_M}^*= \max_{X\neq0} \frac{||AX||_v^*}{||X||_v^*} = \max_{X\neq0} \frac{||QAX||_v}{||QX||_v}$.

• ${||QAQ^{-1}||_M}= \max_{X\neq0} \frac{||QAQ^{-1}X||_v}{||X||_v}$.

Let $Y=QX$. Then $Q^{-1}Y=X$, so:

${||A||_M}^*=\max_{X\neq0} \frac{||QAX||_v}{||QX||_v} =\max_{Y\neq0} \frac{||QAQ^{-1} Y||_v}{||Y||_v} = {||QAQ^{-1}||_M}$.