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The Jordan condition number of a matrix $A$ is defined to be $\min_{V}\kappa(V)$, where $V$ ranges over complex matrices that satisfy $A = VJV^{-1}$ for $J$ being the unique Jordan normal form matrix of $A$. ($\kappa$ is the condition number, i.e., $\kappa(V) = \|V\|\|V^{-1}\|$).

Clearly, if $A$ is normal then the Jordan condition number of $A$ is $1$, and for arbitrary matrices it is unbounded.

Let $A$ be an $n \times n$ matrix with bounded integer entries. Say, even, a 0-1 matrix. How large can the Jordan condition number of $A$ be? Can it be larger, than, say $n^{n}$? As, e.g., answered here, this is not true in general. However, maybe the integrality helps. For example, if an invertible matrix $X$ has integer values between $[-n,n]$, its condition number cannot be larger than $2n^{n}$.

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  • $\begingroup$ The min in defining $V$ is over complex matrices? $\endgroup$
    – YCor
    Commented May 20, 2018 at 8:40
  • $\begingroup$ @YCor Yes, thanks for the clarification. $\endgroup$
    – Daniel86
    Commented May 20, 2018 at 9:01
  • $\begingroup$ Why is the Jordan condition number unbounded? $\endgroup$
    – wlad
    Commented Dec 24, 2022 at 17:11

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