Linear algebra - Kernal and Range

1. Mar 15, 2010

sweetiepi

1. The problem statement, all variables and given/known data

Let V be a finite-dimensional vector space and let T: V -> V be a linear transformation. Prove that there exists a natural number n so that

Ker(T^n) (intersection) Range(T^n) = {0}

Here, T^n represents the n-fold composition of T o T o ... o T

2. Relevant equations

3. The attempt at a solution

I can prove that Ker(T) (intersection) Range(T) = {0} by showing that if an element is in the intersection of the two spaces, it must be zero since would produce a linear combination of one basis equal to the other, and subtracting so they are on both sides gives a linear combination of both bases. But the sum of the spans of the two bases are linearly independent, so the coefficients must all be zero, and thus they can have only the zero vector in common. Now I'm unsure of how to relate this to T^n.

2. Mar 15, 2010

Dick

Your proof of ker(T) (intersect) range(T)={0} can't possibly be correct. Let T:R^2->R^2 be given by the matrix whose first row is [0,1] and whose second row is [0,0]. Then the column vector v=[1,0] is in ker(T) since T(v)=0 and v is in range(T) since T(u)=v, where u is the column vector [0,1]. Put that into your proof. Do you see what's wrong?

3. Mar 15, 2010

sweetiepi

Sorry, in the problem where I showed that Range(T) (intersection) Ker(T) = {0} we were also given that Range(T) + Ker(T) = V. Does that make my explanation make more sense? I was trying not to have to replicate the whole proof but can see where that would have caused confusion. The proof has already been looked at by my professor so I am comfortable with it. The problem I am trying to prove now uses T^n and I don't know how to make that work.

4. Mar 15, 2010

Dick

If Range(T)+Ker(T)=V, then your proof probably makes perfect sense. But that's not generally true for any T. I just gave you an example where it's not true. But it's still true that Ker(T^n) (intersection) Range(T^n) = {0} for some n for ANY T, if V is finite dimensional. Range(T)+Ker(T)=V is just a special case. That's the n=1 case, I gave you an example of the n=2 case. Don't confuse that with dim(range(T))+dim(ker(T))=dim(V) which is always true.

Last edited: Mar 15, 2010
5. Mar 15, 2010

Dick

Ok, here's a hint. How does dim(T(range(T)+ker(T))) compare with dim(range(T)+ker(T))?

6. Mar 15, 2010

sweetiepi

Well, if you take T(range(T) + ker(T)) because T is linear you can compute T(range(T)) + T(ker(T)) and because the kernel of T maps to the zero vector you are left with T(range(T)) so the dim(T(range(T))) would be the number of vectors in the basis of range(T)? Whereas the dim(range(T)+ker(T)) would be the dimension of the number of vectors in the basis for the range plus the number of vectors in the kernel? I'm not really sure if that is right...

7. Mar 16, 2010

Dick

And I'm not sure what my 'hint' is exactly supposed do. It seemed to make sense at the time. Try this. You know range(T^(n+1)) is a subspace of range(T^n), right? And ker(T^n) is a subspace of ker(T^(n+1)). Given that V is finite dimensional, what can you say about the sequences of subspaces range(T^n) and ker(T^n)?

8. Mar 16, 2010

sweetiepi

I actually worked on this problem with a friend earlier today and we got it all figured out. Thank you for your help though!