Yes. The rank of is the image of its image, which being a subspace of is at most .
the rank of T (where T is a linear transformation of a finite-dimensional vector space into a finite-dimensional vector space) is defined to be the rank of any matrix A for T.
if the domain of T is Rn, and T maps into Rm, such a matrix will be an mxn matrix. this has only m rows, so only m (at most) of them can be non-zero when we row-reduce it.
alternatively (and equivalently) if you define rank(T) = dim(im(T)) it should be clear im(T) = T(Rn) ⊆ Rm, so any basis for im(T) can at most possess m elements.
the nifty thing about rank is we can choose any bases for Rn and Rm we like to calculate the matrix A for T. some bases are easier to work with than other ones.