# Rank(T) = Rank(L_A) proof?

1. Jul 15, 2010

### Buri

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).

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

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].

*********
!!!!!PROOF!!!!!
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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

2. Jul 16, 2010

### marcin w

By definition of [T], the column space of [T] = A is the range of T. Similarly, by definition of L_A, it's range is the column space of A. So range of T is the same space as range of L_A and they must have same dimension. That's it.

Last edited: Jul 16, 2010
3. Jul 16, 2010

### Buri

At this point in the text, the column space of a matrix hasn't been defined. So its meant to be done without it. I'm mainly concerned about if whether the above proof is correct or not and not really whether there are quicker ways of doing it.

4. Jul 16, 2010

### marcin w

OK, without column spaces - Obviously [A] is the matrix representation of L_A in bases F^m and F^n. If c is any vector in V and [c] is its representation in F^n, then [Tc] = [T][c] = [A][c] = (L_A)[c] . So you have Tc is in range of T if and only if (L_A)[c] is in range of L_A. So those two subspaces are equal and have same dimension.

Last edited: Jul 16, 2010
5. Jul 17, 2010

### Buri

I believe that only shows that φ_γ(R(T)) = R(L_A) which is what I showed above.

It seems that I have no way to connect Rank(T) and Rank(L_A) unless I use the theorem:

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)).

By applying the theorem to V0 = R(T) and using the isomorphism φ_γ, I have that dim(φ_γ(R(T))) = dim(R(T)) and hence, by what we have both shown, also equal to dim(R(L_A)).

You see what I mean?

Last edited: Jul 18, 2010
6. Jul 20, 2010

### Buri

Can anyone else verify?

7. Jul 20, 2010

### marcin w

An isomorphism is a very strong relationship between structures and in essence states that the two structures are identical in their behavior under operations, that sets of images have the same characteristics, etc. You acknowledge that φ_γ mapping is an isomorphism, so it maps bases of R(T) onto bases of F^m, hence same dimension. No further work is required.

8. Jul 21, 2010

### Buri

So you're technically using the result I had proved earlier. This only gives me dim(R(T)) = dim(φ_γ(R(T)) and from "...[Tc] = [T][c] = [A][c] = (L_A)[c]" I get φ_γ(R(T)) = R(L_A) and I'm done.

Anyways, thanks for taking a look at this. It was really clustered so I thought no-one was going to take a look. So thanks a lot!