Problem. Let \(V\) be a vector space with \(\dim V=n\), and let \(T\colon V\to V\) be a linear map such that \(\operatorname{Im}(T)\cap\ker(T)\ne\{0_{V}\}\). Prove that \(\operatorname{rk}(T\circ T)\le n-2\).
Proof. \[\text{rk}(T\circ T)=\dim\left(\text{Im}(T\circ T)\right)=\dim\left(\left\{ T\left(T\left(v\right)\right)\mid v\in V\right\} \right)=\dim\left(\left\{ T(v)\mid v\in\text{Im}T\right\} \right)\]
Let us look at the function \(T':\operatorname{Im}(T)\rightarrow V\) which is defined by \(\forall x\in\operatorname{Im}(T),T'(x)=T(x)\) (It’s also a linear transformation from inheritance)
\[x\in\text{ker}(T')\iff x\in\operatorname{Im}(T)\wedge T'(x)=0\iff x\in\operatorname{Im}(T)\wedge T(x)=0\iff x\in\operatorname{Im}(T)\wedge x\in\ker(T)\]
Then \[\text{ker}(T')=\operatorname{im}(T)\cap\ker(T)\ne\{0_{V}\}\] Which means \[\dim\left(\text{ker}(T')\right)\ge1\]
Furthermore, we know that \(\operatorname{Im}(T)\cap\ker(T)\ne\{0_{V}\}\) so in particular \(\ker(T)\ne\{0_{V}\}\) and \(\dim\left(\text{ker}(T)\right)\ge1\) as well.
According to the second dimensions theorem,
\[\dim\left(\text{Im}(T)\right)=\dim\left(V\right)-\dim\left(\text{ker}(T)\right)\le n-1\]
and
\[\dim\left(\text{Im}(T')\right)=\dim\left(\operatorname{Im}(T)\right)-\dim\left(\text{ker}(T')\right)\le\dim\left(\operatorname{Im}(T)\right)-1\le n-1-1=n-2\]
Let us notice that
\[\operatorname{Im}(T')=\left\{ T'(x)\mid x\in\operatorname{Im}(T)\right\} =\left\{ T(x)\mid x\in\operatorname{Im}(T)\right\} =\left\{ T\left(T\left(x\right)\right)\mid x\in V\right\} =\text{Im}(T\circ T)\] Therefore
\[\text{rk}(T\circ T)=\dim\left(\text{Im}(T\circ T)\right)=\dim\left(\text{Im}(T')\right)\le n-2\] ∎
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.