# Matrix norm $|| g^{-1}||$

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?

## 1 Answer

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.

• 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 Sep 3, 2019 at 1:35
• Sorry, brain fart. I'll edit a fix in a sec Sep 3, 2019 at 2:04
• This proof should be better, sorry about that Sep 3, 2019 at 5:06