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More linear maps

  • Thread starter *melinda*
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86
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Homework Statement
Prove that if there exists a linear map on V whose null space and range are both finite dimensional, then V is finite dimensional.

The attempt at a solution
I *think* the following is true: For all v in V, T(v) is in range(T), otherwise T(v) = 0 which implies v is in null (T).

Other than that, I know I can write a basis {v_1, ..., v_n} for null(T) and a basis {T(u_1), ..., T(u_m)} for range(T), where range(T) = {T(u) : u is in V}. But since this is a linear map {u_1, ..., u_m} should also be a basis for some U such that U is a subspace of V.

Does anyone know if these assumptions are heading in the right direction?
 

HallsofIvy

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Homework Statement
Prove that if there exists a linear map on V whose null space and range are both finite dimensional, then V is finite dimensional.

The attempt at a solution
I *think* the following is true: For all v in V, T(v) is in range(T), otherwise T(v) = 0 which implies v is in null (T).
No, that's not true. T(v)= 0 even when T(v) is in range(T) because 0 is in range(T) (range(T) is a subspace). For a simple example, let V= R2, T<x,y>= <x-y,x-y>. null(T) consists of all vectors of the form <x,x>. Range(T) is exactly null(T).

Other than that, I know I can write a basis {v_1, ..., v_n} for null(T) and a basis {T(u_1), ..., T(u_m)} for range(T), where range(T) = {T(u) : u is in V}. But since this is a linear map {u_1, ..., u_m} should also be a basis for some U such that U is a subspace of V.

Does anyone know if these assumptions are heading in the right direction?
You seem to be assuming that V is the direct sum of range(T) and null(T) and that's not true.
 
86
0
ok, but can I at least write a basis for null(T) and range(T)? I can't see how to prove this without defining something, because I know I can't prove this by only refering to the finite dimensions of null and range.
 

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