Rank of a linear map equals rank of matrix representation? Let $ T:V \rightarrow W $ be a linear map where $V,W$ are finite dimensional.
Let $\beta,\gamma$ be arbitrary bases for $V,W$ respectively.
Is it true that $rank(T) = rank([T]^{\gamma}_{\beta})$ ? If so, how would I prove this?
Note: I define the rank of a matrix $A$ to be the rank of the linear map corresponding to left multiplication by $A$, namely $L_{A}$. Thus, $rank(A) = rank(L_{A})$.
I define the rank of a linear map to be the dimension of its range (this definition is standard I believe.)

This is true, and in your approach (defining rank for linear maps first, then for matrices in terms of linear maps) there is absolutely nothing to prove. The rank of a linear map is independent of any choice of basis simply because there are no bases involved in the definition. The rank does not change under left of right composition with an isomorphism of vector spaces; this is the only point of the argument where the definition of rank is actually used, but it is quite obvious.
Now choosing an ordered basis in a finite dimensional $K$-vector space $V$ fixes an isomorphism $V\leftrightarrow K^n$ when $n=\dim(V)$ (expressing in coordinates on the basis, in the left to right direction), and the $n\times m$ matrix $A=[T]_\beta^\gamma$ is defined to be so that left multiplication by $A$ gives the linear map $K^m\to K^n$ defined by$~T$ composed left and right with the (inverse) coordinate isomorphisms. What you are asking is whether $T$ and this composite map have the same rank, and this is true because one has composed with isomorphisms.
