This is a T/F - prove type of question: A is m x n, M is matrix of TA with respect to bases B of R^m and B' of R^n. Then rank of A = rank of M. My reasoning is that it is true, since the lin. transf. is R^n->R^m, which means that in this formula: M = CB' A PB (CB' (coord matrix) is inverse of PB', and PB or PB' stands for point matrix with respect to the B or B' basis, respectively) PB has dimension n and PB' has dimension m, and CB' has the same dimension as PB' (inverting a matrix should not change dimension, right?). And it also should be true that both PB and CB' are square matrices (?) because they are invertible and we can invert only square matrices (?), so what I have is following composition of dimensions of the matrices in the formula for M: (m x m) (m x n) (n x n) = (m x n) Is this correct? Thanks in advance.