If V is a vector space why is T^2(V) = T(V) iff ker(T^2) = ker(T)?

[Eclair_de_XII]

Homework Statement

Let $V$ be a finite-dimensional vector space. Prove that for a linear operator $T$, that $T^2(V)=T(V)$ iff $ker(T^2)=ker(T)$.

Relevant Equations

$ker(T)$: The set of $v\in V$ such that $T(v)=0$
$T(V)$: The image of $v\in V$ through $T$.

--$ker(T^2)=ker(T)$ if $T(V)=T^2(V)$--

Suppose that $T^2(V)=T(V)$. So $T:T(V)\mapsto T^2(V)=T(V)$. Hence, $T$ is one-to-one and so $ker(T)=\{0\}$. Suppose that $T^2(w)=0$ for some $w\in ker(T^2)$. Then $T^2(w)=T(T(w))=0$ which implies that $T(w)\in ker(T)$ and so $T(w)=0$, so $w=0$. Hence, $ker(T^2)=ker(T)=\{0\}$.

--$T^2(V)=T(V)$ if $ker(T^2)=ker(T)$--

Suppose that $ker(T^2)=ker(T)$. Let $v\in ker(T^2),ker(T)$ so that $T^2(v)=0$. We have that $T^2:ker(T)\mapsto \{0\}$ and that $T:T(ker(T))=\{0\}\mapsto \{0\}$. Hence, since $ker(T)=ker(T^2)$, we have that $ker(T)=\{0\}$. In turn, $T$ is surjective. And so $T(V)=V$, and furthermore, $T^2(V)=T(V)=V$.

All in all, I'm not really too sure about the last step.

 

[Orodruin]

Your conclusion that $\operatorname{ker}(T) = \{0\}$ is wrong, $T(V)$ is not necessarily $V$. All that $T^2(V) = T(V)$ tells you is that they are the same subspace of $V$. All you can say is that $T$ is bijective from $T(V)$ to $T(V)$.

As a counter example, consider the zero mapping $T(x) = 0$ for all $x \in V$. This map is clearly linear and its kernel is all of $V$. You also have $T(T(x)) = T(0) = T(x)$ for all $x$ so $T^2 = T$ and they therefore obviously have the same kernel.

 

[Eclair_de_XII]

All you can say is that $T$ is bijective from $T(V)$ to $T(V)$.

Does that mean that $\ker(T|_{T(V)})=\{0\}$? If so, then I don't know how that will relate to $T$ or its kernel on the whole domain of $V$.

 

[pasmith]

By definition, if $v \in \ker T^2$ then $T^2(v) = 0$. Now that could be because $T(v) = 0$ so that $v \in \ker T$, but could also be because $T(v)$ is some other member of $\ker T$ so that $v \notin \ker T$.

 

[Eclair_de_XII]

Okay, let me think, here...

Let $v\in \ker(T)$. Then $T(v)=0$. Also, $T^2(v)=T(T(v))=T(0)=0$, which means that $v\in \ker(T^2)$. Hence $\ker(T)\subset \ker(T^2)$.

Let $v \in \ker(T^2)$. Then $T^2(v)=T(T(v))=0$, which can mean either that:

(1) $T(v)=0$ so that $v\in \ker(T)$, and $\ker(T^2)\subset \ker(T)$, which means that we are done in this case.
(2) $T(v)\in \ker(T)$.

In the latter case, it is implied that $T(V)=T^2(V)\subset \ker(T)$. So $T(T^2(u))=0$ for some $u\in V$. So my question then becomes that of figuring out if there is a $u\in V$ such that $T^2(u)=v$.

 

[StoneTemplePython]

you should be able to streamline this by observing that T + its complement creates the identity map. This is a general point about idempotence. (equivalently the identity minus T is the complement...)

 

[Eclair_de_XII]

T + its complement creates the identity map

Okay, let me take a quick stab at this...

Let $T^c$ be the complement of $T$ such that $T+T^c=I_V$. Let $v\in \ker(I_V-T^c)$. With this equality, we have that: $T=I_V-T^c$. And so if we apply this to $v$, $T(v)=(I_V-T^c)(v)=0$, so $v\in \ker(T)$. On the other hand, $T^2(V)=T(V)$, so $T^2=I_V-T^c$, and $T^2(v)=(I_V-T^c)(v)=0$. Hence, $v\in \ker(T^2)$. Equality follows, I think?