Invertible Matrix and matrix norms

Question

I am studying numerical analysis and I have a problem with these questions:

$1.$ Prove that if a square matrix satisfies and inequality $||Ax||\geq\theta||x||$ for all $x$ with $\theta>0$, the $A$ is nonsingular, and $||A^{-1}||\le\theta^{-1}$. This is valid for any vector norm and its subordinate matrix norm. The second part I proved, but I don't know how to proceed proving $A$ is nonsingular.

$2.$ Also I have to prove if $A$ is diagonally dominant, then it will have the previous property.

I would like some clues for proceeding.

Thank you.

Do you know that a square matrix is nonsingular if and only if it has a trivial kernel, i.e. $Ax=0$ iff $x=0$? — Ian, Oct 17 '16 at 02:06
Thank you. I don't know why I didn't realize that. I proved by contradiction. — Laura Cadavid, Oct 17 '16 at 11:03

score 1 · Accepted Answer · answered Oct 17 '16 at 07:34

1

Use $$ \|x\|=\|A^{-1}Ax\|\le\|A^{-1}\|·\|Ax\|. $$

A diagonally dominant matrix is always invertible. Use the perturbation theorem in the row- or column sum norm or use Gershgorin's circle theorem on the eigenvalues.

answered Oct 17 '16 at 07:34

Lutz Lehmann

126,666

Invertible Matrix and matrix norms

1 Answers1