Is it possible to upper bound this family of matrices in operator norm?

Let

$$\mathcal E = \{A \in \mathcal M(n \times n; \mathbb C): \|A\|_2 \le \|A_0\|_2\}$$

where $$A_0$$ is some fixed matrix and $$\|\cdot\|_2$$ denotes the induced $$2$$-norm. We also have for every $$A \in \mathcal E$$, $$\rho(A)< 1$$ where $$\rho(\cdot)$$ denotes the spectral radius and $$\rho(A_0) < 1$$. Is it possible to give an upper bound $$C$$ in terms of $$\|A_0\|_2$$ such that $$\|(I-A)^{-1}\|_2 \le C$$ for all $$A \in \mathcal E$$?

• Notably, $\mathcal E$ is compact. So, there necessarily exists a stronger upper bound $r < 1$ such that $\rho(A) \leq r$ for all $A \in \mathcal E$. – Omnomnomnom Jun 21 '18 at 22:40
• Does $\|A\|_2$ denote the Frobenius norm or the induced 2-norm? – Omnomnomnom Jun 21 '18 at 22:41
• @Omnomnomnom: It is induced $2$-norm. – user1101010 Jun 21 '18 at 22:42

The answer is yes. The upper bound I come up with below is $\frac{1}{1 - \|A_0\|_2}$.

It doesn't seem like there's any need to consider the matrix $A_0$ itself. In the below, we will take $M = \|A\|_0 \geq 0$, since no other information about $A_0$ will be used.

Suppose that $M \geq 1$. Then $\mathcal E$ includes the identity matrix, and so we fail to have $\rho(A) < 1$.

Suppose that $M < 1$. We note that $$\|(I-A)^{-1}\|_2 = \left\|\sum_{k=0}^\infty A^k\right\|_2 \leq \sum_{k=0}^\infty \left\|A^k\right\|_2 \leq \sum_{k=0}^\infty \left\|A\right\|_2^k \leq \sum_{k=0}^\infty M^k = \frac{1}{1 - M}$$

• This answer simplifies to the simplest answer to the question: Yes! (Sorry, couldn't help myself.) – Robert van de Geijn Jun 21 '18 at 22:50
• @ulaff.net I agree that any answer to a well formed yes/no question should include either a yes or a no; I've edited accordingly. – Omnomnomnom Jun 21 '18 at 22:53
• @Omnomnomnom: Thanks. I realized I formulated my question in a wrong way. Indeed, the family of interest is for the case $M > 1$ and the family $\mathcal E = \{A \in \mathcal M(n \times n; \mathbb C): \|A\|_2 \le M \text{ and } \rho(A) < 1\}$. Anyway, your answer does addresses the question I asked. – user1101010 Jun 21 '18 at 22:59
• @iris2017 you should make a new post with the correctly formulated question, if you haven't done so already – Omnomnomnom Jun 21 '18 at 23:02
• @iris2017 also, I'm fairly confident that you'll only get the upper bound you want if you define your family as $$\mathcal E = \{A : \|A\|_2 \leq M \text{ and } \rho(A) \leq r\}$$ for some fixed $r$ with $0 \leq r < 1$. For this set, it is notable that an upper bound must exist since $\mathcal E$ is compact and the inverse function is continuous. – Omnomnomnom Jun 21 '18 at 23:04