Prove that T^2 = T0 IFF R(T) C N(T)

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SUMMARY

The discussion centers on proving that for a linear transformation T from vector space V to itself, T^2 = T0 (the zero mapping) if and only if the range R(T) is contained in the null space N(T). Participants clarify that R(T) does not have to be zero and that T is not exclusively the zero transformation. A counterexample involving the shift operator on R^2 is suggested to illustrate the concept further.

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Let T be a linear transformation from V to V. Prove that T^2 = T0 (with T0 the zero mapping) IFF R(T) C N(T). ( "is contained in" = 'C'. )

Comments:

It seems clear that T is also the zero transformation. IF this is wrong can someone give me a counterexample. If it is true, then R(T) = { 0 } and N(T) = {v | v in V}, and clearly R(T) C N(T). That would show half, and if this part is right, then I can finish the second half. But is my reasoning for this part correct?--it almost seems too easy.
 
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R(T) is the range and N(T) is the null space, correct?

R(T) need not be zero, and T is certainly not just the zero map.

T^2(x) = 0 for all x if and only if T(T(x))=0, ie if and only if T(x) is in...
 
consider the shift operator, on R^2, taking (1,0) to (0,1) and (0,1) to (0,0).
 

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