Let \(A\) be an \(m\times n\) matrix with entries in a field \(\mathbf F\). Define the row rank and the column rank of \(A\) and show from first principles that they are equal.
Solution
1. The row rank of \(A\) is the dimension of the subspace of \(\mathbf F^n\) spanned by the rows of \(A\), and the column rank of \(A\) is the dimension of the subspace of \(\mathbf F^m\) spanned by the columns of \(A\).
These are the definitions of the row space, column space, row rank, and column rank.
2. For every matrix \(A\),\[ \operatorname{columnrank}(A)\leq\operatorname{rowrank}(A). \]
Let \[ r=\operatorname{rowrank}(A). \] If \(r=0\), every row of \(A\) is zero, hence \(A=0\) and its column rank is also \(0\).
Assume \(r>0\). Choose rows \[ R_1,\ldots,R_r \] of \(A\) forming a basis of its row space, and let \(B\) be the \(r\times n\) matrix whose rows are \(R_1,\ldots,R_r\). Every row \(S_i\) of \(A\) has a unique expression \[ S_i=\sum_{k=1}^r\lambda_{ik}R_k. \] Let \[ \Lambda=(\lambda_{ik}) \] be the resulting \(m\times r\) matrix. Row by row, the preceding equations say exactly that \[ A=\Lambda B. \]
For each column index \(j\), let \(b_j\in\mathbf F^r\) be the \(j\)th column of \(B\). Then the \(j\)th column of \(A\) is \[ \Lambda b_j, \] which is a linear combination of the \(r\) columns of \(\Lambda\). Hence every column of \(A\) lies in the span of those \(r\) columns. Therefore the column space of \(A\) has dimension at most \(r\), proving \[ \operatorname{columnrank}(A)\leq r=\operatorname{rowrank}(A). \]
3. For every matrix \(A\),\[ \operatorname{rowrank}(A)\leq\operatorname{columnrank}(A). \]
Apply step 2 to the transpose \(A^{\mathsf T}\). This gives \[ \operatorname{columnrank}(A^{\mathsf T}) \leq \operatorname{rowrank}(A^{\mathsf T}). \] The columns of \(A^{\mathsf T}\) are exactly the transposes of the rows of \(A\), so \[ \operatorname{columnrank}(A^{\mathsf T}) = \operatorname{rowrank}(A). \] Similarly, \[ \operatorname{rowrank}(A^{\mathsf T}) = \operatorname{columnrank}(A). \] Substitution gives the required inequality.
4. The row rank and column rank of \(A\) are equal.
Steps 2 and 3 give the two opposite inequalities, hence \[ \boxed{ \operatorname{rowrank}(A)=\operatorname{columnrank}(A) }. \]
5. Q.E.D.
Step 1 gives the requested definitions, and step 4 proves their equality.