Operator inverse as a geometric series

Question

Let us consider an operator in the form

$$ \left( \mathbf{1} - \lambda \hat{O} \right) \,. $$

Under which circumstances I can write its inverse as

$$ \left( \mathbf{1} - \lambda \hat{O} \right)^{-1} = \sum_{k=0}^{+\infty} \lambda^n\hat{O}^n \,? $$

I'm particularly interested in the finite-dimensional case, or NxN complex matrices, but also in the general one. Is it enough to say "The highest eigenvalue of $\hat{O}$ should be smaller than 1?

Answer 1

Just for reference, we have:

Theorem. For a $d\times d$ complex matrix $A$ with eigenvalues $\lambda_1, \ldots, \lambda_d$, the followings are equivalent:

1. $\sum_{n=0}^{\infty} A^n$ converges.
2. $A^n \to 0$.
3. $\rho(A) < 1$, where $\rho(A) = \max_j |\lambda_j|$ is the spectral radius of $A$.

Proof.

$(1) \implies (2)$ : Trivial.

$(2) \implies (3)$ : Let $(\lambda_j, \mathbf{v})$ be any eigenvalue-eigenvector pair of $A$. Then $ \lambda_j^n\mathbf{v} = A^n\mathbf{v} \to 0$ and hence $\lambda_j^n \to 0$, which implies $|\lambda_j| < 1$. Since this is true for any eigenvalues $\lambda_j$ of $A$, we get $\rho(A) < 1$.

$(3) \implies (1)$ : From the Jordan decomposition of $A$, we can write $A = P(D+N)P^{-1}$, where

• $P$ is invertible,
• $D = \operatorname{diag}(\lambda_1, \ldots, \lambda_d)$, and
• $N$ is a nilpotent matrix with $N^d = 0$ and $ND = DN$.

Then $A^n = P(D + N)^n P^{-1} = P \left( \sum_{k=0}^{d-1} \binom{n}{k} D^{n-k}N^k \right) P^{-1}$. Since $\binom{n}{k}$ grows only polynomially and each entry of $D^{n-k} = \operatorname{diag}(\lambda_1^{n-k}, \ldots, \lambda_d^{n-k})$ decays exponentially fast as $n\to\infty$, it follows that $\sum_{n=0}^{\infty} A^n \to 0$ converges. $\square$