Proving the Direct Sum Decomposition of V using Linear Maps

[*melinda*]

Homework Statement

Suppose that T is a linear map from V to F, where F is either R or C. Prove that if u is an element of V and u is not an element of null(T), then

V = null(T) (direct sum) {au : a is in F}.

2. Relevant information
null(T) is a subspace of V
For all u in V, u is not in null(T)
For all n in V, n is in null(T)
T(n) = 0, T(u) not= 0

The Attempt at a Solution

I think I should let U = {au : a is in F} and show that it's a subspace of V. Then I can show that each element of V can be written uniquely as a sum of u + n. Should I do this by showing that (u, n) is a basis for V?

 

[Dick]

What's the dimension of null(T) in terms of the dimension of V?

 

[*melinda*]

if V were finite dimensional then I could say, dim{null(T)} = dim(V) - dim{range(T)}.

But nothing given in the problem statement will let me assume V is finite.

 

[Dick]

Ok, I think I was thinking about this wrong. Suppose x=a*u+n and x=a'*u+n' where n and n' are in the null space. Then T(x)=a*T(u)=a'*T(u). Since T(u) is nonzero, a=a'. Right? So x=a*u+n and x=a*u+n'. Can n and n' be different?

 

[*melinda*]

n and n' could definitely be different, but I don't think it matters much since they both get mapped to zero.

Is the result of a = a' is enough to prove uniqueness for a direct sum?

 

[Dick]

It does matter in V. But if x=a*u+n and x=a*u+n' (after you've shown a=a') then n=x-a*u and n'=x-a*u. Conclusion?

 

[*melinda*]

Gosh, I must be getting sleepy to overlook the importance of n being unique.

So, I can show that each element of V can be written uniquely as a sum of u + n.

Should I also prove U = {au : a is in F} is a subspace of V

 

[Dick]

Gosh, I must be getting sleepy to overlook the importance of n being unique.

So, I can show that each element of V can be written uniquely as a sum of u + n.

Should I also prove U = {au : a is in F} is a subspace of V

Even being sleepy, I think you could, right? I think it could actually be considered as 'obvious' and not deserving of proof.

 

[*melinda*]

:zzz:

I should be able to stay awake long enough to write down my solution.

Thanks for the help!

 

[Dick]

Very, very welcome.