# Proof of subordinate matrix norm equality

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.

• ${||A||_M}^*= \max_{X\neq0} \frac{||AX||_v^*}{||X||_v^*} = \max_{X\neq0} \frac{||QAX||_v}{||QX||_v}$.
• ${||QAQ^{-1}||_M}= \max_{X\neq0} \frac{||QAQ^{-1}X||_v}{||X||_v}$.
Let $Y=QX$. Then $Q^{-1}Y=X$, so:
${||A||_M}^*=\max_{X\neq0} \frac{||QAX||_v}{||QX||_v} =\max_{Y\neq0} \frac{||QAQ^{-1} Y||_v}{||Y||_v} = {||QAQ^{-1}||_M}$.