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 A1=A1

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?

Answer

If A is nonsingular, then AA1=I, so

1=||I||=||AA1||

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}||.

Attribution
Source : Link , Question Author : Gary , Answer Author : clocktower

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