Boundedness of $\log(\|A^{-1}\|)/\log(\|A\|)$ as $A$ ranges over $SL(n,\mathbb R)$ I wonder if the quantity $\frac{\log(\|A^{-1}\|)}{\log(\|A\|)}$  is bounded by a constant number that only depends on the size $n$ of the matrix as $A$ ranges over $SL(n,\mathbb R)$ (the determinant one matrices). I couldn't find a counterexample, nor can I prove it.
Here I require $\|\cdot \|$ as the operator norm of the matrix w.r.t the 2-norm of the Euclidean spaces. I am not sure if the answer is the same regardless of the norm of the matrix. Please let me know.

Not really an answer I want, but this should be true for the spectral norm. If $A\in SL(n,\mathbb R)$, then so is $A^{T} A$, and we sort its spectrum as follows:
$$0< |a_1| \le \cdots \le |a_n|.$$
Since the product $|a_1 \cdots a_n|=1$, we have $|a_1 a_n^{n-1}|\ge 1$. Therefore
$$\frac{\log(\|A^{-1}\|)}{\log(\|A\|)}\le \frac{\log(|a_1^{-1}|)} {\log(|a_n|)}\le 
\frac{\log(|a_1^{-1}|)} {\log(|a_n|)} \le n-1.$$
Let me know of any mistake.
