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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?

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  • $\begingroup$ I'm not sure I can follow your argument for positive definiteness. Why should $c_1$ and $c_2$ be independent of $X$? $\endgroup$
    – Jochen Glueck
    Commented Mar 3, 2019 at 10:24
  • $\begingroup$ 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 $a \to \infty$. $\endgroup$
    – Jonas Adler
    Commented Mar 3, 2019 at 10:47
  • $\begingroup$ 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. $\endgroup$
    – Jochen Glueck
    Commented Mar 3, 2019 at 11:11
  • $\begingroup$ By the way, sub-additivity is probably also going to be a problem since infima do not respect subadditivity, in general. $\endgroup$
    – Jochen Glueck
    Commented Mar 3, 2019 at 11:14
  • $\begingroup$ I updated the proof of positive definiteness, the new one also gives an explicit bound. $\endgroup$
    – Jonas Adler
    Commented Mar 3, 2019 at 11:17

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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.

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    $\begingroup$ 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 $\|x\|+\|Ax\|+\dots+\|A^{m-1} x\|$. $\endgroup$
    – Fedor Petrov
    Commented Mar 3, 2019 at 11:57
  • $\begingroup$ Beautiful proof! Thank you. $\endgroup$
    – Jonas Adler
    Commented Mar 3, 2019 at 12:19
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    $\begingroup$ I forgot to add powers of $r$: $\|x\|+r^{-1}\|Ax\|+\dots+r^{1-m}\|A^{m-1}x\|$. $\endgroup$
    – Fedor Petrov
    Commented Mar 3, 2019 at 17:06

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