Show that if $\operatorname{Rank}T^2=\operatorname{Rank}T$, then $\operatorname{Rank}T^k=\operatorname{Rank}T$ for all $k\geq1$.

Question

Let $V$ be a finite-dimensional vector space and $T:V\to V$ be a linear transformation. Show that if $\operatorname{Rank}T^2=\operatorname{Rank}T$, then $\operatorname{Rank}T^k=\operatorname{Rank}T$ for all $k\geq1$.

Answer 1

Fist, I prove that $\ker(T^k) \subseteq \ker(T^{k+1})$

Let $x \in \ker(T^k) \Leftrightarrow T^k(x) = 0 \Rightarrow T(T^k(x)) = T(0) \Leftrightarrow T^{k+1}(x) = T(0) = 0 \Leftrightarrow x \in \ker(T^{k+1})$, Q.E.D

We have $\dim V = rank(T) + \dim\ker(T)$, since $rankT^2=rankT$ then $\dim \ker(T) = \dim \ker(T^2)$. On the onther hand, $\ker(T) \subseteq \ker(T^2)$ so $\ker(T) = \ker(T^2)$

I will prove that $rankT=rankT^3$ then by induction you can prove $rankT = rankT^k, \forall k \ge 1$

Let $x \in \ker(T^3) \Leftrightarrow T^3(x) = 0 \Leftrightarrow T^2(T(x))=0 \Leftrightarrow T(x) \in \ker(T^2) = \ker(T) \Leftrightarrow T(T(x)) = 0 \Leftrightarrow T^2(x)=0 \Leftrightarrow x \in \ker(T^2).$, Which implies $\ker(T^3)\subseteq \ker(T^2)$, combine with $\ker(T^2) \subseteq \ker(T^3)$ we can conclude that $\ker(T^2)=\ker(T^3).$

Therefore $rankT=\dim V - \dim \ker(T)=\dim V - \dim \ker(T^3)=rankT^3$

Answer 2

Consider the induced homomorphism:

$\matrix{S:T(V) & \rightarrow & T^2(V)\\ x & \mapsto & T(x)}$

We remark that $S(T(V))=T(T(V))=T^2(V)$

Then $\text{rank}(S)=\text{rank}(T^2)$

Moreover

$\text{Ker}(S)=\{x\in T(V) | T(x)=0\}=\text{Ker}(T) \cap T(V)$

However $\text{dim}(\text{Ker}(S))+\text{rank}(S)=\text{dim}(T(V))$

We deduce that

$\text{dim}(\text{Ker}(S))=\text{rank}(T)-\text{rank}(T^2)=0$

So $\text{Ker}(T)\cap T(V)=\text{Ker}(S)=\{0\}$

Let $k\geq 1$, and consider the induced homomorphism:

$\matrix{S_k:T^k(V) & \rightarrow & T^{k+1}(V)\\ x & \mapsto & T(x)}$

$\text{Ker}(S_k)=\{x\in T^k(V) | T(x)=0\}=\text{Ker}(T) \cap T^k(V)$

However $T^{k}(V)\subseteq T(V)$

It follows that

$\text{Ker}(S_k) \subseteq \text{Ker}(T) \cap T(V)=\{0\}$

Thus $\text{Ker}(S_k)=\{0\}$

So $\text{dim}(T^k(V))=\text{rank}(S_k)$

As $S_k(T^k(V))=T(T^{k}(V))=T^{k+1}(V)$

we deduce that

$\text{dim}(T^k(V))=\text{rank}(S_k)=\text{dim}(T^{k+1}(V))$

Hence

$\text{rank}(T^k)=\text{rank}(T^{k+1})$