V is a n-dimensional vector space and $T : V \rightarrow V$ is a linear transformation s.t $rank(T) = rank(T^2)$. Prove that $range(T)\cap ker(T) = \{\mathbb{0}\}$
I am stuck with this problem and don't know how to proceed. My working so far:
$T^2 = T \circ T: V \rightarrow V$
By the rank theorem:
$rank(T) + nullity(T) = dim(V)$
$rank(T^2) + nullity(T^2) = dim(V)$
$\Longrightarrow rank(T) + nullity(T) = rank(T^2) + nullity(T^2)$
$\Longrightarrow nullity(T) = nullity(T^2)$ (By hypothesis)
Where can I go from here?
Best Answer
You're almost there with what you have. We know that $\ker(T)\subseteq\ker(T^2)$, so by the nullity result we must have $\ker(T)=\ker(T^2)$. Now if $y=Tx\in\ker(T)$, then $T^2x=Ty=0$, so $x\in\ker(T^2)=\ker(T)$, so $y=Tx=0$.