Question

Here is the problem: Let T: V --> W be linear, with dimV = n and dimW = m and ordered bases beta and gamma, respectively. Prove that rank(T) = rank(L_A) where A is the matrix representation of T in the bases beta and gamma. L_A is the transformation defined by multiplying A by a vector in F^n (F is any arbitrary field).

The problem suggests that I use a previously-derived result, that the dimension of a subspace is the same as the dimension of the range of an isomorphism applied to the subspace.

Answer 1

I believe I have a solution, but I feel a little uneasy about it, so I was hoping someone could look it over. I proceed, in outline, by showing that \[rank(T)=dim(R(T))=dim(\phi_{\gamma}(R(T)))\] where phi_gamma (as used in the above equation) is the isomorphism which maps a vector to its coordinate vector in the ordered basis gamma. This part is trivial. The first equality is due to the definition of rank. The second equality is due to the previous result. The result applies because the range of a transformation is a subspace of the space into which it's mapped, and because I've previously shown that phi_gamma is an isomorphism.

From here I seek to show \[dim(\phi_{\gamma}(R(T)))=dim(R(L_{A}))=rank(L_{A})\] which I'll try to accomplish by showing \[\phi_{\gamma}(R(T))=R(L_{A})\] by means of showing mutual set inclusion. The last equality will follow immediately from the definition of rank.

Now to show mutual set inclusion, start by supposing x is some vector such that \[x\in \phi_{\gamma}(R(T))\] and since phi_gamma maps vectors to F^n, then x is a vector in F^n such that there exists some y in V and \[[T(y)]_{\gamma}=x\] By previous theorems about the matrix representations of linear transformations, the left-hand side is equal to \[[T]_{\beta}^{\gamma}[y]_{\beta}\] and since we defined \[A=[T]_{\beta}^{\gamma}\] then we have \[A[y]_{\beta}=x\] But since y is a vector in V then \[ [y]_{\beta}\in F^{n}\].

On the other hand, \[L_{A}\] is the transformation defined by multiplying vectors by A, and therefore x is in the range of this transformation. Thus we have shown set inclusion in one direction. Showing set inclusion in the opposite direction is essentially just working backwards along this same path.