# Matrix Norm Inequalities.

Let N(.) be a norm on $$\mathbb{R}^n$$. For a $$n \times n$$ matrix, A nor is defined as $$||A|| = \underset{x \neq 0}{sup}\ \frac{N(Ax)}{N(x)}$$

Assume $$||A|| < 1$$. Show that $$I-A$$ and $$I+A$$ are non singular. Also show that $$\frac{1}{1+||A||} \leq ||(I-A)^{-1}|| \leq \frac{1}{1-||A||}$$

My Approach: I have already proved that the given matrix norm is the norm along with some other results like

• $$N(Ax) \leq ||A||N(x)$$ for all $$x$$.
• $$||AB|| \leq ||A||\ ||B||$$

Let $$B = (I-A)^{-1}$$ and assume it exists.

$$||B|| \geq \frac{1}{||I-A||}$$ $$||B|| \geq \frac{N(x)}{N((I-A)x)}$$ $$||B|| \geq \frac{N(x)}{N(x-Ax)}$$ $$||B|| \geq \frac{N(x)}{N(x) + N(Ax)}$$ $$||B|| \geq \frac{1}{1 + ||A||}$$

For II inequality $$B = (I-A)^{-1}$$ $$(I-A)B = (I-A)(I-A)^{-1}$$ $$B - AB = I$$ $$B = I + AB$$ Take matrix norm both side $$||B|| = ||I + AB||$$ $$||B|| \leq ||I|| + ||AB||$$ $$||B|| \leq 1 + ||A||\ ||B||$$ $$||B|| \leq \frac{1}{1-||A||}$$ I am able to prove the bounds of the norm. How can I prove that $$(I-A)$$ and $$(I+A)$$ are non-singular?

Because $$N(Ax) \le \|A\| N(x) < N(x)$$ it is impossible for $$Ax=x$$ or $$Ax=-x$$ to occur, so $$A-I$$ and $$A+I$$ are nonsingular.
By the sub-multiplicative property $$||AB\| \le \|A\| \|B\|$$, we have $$\|(I - A)^{-1}\| \ge \frac{1}{\|I-A\|} \ge \frac{1}{1 + \|A\|}$$ where the last inequality is due to the triangle inequality for the matrix norm ($$\|A+I\| \le \|A\| + \|I\|$$).
For the other direction, for any $$x \ne 0$$, $$N(x) = N((I-A)(I-A)^{-1} x) \ge N((I-A)^{-1} x) - \|A\| N((I-A)^{-1} x) = (1-\|A\|)N((I-A)^{-1} x).$$ Dividing both sides by $$N(x)$$ and taking the supremum over $$x \ne 0$$ proves the other inequality.
• A few issues: a) $\|B\| \ne \frac{1}{\|I-A\|}$, b) where did the supremum go? c) What is $x$ and why is it appearing sometimes and disappearing other times? Commented Aug 12, 2020 at 6:35
• agree with your first point, I had to use the inequality. $x$ is appearing because I am using the definition of the matrix norm defined in the first few lines of the question. I Commented Aug 12, 2020 at 6:44