A concise proof of row rank == column rank

As I was reading up on linear algebra lately, I was looking for a more intuitive proof of row rank == column rank than the one in Axler’s book. This led me to Wardlaw’s concise proof [1], which happens to use a matrix factorization.

For this proof, it will be helpful to use the notation that for a matrix \(S\), \(s_{i, \cdot}\) represents the ith row and \(s_{\cdot, j}\) represents jth column.

Consider a \(m \times n\) matrix \(A\):

\[A = \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{pmatrix}\]

with \(\textrm{rank}_\textrm{col}(A) = k\).

Now consider the \(m \times k\) matrix \(C\), whose columns form a basis of the column space of \(A\). Then each column of \(A\) can be formed by taking a linear combination the columns of \(C\), \(c_{\cdot,1},..,c_{\cdot,k}\). Let \(R\) be the \(k \times n\) matrix whose ith column, \(r_{\cdot,i}\) contains the coefficients of the linear combination to form the ith column of \(A\). More concisely, \(A=CR\):

\(A = \color{#ff8c1a}C \color{#0066ff}R \color{black}= \begin{pmatrix} \color{#ff8c1a} c_{1,\cdot} \\ \color{#ff8c1a} c_{2,\cdot} \\ \vdots \\ \color{#ff8c1a} c_{m,\cdot} \end{pmatrix} \color{#0066ff}R \color{black} = \begin{pmatrix} \color{#ff8c1a} c_{1,\cdot} \color{#0066ff}R \\ \color{#ff8c1a} c_{2,\cdot} \color{#0066ff}R \\ \vdots \\ \color{#ff8c1a} c_{m,\cdot} \color{#0066ff}R \end{pmatrix} \\ = \begin{pmatrix} \color{#ff8c1a} c_{1,1} \color{#0066ff} r_{1,\cdot} \color{black} + \color{#ff8c1a} c_{2,1} \color{#0066ff} r_{2,\cdot} \color{black} + \cdots + \color{#ff8c1a} c_{k,1} \color{#0066ff} r_{k,\cdot} \\ \color{#ff8c1a} c_{1,2} \color{#0066ff} r_{1,\cdot} \color{black} + \color{#ff8c1a} c_{2,2} \color{#0066ff} r_{2,\cdot} \color{black} + \cdots + \color{#ff8c1a} c_{k,2} \color{#0066ff} r_{k,\cdot} \\ \vdots \\ \color{#ff8c1a} c_{1,m} \color{#0066ff} r_{1,\cdot} \color{black} + \color{#ff8c1a} c_{2,m} \color{#0066ff} r_{2,\cdot} \color{black} + \cdots + \color{#ff8c1a} c_{k,m} \color{#0066ff} r_{k,\cdot} \\ \end{pmatrix}\).

At this point, it is clear that the rows of \(A\) are linear combinations of the \(k\) rows of \(R\), which implies that the row space of \(A\) has dimension at most \(k\). It follows that \(\textrm{rank}_\textrm{row}(A) \leq \textrm{rank}_\textrm{col}(A)\). Following the same steps with the transpose of \(A\) leads to the reverse inequality, thus completing the equality and the proof.

[1] Wardlaw, William P. “Row rank equals column rank.” Mathematics Magazine 78, no. 5 (2005): 316-318. DOI PDF