# Is my linear algebra proof correct?

1. Jul 15, 2010

### 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.

1. The problem statement, all variables and given/known data

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].

2. Relevant 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).

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

3. 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!

Last edited: Jul 16, 2010
2. Jul 15, 2010

### 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.

3. Jul 15, 2010

### 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.

4. Jul 15, 2010

### 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.

5. Jul 15, 2010

### 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.

6. Jul 15, 2010

### Newtime

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.

7. Jul 15, 2010

### 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.