Rank(T) = Rank(L_A) proof?
I'd like to have someone check whether my proof of the following is correct. First, a couple of definitions to make sure we're all on the same page:
L_A is the mapping L_A: F^n → F^m defined by L_A (x) = Ax (the matrix product of A and x) for each column vector x in F^n.
The standard representation of V with respect to β is the function φ_β: V → F^n defined by φ_β(x) = [x]_β (i.e. the vector of x relative to β).
A result I've proved already:
Let V and W be finite-dimensional vector spaces and T: V → W be an isomorphism. Let V0 be a subspace of V. Then T(V0) is a subspace of W and dim(V0) = dim(T(V0)).
Let V and W be a vector spaces of dimension n and m, respectively, and let T: V → W be a linear transformation. Define A = [T] (i.e. the matrix representation of T in the ordered bases β and γ). So we have the following relationship (please ignore the dotted lines):
F^n -------L_A---> F^m
So to the quesiton finally:
Let T: V → W be a linear transformation from an n-dimensional vector space V to an m-dimensional vector space W. Let β and γ be ordered bases for V and W, respectively. Prove that rank(T) = rank(L_A), where A = [T].
I'll show that φ_γ(R(T)) = R(L_A) (R here is talking about the range..). From which then the result I'd proved already I'll have dim(R(T)) = dim(φ_γ(R(T)) = dim(R(L_A)) (remember that φ_γ(R(T) and R(L_A) are both in F^m).
So let x be in φ_γ(R(T)).
This means for some T(y) in R(T) I have x = φ_γ(T(y)) which means x = [T(y)]_γ = [T][y]_β = Ay which is in R(L_A). Since its arbitrary I've shown that φ_γ(R(T))⊂ R(L_A).
Further, let z in R(L_A). Then z = Ax for some x in F^n. Therefore, z = [T][x]_β = [T(x)]_γ which is in φ_γ(R(T)). Therefore, R(L_A) ⊂ φ_γ(R(T)) and hence, R(L_A) = φ_γ(R(T)). So by the result I'd proved earlier dim(R(T)) = dim(φ_γ(R(T))) = dim(R(L_A)).
Was this correctly done? Or am I at least on the right track? I'd really appreciate the help. I'm taking a look at linear algebra on my own this summer, so any help is REALLY appreciated! Thanks