Let $V$ be a finite-dimensional vector space and let $T:V\to V$ be linear.

Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$, then $R(T) \cap N(T) = \{0\}$.

I don't see this implication, at all. Please give hints and explain conceptually.


2 Answers 2


We have that $T^2(V)\subseteq T(V)$, so $R(T)=R(T^2)$ implies that $T^2(V)=T(V)$.

Now, for the rank nullity theorem we know that $\dim V= R(T) + N(T)=R(T^2)+N(T^2)$, so $N(T)=N(T^2)$, but similarly to before, $\ker T \subseteq \ker T^2 $ and $\ker(T)=\ker(T^2)$.

Let $x \in \ker (T) \cap T(V)$. The fact that $x\in T(V)$ implies that there exists $x' \in V$ such that $T(x')=x$, then $T(T(x'))=T(x)=0$ (because $x \in \ker (T)$).

Then $x' \in \ker (T^2)=\ker (T)$, so $T(x')=0=x$.

We conclude that $\ker (T) \cap T(V)=\{ 0\}$.

  • $\begingroup$ Great solution. I just loved it. $\endgroup$
    – Boka Peer
    Dec 7, 2020 at 0:39

You have $T^2(V)\subseteq T(V)$, and the dimension of these two subspaces is equal so $T(V) = T^2(V)$. Hence, $T$ is 1-1 on $T(V)$.

  • $\begingroup$ How do justify the claim that $T(V) \subseteq T^2(V)$? $\endgroup$ Oct 10, 2013 at 1:34
  • $\begingroup$ Thanks for the catch. I had the subset relationship backwards. $\endgroup$ Oct 10, 2013 at 1:35
  • $\begingroup$ So $T$ being 1-1 implies the intersection between the range of T and the nullspace is $\{0\}$? I don't understand that. Could you explain please. $\endgroup$ Oct 10, 2013 at 2:03
  • 1
    $\begingroup$ If $T$ is 1-1 on $T(V)$ its kernel when restricted to $T(V)$ is $\{0\}$. Hence, $T(V)\cap \ker(T) = \{0\}$. $\endgroup$ Oct 11, 2013 at 0:10
  • 2
    $\begingroup$ Let $x\in T^2(V)$. Then you can choose $v\in V$ so that $T(T(v)) = x$. Since $T(v)\in V$, $x\in T(V)$. $\endgroup$ Oct 13, 2013 at 1:55

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