Dimension of kernels of powers of operators

Question

Let $A$ be a closed operator on a Banach space $E$ and $\lambda$ some eigenvalue of $A$ such that $n = \dim \ker(\lambda - A) < \infty$. Is it than necessarily true that $\dim \ker(\lambda - A)^k \leq nk$ for all $k \in \mathbb N$?

Answer 1

WLOG $\lambda=0$. The exact sequence $0\to\ker A\to\ker A^{k+1}\xrightarrow{A}\ker A^k$ gives $\dim\ker A^{k+1}\leq\dim\ker A+\dim\ker A^{k}$ from which the result is immediate by induction.

Answer 2

The statement is true. It can be shown by induction on $k$.

The statement is true for $k=1$ by $n = \dim (\ker (\lambda I-A))$. Now let $k\in\mathbb{N}$. By the induction hypothesis, $\ker (\lambda I - A)^{k}$ has a basis $F$ of cardinality less than or equal to $nk$. As $\ker (\lambda I-A)^{k}$ is a vector subspace of $\ker (\lambda I-A)^{k+1}$, $F$ can be extended to a basis $F_{0}$ of $\ker (\lambda I - A)^{k+1}$. If $x\in F_{0}\setminus F$, then $(\lambda I-A)^{k}(x)\in \ker (\lambda I - A)$. In addition, as $F_{0}\setminus F$ is a subset of the linearly independent set $F_{0}$, $F_{0}\setminus F$ is linearly independent.

I claim that $$G := \{(\lambda I - A)^{k}(x) : x\in F_{0}\setminus F\}$$ is a linearly independent subset of $\ker (\lambda I-A)$ with the same cardinatlity as $F_{0}\setminus F$. For if this has been shown, then the cardinality of $G$ is less than or equal to $n$ because $\dim (\ker (\lambda I-A))$, and then as $F_{0}$ is equal to $F\cup (F_{0}\setminus F)$, the cardinality of $F_{0}$ is less than or equal to $nk + n = n(k+1)$, proving the statement for $k+1$ and completing the proof by induction.

To show that $G$ has the same cardinality as $F_{0}\setminus F$, it is sufficient to show that distinct elements of $F_{0}\setminus F$ correspond to distinct elements of $G$. Suppose for a contradiction that this fails, so that there are $x,y\in F_{0}\setminus F$ with $x\neq y$ such that $(\lambda I-A)^{k}(x) = (\lambda I-A)^{k}(y)$. Then $x-y\in \ker (\lambda I-A)^{k}$, but as $\ker (\lambda I - A)^{k} = \text{span}(F)$ and $x-y\not\in \text{span}(F)$, this leads to a contradiction. Therefore, each $x\in F_{0}\setminus F$ corresponds to a unique element of $G$, which shows that $F_{0}\setminus F$ and $G$ have the same cardinality.

To show that $G$ is a linearly independent subset of $\ker (\lambda I-A)$, suppose for a contradiction that it is linearly dependent. Then using that distinct elements of $F_{0}\setminus F$ correspond to distinct elements of $G$, there is a finite index set $\mathcal{F}$ such that $\{x_{i}:i\in\mathcal{F}\} \subseteq F_{0}\setminus F$ with $x_{i} \neq x_{j}$ for $i\neq j$ and a collection of scalars (not necessarily distinct) $\{\alpha_{i}:i\in\mathcal{F}\}$ which are not all zero, such that

$$\sum_{i\in \mathcal{F}}\alpha_{i}(\lambda I-A)^{k}(x_{i}) = 0.$$

But by the linearity of $(\lambda I-A)^{k}$, it follows that

$$(\lambda I-A)^{k}\left( \sum_{i\in\mathcal{F}}\alpha_{i}x_{i} \right) = 0.$$

This implies that $\sum_{i\in\mathcal{F}}\alpha_{i}x_{i} \in \ker (\lambda I-A)^{k}$. But $\ker (\lambda I - A)^{k} = \text{span}(F)$ and $\sum_{i\in\mathcal{F}}\alpha_{i}x_{i}\not\in \text{span}(F)$, which is a contradiction. Therefore, $G$ is linearly independent. This completes the proof.

Note that this proof only used linear algebra. It didn't use $A$ being a closed operator or $E$ being a Banach space anywhere.