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In first line of the proof to theorem 1.2 of this paper https://link.springer.com/article/10.1007/BF00537227. The author stated

If $g\in SL(m,\mathbb{R})$, and $|| g||=m \max_{i,j} |g_{ij}|$ then $|| g^{-1}||\le || g||^{m-1}$.

How do we show this implication is true?

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1 Answer 1

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Edit: last proof was totally wrong, here is a revised one:

Since $g \in SL(m,\mathbb{R})$, $\det(g) = 1$ and so $g^{-1} = \frac{\text{adj}(g)}{\det(g)} = \text{adj}(g)$ where adj denotes the adjugate matrix. This shows that the norms of the entries of $g^{-1}$ are exactly the norms of the first minors $M_{ij}$ of $g$, and so to bound $||g^{-1}||$ it is enough to bound the latter. We know that $M_{ij}$ is the determinant of a $(m-1)\times(m-1)$ sub-matrix $A$ of $g$, so that the norm of every entry of $A$ is bounded by $\frac{||g||}{m}$. Thus by the Leibniz formula for determinants, $M_{ij}$ is bounded by $(m-1)!\cdot\left(\frac{||g||}{m}\right)^{m-1}$ and so $$ ||g^{-1}|| \leq m!\cdot\left(\frac{||g||}{m}\right)^{m-1} = \frac{m!}{m^{m-1}}\cdot||g||^{m-1}. $$ Noting that $m! \leq m^{m-1}$ for every $m \geq 1$ completes the proof.

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  • $\begingroup$ Why is this inequality true$$\frac{1}{|a_0|}(||g^{m-1}|| + \ldots + ||a_1||) \leq \frac{||g^{m-1}||}{|a_0|}$$ when $||g^i||$ are all nonnegative $\endgroup$ Sep 3, 2019 at 1:35
  • $\begingroup$ Sorry, brain fart. I'll edit a fix in a sec $\endgroup$
    – Andrew
    Sep 3, 2019 at 2:04
  • $\begingroup$ This proof should be better, sorry about that $\endgroup$
    – Andrew
    Sep 3, 2019 at 5:06

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