Editing Rank (linear algebra) (section)

===Proof using linear combinations===
Let {{mvar|A}} be an {{math|''m'' × ''n''}} matrix. Let the column rank of {{mvar|A}} be {{mvar|r}}, and let {{math|'''c'''<sub>1</sub>, ..., '''c'''<sub>''r''</sub>}} be any basis for the column space of {{mvar|A}}. Place these as the columns of an {{math|''m'' × ''r''}} matrix {{mvar|C}}. Every column of {{mvar|A}} can be expressed as a linear combination of the {{mvar|r}} columns in {{mvar|C}}. This means that there is an {{math|''r'' × ''n''}} matrix {{mvar|R}} such that {{math|1=''A'' = ''CR''}}. {{mvar|R}} is the matrix whose {{mvar|i}}th column is formed from the coefficients giving the {{mvar|i}}th column of {{mvar|A}} as a linear combination of the {{mvar|r}} columns of {{mvar|C}}. In other words, {{mvar|R}} is the matrix which contains the multiples for the bases of the column space of {{mvar|A}} (which is {{mvar|C}}), which are then used to form {{mvar|A}} as a whole. Now, each row of {{mvar|A}} is given by a linear combination of the {{mvar|r}} rows of {{mvar|R}}. Therefore, the rows of {{mvar|R}} form a spanning set of the row space of {{mvar|A}} and, by the [[Steinitz exchange lemma]], the row rank of {{mvar|A}} cannot exceed {{mvar|r}}. This proves that the row rank of {{mvar|A}} is less than or equal to the column rank of {{mvar|A}}. This result can be applied to any matrix, so apply the result to the transpose of {{mvar|A}}. Since the row rank of the transpose of {{mvar|A}} is the column rank of {{mvar|A}} and the column rank of the transpose of {{mvar|A}} is the row rank of {{mvar|A}}, this establishes the reverse inequality and we obtain the equality of the row rank and the column rank of {{mvar|A}}. (Also see [[Rank factorization]].)