Let $M\in \mathcal{M}_n(\Bbb{K})$ a nonsingular matrix, we have $\Vert x\Vert_M=\Vert Mx\Vert$ is a norm over $\Bbb{K}^n$. I have to prove that the subordinate norm is equal to $\Vert A\Vert_M=\Vert MAM^{-1}\Vert$.
I know we have $\Vert MA\Vert\le \Vert M\Vert\Vert A\Vert$ and also $\Vert AM^{-1}\Vert\le \Vert A\Vert\Vert M^{-1}\Vert$ then $$\Vert MAM^{-1}\Vert\le \Vert M\Vert\Vert A\Vert\Vert\Vert M^{-1}\Vert$$
How can I continue?
$$\|A\|_{M} = \sup_{x \in \mathbb{K}^n, x \neq 0} \frac{\|Ax\|_M}{\|x\|_M} $$ $$= \sup_{x \in \mathbb{K}^n, x \neq 0} \frac{\|MAx\|}{\|Mx\|} = \sup_{x \in \mathbb{K}^n, x \neq 0} \frac{\|MAM^{-1}Mx\|}{\|Mx\|}$$
Does this spark some ideas?
Turns out there is slightly simple way to prove this. $$\|A\|_{M} = \sup_{x \in \mathbb{K}^n, x \neq 0} \frac{\|Ax\|_M}{\|x\|_M} $$ $$= \sup_{x \in \mathbb{K}^n, x \neq 0} \frac{\|MAx\|}{\|Mx\|} = \sup_{y \in \mathbb{K}^n, y \neq 0} \frac{\|MAM^{-1}y\|}{\|MM^{-1}y\|} = \|MAM^{-1}\|$$