Show that $\|A^{-1}\|\leq \|A\|^{n-1}$

Question

Suppose that $A$ is a matrix in $SL_n(\mathbb{R})$. Show that $\|A^{-1}\|\leq \|A\|^{n-1}$.

By $\|A\|$, I mean the operator norm $\displaystyle\sup_{\|v\|=1} \|Av\|$.

Answer 1

Here is a proof for the Euclidean norm. This approach is buried in Schäffer. J., "Norms and determinants of linear mappings", Technical report, CMU, Department of Mathematical Sciences, 1970. He does not use the SVD, but the idea is essentially the same.

Let $A=U \Sigma V^*$ be a singular value decomposition of $A$, with $\Sigma=\operatorname{diag} (\sigma_1,...,\sigma_n)$, and $\sigma_1\ge ... \ge\sigma_n$.

Then $\|A\| = \sigma_1, \|A^{-1}\| = {1 \over \sigma_n}$, and $|\det A| = \sigma_1 \cdots \sigma_n$.

Hence $|\det A| \le \sigma_1 \cdots \sigma_{n-1} {1 \over \|A^{-1} \|} \le \|A\|^{n-1} {1 \over \|A^{-1} \|}$, from which we obtain $|\det A| \|A^{-1} \| \le \|A\|^{n-1}$.

The result is not true for general operator norms, for example, with $A=\begin{bmatrix} {1 \over 2} & 1 \\ 0 & 2 \end{bmatrix}$, we have $\det A = 1$, $\|A^{-1} \|_\infty = 3, \|A\|_\infty = 2$ and so $3=\|A^{-1} \|_\infty \not< \|A\|_\infty^{2-1} = 2$.

As an aside, it is worth noting the related Hadamard's inequality, $|\det A| \le \|Ae_1\| \cdots \|A e_n\|$ (Euclidean norm).