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

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 $\displaystyle\sup_{\|v\|=1} \|Av\|$.

• @SangchulLee the "largest modulus of eigenvalues" function is not a matrix norm (that is, it isn't submultiplicative). Aug 16, 2016 at 9:58
• @Omnomnomnom, You are right. I guess my intuition is always based on some nice matrices (such as normal matrices) and often I do silly mistakes like this :(... Aug 16, 2016 at 10:01

## 1 Answer

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).