Editing Rank (linear algebra) (section)

===Proof using orthogonality===
Let {{mvar|A}} be an {{math|''m''&thinsp;×&thinsp;''n''}} matrix with entries in the [[real number]]s whose row rank is {{mvar|r}}. Therefore, the dimension of the row space of {{mvar|A}} is {{mvar|r}}.  Let {{math|'''x'''<sub>1</sub>, '''x'''<sub>2</sub>, …, '''x'''<sub>''r''</sub>}} be a [[basis (linear algebra)|basis]] of the row space of {{mvar|A}}. We claim that the vectors {{math|''A'''''x'''<sub>1</sub>, ''A'''''x'''<sub>2</sub>, …, ''A'''''x'''<sub>''r''</sub>}} are [[linearly independent]]. To see why, consider a linear homogeneous relation involving these vectors with scalar coefficients {{math|''c''<sub>1</sub>, ''c''<sub>2</sub>, …, ''c<sub>r</sub>''}}:
<math display="block">0 = c_1 A\mathbf{x}_1 + c_2 A\mathbf{x}_2 + \cdots + c_r A\mathbf{x}_r = A(c_1 \mathbf{x}_1 + c_2 \mathbf{x}_2 + \cdots + c_r \mathbf{x}_r) = A\mathbf{v}, </math>
where {{math|1='''v''' = ''c''<sub>1</sub>'''x'''<sub>1</sub> + ''c''<sub>2</sub>'''x'''<sub>2</sub> + ⋯ + ''c<sub>r</sub>'''''x'''<sub>''r''</sub>}}.  We make two observations: (a) {{math|'''v'''}} is a linear combination of vectors in the row space of {{mvar|A}}, which implies that {{math|'''v'''}} belongs to the row space of {{mvar|A}}, and (b) since {{math|1=''A'''''v''' = 0}}, the vector {{math|'''v'''}} is [[orthogonal]] to every row vector of {{mvar|A}} and, hence, is orthogonal to every vector in the row space of {{mvar|A}}. The facts (a) and (b) together imply that {{math|'''v'''}} is orthogonal to itself, which proves that {{math|1='''v''' = 0}} or, by the definition of {{math|'''v'''}},
<math display="block">c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_r \mathbf{x}_r = 0.</math>
But recall that the {{math|'''x'''<sub>''i''</sub>}} were chosen as a basis of the row space of {{mvar|A}} and so are linearly independent. This implies that {{math|1=''c''<sub>1</sub> = ''c''<sub>2</sub> = ⋯ = ''c<sub>r</sub>'' = 0}}.  It follows that {{math|''A'''''x'''<sub>1</sub>, ''A'''''x'''<sub>2</sub>, …, ''A'''''x'''<sub>''r''</sub>}} are linearly independent.

Now, each {{math|''A'''''x'''<sub>''i''</sub>}} is obviously a vector in the column space of {{mvar|A}}. So, {{math|''A'''''x'''<sub>1</sub>, ''A'''''x'''<sub>2</sub>, …, ''A'''''x'''<sub>''r''</sub>}} is a set of {{mvar|r}} linearly independent vectors in the column space of {{mvar|A}} and, hence, the dimension of the column space of {{mvar|A}} (i.e., the column rank of {{mvar|A}}) must be at least as big as {{mvar|r}}. This proves that row rank of {{mvar|A}} is no larger than the column rank of {{mvar|A}}.  Now apply this result to the transpose of {{mvar|A}} to get the reverse inequality and conclude as in the previous proof.