Is my linear algebra proof correct?

[Buri]

I'd like to have someone check whether my proof of the following is correct. The definitions of all the stuff I use is in part 2, just to make sure we're all on the same page.

Homework Statement

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].

Homework Equations

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):

V------T--------->W
|...... |
|......|
φ_β......φ_γ
|......|
F^n -------L_A---> F^m

The Attempt at a Solution

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!

 

[Newtime]

What is "F^m" are "F^n"? Correspondingly, what is L_A? I haven't read the last part (main part) of your proof, but I suspect whatever you're trying to prove can be easily done using the simple fact that the rank of a transformation is defined as the dimension of it's image.

 

[Buri]

F^n is the set of n-tuples with entries in a field F, likewise F^m. The definition of L_A is in Part 2.

 

[Newtime]

Then I would have just said something like rank(T) = rank([T]) almost by definition, and then used the fact that F^n is isomorphic to V and F^m is isomorphic to W. The result follows immediately. Your proof seems correct on first reading, but also seems a bit convoluted. Very sophisticated work for a first run at linear algebra though.

 

[Buri]

Ranks of matrices haven't been defined in the text at this part. That's why I'm applying this result:

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)).

...to the diagram I attempted to show above.

 

[Newtime]

Ranks of matrices haven't been defined in the text at this part. That's why I'm applying this result:

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)).

...to the diagram I attempted to show above.

Ahh ok then. Out of curiosity, what book are you using? Usually if you're working with matrix representations of transformations, you've learned to apply the ideas of rank to them.

 

[Buri]

The text I'm using is Linear Algebra by Friedberg, Insel & Spence. Its the text they use at University of Toronto for first year algebra in the pure mathematics program.