- #1

Eclair_de_XII

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- 91

- 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.

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.