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

Question

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.

score 12 · Answer 1 · answered Apr 28 '15 at 01:01

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

Great solution. I just loved it. – Matha Mota Dec 07 '20 at 00:39 — Matha Mota, Dec 07 '20 at 00:39

ncmathsadist · Accepted Answer · 2013-10-10T01:38:19.520

6

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 $T(V)$.

edited Oct 10 '13 at 01:38

answered Oct 10 '13 at 01:28

ncmathsadist

49,383

How do justify the claim that $T(V) \subseteq T^2(V)$? – JohanLiebert Oct 10 '13 at 01:34
Thanks for the catch. I had the subset relationship backwards. – ncmathsadist Oct 10 '13 at 01:35
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. – JohanLiebert Oct 10 '13 at 02:03
1

If $T$ is 1-1 on $T(V)$ its kernel when restricted to $T(V)$ is ${0}$. Hence, $T(V)\cap \ker(T) = {0}$. – ncmathsadist Oct 11 '13 at 00:10
@ncmathsadist Excuse me, could you please tell me why $T^2(V) \subseteq T(V)$ and why $T(V)=T^2(V)$ implies that $T$ is injective? – Twnk Oct 13 '13 at 01:48
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Let $x\in T^2(V)$. Then you can choose $v\in V$ so that $T(T(v)) = x$. Since $T(v)\in V$, $x\in T(V)$. – ncmathsadist Oct 13 '13 at 01:55
Secondly you have that $T^2(V)$ is a full-dimensional subspace of $T(V)$. – ncmathsadist Oct 13 '13 at 01:56
+1 @ncmathsadist Thanks. And why does that imply $T$ is injective? – Twnk Oct 13 '13 at 01:59
$T$ is a mapping of an $n$-dimensional space onto an $n$-dimensional space. By the dimension theorem, $T$ is 1-1. – ncmathsadist Oct 13 '13 at 02:01
@ncmathsadist But to conclude that we'd have to know that $T$ is surjective – Twnk Oct 13 '13 at 02:04
This is true by the rank condition. – ncmathsadist Oct 13 '13 at 02:05

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

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