$\operatorname{rank}(T^3) + \operatorname{rank}(T) \geq 2\operatorname{rank}(T^2)$ for finite-dimensional $V$

Question

Consider a finite-dimensional vector space $V$ over the field $\mathbb{F}$ with a linear operator $T.$ Prove that we have $\operatorname{rank}(T^3) + \operatorname{rank}(T) \geq 2 \operatorname{rank}(T^2).$

First, let us assume that $\dim_{\mathbb{F}}(V) = n.$ Our immediate observations include that $T^3(V) \subseteq T^2(V) \subseteq T(V)$ so that $\operatorname{rank}(T^3) \leq \operatorname{rank}(T^2) \leq \operatorname{rank}(T)$ and $\operatorname{rank}(T^3) \geq \operatorname{rank}(T) + \operatorname{rank}(T^2) - n$ by Sylvester's Rank Inequality. Can we do some manipulations with what we have to get the desired inequality, or do we require something else?

Answer 1

Define the linear map $T':T(V)\rightarrow T^2(V)/T^3(V)$ by $T'(v)=T(v)+T^3(V)$.

The quotient vector space $T^2(V)/T^3(V)$ has dimension $\text{rank}(T^2)-\text{rank}(T^3)$ as $T^3(V)$ is a proper subspace of $T^{2}(V)$.

Now, notice that $T'$ is surjective and that $T^2(V)\subset \text{Ker}(T')$, therefore from rank nullity theorem we obtain $\text{rank}(T)-\text{rank}(T^2)\geq \text{rank}(T^2)-\text{rank}(T^3)$. By rearranging the previous inequality we conclude the result.