I am trying to prove the following theorem:
If $L:V \to V$ is a linear operator and $V$ is a finite dimensional vector space, then $\text{rank}(L^k) = \text{rank}(L^{k+1})$ for some $k$ and $V = \text{ker}(L^k) \oplus \text{ran}(L^k)$.
Can anyone help me get started? I am not sure where to begin.
Best Answer
To get you started, show that $$\operatorname{ran}(L^k) = \operatorname{ran} (L^{k+1}) = \operatorname{ran} (L^{k + 2}) = \ldots$$ In particular, try using the fact that $\operatorname{ran} (L^k) = \operatorname{ran} (L^{2k})$. This means that, given any $v \in V$, there exists a $w \in V$ such that $L^kv = L^{2k}w$. Try using $w$ to find a decomposition of $v$ into a sum from the kernel and range of $L^k$.
Then, try showing that the intersection of the kernel and range is trivial. For this, you'll want to use the fact that $\operatorname{ker} (L^k) = \operatorname{ker} (L^{2k})$.
EDIT:
Because $\operatorname{rank}(L^k) = \operatorname{rank} (L^{k+1})$ and $\operatorname{ran}(L^k) \supseteq \operatorname{ran} (L^{k+1})$, we must have $\operatorname{ran}(L^k) = \operatorname{ran} (L^{k+1})$. By the rank-nullity theorem, we also have $\operatorname{null}(L^k) = \operatorname{null} (L^{k+1})$. Since we have $\operatorname{ker}(L^k) \subseteq \operatorname{ker} (L^{k+1})$, we similarly get $\operatorname{ker}(L^k) = \operatorname{ker} (L^{k+1})$.
Moreover, this equality continues to hold as we increase $k$. This is a standard result in linear algebra. Again, we have that $\operatorname{ran}(L^{k+1}) \supseteq \operatorname{ran}(L^{k+2})$, so we just need to show that, given a vector $v \in \operatorname{ran}(L^{k+1})$, then $v \in \operatorname{ran}(L^{k+2})$.
If $v \in \operatorname{ran}(L^{k+1})$, then there exists some $w \in V$ such that $v = L^{k+1}w$. Note that $$L^k w \in \operatorname{ran}(L^k) = \operatorname{ran}(L^{k+1}),$$ so there must exist some $u \in V$ such that $L^k w = L^{k+1} u$. Therefore, $$v = L^{k + 1} w = L(L^k w) = L(L^{k + 1} u) = L^{k + 2} u \in \operatorname{ran}(L^{k+2}),$$ as required. Therefore, $\operatorname{rank}(L^{k+1}) = \operatorname{rank} (L^{k+2})$, and so, by the previous logic, we also get $\operatorname{ker}(L^{k+1}) = \operatorname{ker} (L^{k+2})$.
You can continue this by induction to show that \begin{align*} \operatorname{ran}(L^k) &= \operatorname{ran} (L^{k+1}) = \operatorname{ran} (L^{k+2}) = \ldots \\ \operatorname{ker}(L^k) &= \operatorname{ker} (L^{k+1}) = \operatorname{ker} (L^{k+2}) = \ldots \end{align*}
In particular, as I said, we need the fact that $\operatorname{ran}(L^k) = \operatorname{ran} (L^{2k})$ and $\operatorname{ker}(L^{2k}) = \operatorname{ker} (L^{2k})$.
Now we show $V = \operatorname{ker}(L^k) \oplus \operatorname{ran}(L^k)$. Let $v \in V$. Note that $$L^k v \in \operatorname{ran}(L^k) = \operatorname{ran}(L^{2k})$$ so there must exist some $w \in V$ such that $L^{2k}w = L^k v$. Then, we have $$v = (v - L^k w) + L^k w,$$ where $L^k w \in \operatorname{ran}(L^k)$ and $v - L^k w \in \operatorname{ker}(L^k)$, as $$L^k(v - L^k w) = L^k v - L^{2k} w = 0.$$ Therefore, $V = \operatorname{ker}(L^k) + \operatorname{ran}(L^k).$
To show the sum is direct, it suffices to show $\operatorname{ker}(L^k) \cap \operatorname{ran}(L^k) = \{ 0 \}$. Suppose $v \in \operatorname{ker}(L^k) \cap \operatorname{ran}(L^k)$. Then $v = L^k w$ for some $k$ and $0 = L^k v = L^{2k} w$. Therefore $w \in \operatorname{ker}(L^{2k}) = \operatorname{ker}(L^k)$, so $0 = L^k w = v$, as required.