[Math] Upper bound for the norm of a matrix inverse

Question

Prove the following inequality $$\|A^{-1}\| \le \dfrac{\|A\|^{n-1}}{|\det(A)|} $$
Where A is an $n\times n$, non-singular matrix.

The approach I've taken so far is to use the upper bound on the eigenvalues of the matrix A, as follows

$ |\lambda_k| \le \|A\|$

where $\lambda_k$ is the $k$-th Eigenvalue. Now, for all the eigenvalues, we can write

$|\prod_{k=1}^n \lambda_k| \le \|A\|^n$

= $|\det(A)| \le \|A\|^n$

Since the product of the eigenvalues can be written as the determinant of the matrix $A$.

I feel like I'm just missing one step. How do I proceed after this? I'm unable to introduce $\|A^{-1}\|$ into the mix.

Edit: The norm being used is $||A||_2 = \sigma_{max}(A)$ (the greatest singular value of the matrix A)

Answer 1

One approach is as follows: it suffices to note that

• $|\det (A)| = \prod_k \sigma_k(A)$,
• $\|A\| = \sigma_{\max}(A)$,
• $\|A^{-1}\| = 1/{\sigma_{\min}(A)}$.

In particular, we have $$ \frac{\|A\|^{n-1}}{|\det(A)|} = \frac{\sigma_{\max}(A)^{n-1}}{\prod_{k=1}^n \sigma_k(A)} = \left(\prod_{k=1}^{n-1} \frac{\sigma_{\max}(A)}{\sigma_k}\right) \cdot \frac{1}{\sigma_{\min}(A)} \geq 1/\sigma_{\min}(A) = \|A^{-1}\| $$