Let $X$ be a finite dimensional Banach space and define a matrix norm $\| \cdot \|\_{X}$ by


$$
\| A \|\_{X} = \sup\_{x \ne 0} \frac{\|A x\|\_{X}}{\|x\|\_{X}}
$$


where the matrix $A$ is interpreted as an operator $X \to X$ in the obvious way.


I'm looking to define a minimal norm, namely


$$
f(A)
=
\inf\_{X}
\| A \|\_{X}
$$


What can be said about $f$? Does it define a norm? Can it be computed?



Minimal value of matrix norm induced by a norm
For finite matrices, your 'norm' is the spectral radius of $A$. Indeed, one can construct for each matrix $A$ a matrix norm induced by a vector norm such that $\|A\| \leq \rho(A) + \varepsilon$ for each $\varepsilon>0$. (And, on the other hand, $\|A\|\geq \rho(A)$ for each norm induced by a vector norm).


1. for each matrix $A$ you can construct a Jordan-like form $A=VJV^{-1}$ in which the strictly upper diagonal part contains $\varepsilon$'s instead of 1 (you can get it by taking the Jordan form and conjugating by $\operatorname{diag}(1,\varepsilon,\varepsilon^2,\dots,\varepsilon^{n-1})$).
2. then define the norm $\|M\|:=\|V^{-1}MV\|\_2$ (where $\|K\|\_2$ is the operator norm of $K$), which is induced by the vector norm $\|x\| = \|V^{-1}x\|$. Then $\|A\| = \rho(A) + O(\varepsilon)$. On the other hand, each norm is greater than the spectral radius.


This is a 'standard' proof trick that I learned in my courses.


