norm of inverse less than 1

Question

I just wanna ask if what I am doing here make sense:

Let $\epsilon$ be arbitrary positive number. Choosing $\epsilon$ and let it approaches 0, I would like to have $||(I-\epsilon A)^{-1}|| < 1$. First, I consider $\epsilon_0$ such that $\epsilon_0 < \frac{1}{||A||}$ then I can use the Taylor expansion to get:

$||(I-\epsilon_0 A)^{-1}|| = ||I + \epsilon_0 A + ...+ \epsilon_0^n A^n|| \leq ||I||+\epsilon_0 ||A ||+ \epsilon_0^2 ||A||^2 + ...$

And I can say, for all $\epsilon < \epsilon_0$,

$||(I-\epsilon A)^{-1}|| < ||I||+\epsilon ||A ||+ \epsilon^2 ||A||^2 + ...$

And when taking $\epsilon$ approaches zero, $||(I-\epsilon A)^{-1}|| <1$.

Answer 1

No, it doesn't make sense. Your work is correct up to the line $$\|(I-\epsilon A)^{-1}\|<\|I\|+\epsilon\|A\|+\epsilon^2\|A\|^2+\ldots \left(= f(\epsilon),\ \text{ say.}\right)\tag{1}$$ However, you cannot infer from $\|(I-\epsilon A)^{-1}\|<f(\epsilon)<f(\epsilon_0)$ that $\|(I-\epsilon A)^{-1}\|<1$ because $f(\epsilon)$ is not smaller than or equal to $1$.

Actually, $\|(I-\epsilon A)^{-1}\|$ can be greater than $1$ for all $\epsilon\neq0$. For instance, consider $A=\pmatrix{0&1\\ 0&0}$ with the operator $2$-norm (i.e. the largest singular value).

Answer 2

Using your notation $f(\epsilon)$, I don't think I meant: $||(I−\epsilon A)^{-1}||<f(\epsilon)<f(\epsilon_0)$, rather

Since $\epsilon_0 < \frac{1}{max |\lambda_i|}$, we have:

$||(I−\epsilon_0 A)^{-1}||< f(\epsilon_0)$ holds,

and then for all $\epsilon \leq \epsilon_0$, (which implies $\epsilon < \frac{1}{max |\lambda_i|}$), we also have

$||(I−\epsilon A)^{-1}||< f(\epsilon)$, Now, take $\epsilon$ tends to 0, $f(\epsilon ) \rightarrow 1$ since $||I||=1$