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 R^{n}, and T maps into R^{m}, 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.
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alternatively (and equivalently) if you define rank(T) = dim(im(T)) it should be clear im(T) = T(R^{n}) ⊆ R^{m}, so any basis for im(T) can at most possess m elements.
the nifty thing about rank is we can choose any bases for R^{n} and R^{m} we like to calculate the matrix A for T. some bases are easier to work with than other ones.