information:filtering, structural:expansion
Proof using linear combinations

Let A be an m × n matrix. Let the column rank of A be r, and let c,..., c be any basis for the column space of A. Place these as the columns of an m × r matrix C. Every column of A can be expressed as a linear combination of the r columns in C. This means that there is an r × n matrix R such that A = CR. R is the matrix whose i th column is formed from the coefficients giving the i th column of A as a linear combination of the r columns of C. In other words, R is the matrix which contains the multiples for the bases of the column space of A (which is C), which are then used to form A as a whole. Now, each row of A is given by a linear combination of the r rows of R. Therefore, the rows of R form a spanning set of the row space of A and, by the Steinitz exchange lemma, the row rank of A cannot exceed r. This proves that the row rank of A is less than or equal to the column rank of A. This result can be applied to any matrix, so apply the result to the transpose of A. Since the row rank of the transpose of A is the column rank of A and the column rank of the transpose of A is the row rank of A, this establishes the reverse inequality and we obtain the equality of the row rank and the column rank of A. (Also see Rank factorization.)