# Comparing matrix norm with the norm of the inverse matrix

I need help understanding and solving this problem.

Prove or give a counterexample: If $A$ is a nonsingular matrix, then $\|A^{-1}\| = \|A\|^{-1}$

Is this just asking me to get the magnitude of the inverse of Matrix $A$, and then compare it with the inverse of the magnitude of Matrix $A$?

If $A$ is nonsingular, then $AA^{-1} = I$, so

In general, then $1 \leqslant ||A||\cdot||A^{-1}|| \implies ||A||^{-1} \leqslant ||A^{-1}||$.

Equality is thus not necessarily guaranteed for arbitrary nonsingular $A$; however, the inequality above implies that equality may occur. Consider an example.

Example:

$A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 0.5 \end{bmatrix}$

$\implies ||A||_{1,2,\infty} = 2$

$\implies ||A^{-1}||_{1,2,\infty} = 1$

$\implies \frac{1}{2} = ||A||_{1,2,\infty}^{-1} \neq ||A^{-1}||_{1,2,\infty} = 1 \implies ||A||^{-1} \neq ||A^{-1}||.$