Minimal value of matrix norm induced by a norm

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?

• I'm not sure I can follow your argument for positive definiteness. Why should $c_1$ and $c_2$ be independent of $X$? – Jochen Glueck Mar 3 at 10:24
• You are right, perhaps that is not at all true. I do think we could find such $c_1$, $c_2$ though, I'll have to look around for a proof. We would also have to "normalize" away a constant since the proof obviously fails by considering the family of norms $\|x\| = a \|x\|_2$ as $a \to \infty$. – Jonas Adler Mar 3 at 10:47
• Well, I think the situation is actually a bit more involved. Multiplying the norm by a constant $a > 0$ is not really a problem since the constant $a$ cancels in the definition of the induced matrix norm. – Jochen Glueck Mar 3 at 11:11
• By the way, sub-additivity is probably also going to be a problem since infima do not respect subadditivity, in general. – Jochen Glueck Mar 3 at 11:14
• I updated the proof of positive definiteness, the new one also gives an explicit bound. – Jonas Adler Mar 3 at 11:17

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.
• Jordan form is too heavy machinery for this question. And the real case is not covered, I am afraid. Instead, we may fix any $r>\rho$, choose $m$ for which $\|A^m\|<r^m$ and define new norm as $\|x\|+\|Ax\|+\dots+\|A^{m-1} x\|$. – Fedor Petrov Mar 3 at 11:57
• I forgot to add powers of $r$: $\|x\|+r^{-1}\|Ax\|+\dots+r^{1-m}\|A^{m-1}x\|$. – Fedor Petrov Mar 3 at 17:06