Linear transformation from V to V proof

1. Nov 9, 2008

p3forlife

1. The problem statement, all variables and given/known data
Let V be a vector space over a field F and let L(V) be the vector space of linear transformations from V to V. Suppose that T is in L(V). Do not assume that V is finite-dimensional.
a) Prove that T^2 = -T iff T(x) = -x for all x in R(T).
b) Suppose that T^2 = -T. Prove that the intersection of N(T) and R(T) = {0}.

2. Relevant equations

3. The attempt at a solution
The T^2 is throwing me off slightly. Does it just mean take the square of the original linear transformation T?
So if T^2 = -T,
T^2 (x) = -T(x)
Also rank(T^2) = rank (-T)
That's all I have so far. How can I approach this if I can't assume that V is finite-dimensional?
If I could get a few tips on how to start this question, it would be very helpful :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 9, 2008

Dick

You haven't gotten very far. Can you show if T(x)=-x, then T^2(x)=-T(x)? That's the easy direction. For the 'only if', if x is in R(T) then there is a y in V such that x=T(y). Act with T on both sides of that and use T^2=-T.

3. Nov 9, 2008

p3forlife

Hmm....I might be completely missing the point here...

Show that if T(x) = -x, then T^2(x) = -T(x)
If T(x) = -x,
T^2(x) = (-x)^2 = x^2
-T(x) = -(-x) = x
T^2(x) does not equal -T(x)

4. Nov 9, 2008

Dick

You are missing the point. T^2(x) means T(T(x)). You can't really 'square' a vector.

5. Nov 9, 2008

p3forlife

Thanks. I got it.
Just to make sure, for part b), would N(T) = 0 since it is a transformation from V to V? Also would R(T) = x?

6. Nov 9, 2008

Dick

Why would you think N(T)=0 (null space, right?)? And why would R(T) (range, right?) be x, whatever that is? R(T) is a subspace. I think you'd better look up those definitions again.