Let a matrix $A$ satisfy $\| A (t) \| < 1$. Prove the following properties

Question

I need help resolving this exercise, any indication would be of great help to me. If anyone knows which book they belong to, I appreciate the information.

Let a matrix $A$ satisfy $\| A (t) \| < 1$. Prove the following properties:

$$\| (I-A)^{-1} \| \leq 1 + \|A\| + \|A\|^{2} + \cdots = \frac{1}{1-\|A\|}$$

and

$$\| (I-A)^{-1} \| \geq \frac{1}{1+\|A\|}$$

Thank you. I will be attentive to any help.

Answer 1

I think you have to be a little careful here. The matrix $A$ is presumably an element of a $\it{finite }$ dimensional, and so complete vector space, $\mathscr V.$

Then, $\| \sum_{k=0}^n A^k \| \le \sum_{k=0}^n \|A\|^k ,$ which converges because $\|A\|<1.$ This implies that the sequence of matrices $\sum_{k=0}^n A^k$ converges to a matrix $A\in \mathscr V.$

To find out which one, observe that $(I-A) \sum_{k=0}^n A^k = I-A^{n+1}$ and now, on taking the limit $n\to \infty,$ we have $(I-A)^{-1}=\sum_{k=0}^n A^k$.

Now we can proceed to prove the inequalites. For example, for the first one,

$\|(I-A)^{-1}\|=\|I+A+A^2+\cdots +\|\le 1+\|A\|+\|A\|^2+\cdots +=\frac{1}{1-║A║}.$

The second one is just as easy.

Answer 2

You can define the inverse by noting that $(I+A+A^2+\cdots)(I-A)=I$. So $(I-A)^{-1}=I+A+A^2+\cdots$. Now just invoke the triangle inequality and the submultiplicitve property of the euclidean norm: $\|A^k\|\leq \|A\|^k$.