changes neither the row rank nor the column rank. It is immediate that both the row and column ranks of this resulting matrix is the number of its nonzero entries. We present two other proofs of this result. The first uses only basic properties of linear combinations of vectors, and is valid over any field. The proof is based upon Wardlaw (2005). The second uses orthogonality and is valid for matrices over the real numbers; it is based upon Mackiw (1995). Both proofs can be found in the book by Banerjee and Roy (2014). Proof using linear combinations Let be an matrix. Let the column rank of be , and let be any basis for the column space of . Place these as the columns of an matrix . Every column of can be expressed as a linear combination of the columns in . This means that there is an matrix such that . is the matrix whose th column is formed from the coefficients giving the th column of as a linear combination of the columns of . In other words, is the matrix which contains the multiples for the bases of the column space of (which is ), which are then used to form as a whole. Now, each row of is given by a linear combination of the rows of . Therefore, the rows of form a spanning set of the row space of and, by the Steinitz exchange lemma, the row rank of cannot exceed . This proves that the row rank of is less than or equal to the column rank of . This result can be applied to any matrix, so apply the result to the transpose of . Since the row rank of the transpose of is the column rank of and the column rank of the transpose of is the row rank of , this establishes the reverse inequality and we obtain the equality of the row rank and the column rank of . (Also see Rank factorization.) Proof using orthogonality Let be an matrix with entries in the real numbers whose row rank is . Therefore, the dimension of the row space of is . Let be a basis of the row space of . We claim that the vectors are linearly independent. To see why, consider a linear homogeneous relation involving these vectors with scalar coefficients : where . We