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Linear Algebra, Linear Maps

  • Thread starter *melinda*
  • Start date
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1. 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

3. 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?
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Dick

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What's the dimension of null(T) in terms of the dimension of V?
 
86
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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

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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?
 
86
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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

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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?
 
86
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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

Science Advisor
Homework Helper
26,258
618
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.
 
86
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:zzz:

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

Thanks for the help!
 

Dick

Science Advisor
Homework Helper
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618
Very, very welcome.
 

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