Rank(T) = Rank(L_A) proof?

Jun 2010
26
1
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):

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!!!!!
*********

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
 
Feb 2010
422
141
This is correct (I think). Note that this works for any commuting sqaure
\(\displaystyle \begin{array}[c]{ccc}
V&\rightarrow&W\\
\downarrow\scriptstyle{\cong}&&\downarrow\scriptstyle{\cong}\\
V'&\rightarrow&W'
\end{array}\)
where the vertical maps are iso.
 
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