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?
Best Answer
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.
There is a lot of "hand-waving" in your proof of the norm bounds.
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.