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 [itex]v \in \ker T^2[/itex] then [itex]T^2(v) = 0[/itex]. Now that could be because [itex]T(v) = 0[/itex] so that [itex]v \in \ker T[/itex], but could also be because [itex]T(v)[/itex] is some other member of [itex]\ker T[/itex] so that [itex]v \notin \ker T[/itex].

 

[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?