Let \(A\) be an \(n \times n\) matrix over a field, \(A \neq 0\). If \(r\) is any positive integer between 1 and \(n\), an \(r \times r\) submatrix of \(A\) is any \(r \times r\) matrix obtained by deleting \((n - r)\) rows and \((n - r)\) columns of \(A\). The determinant rank of \(A\) is the largest positive integer \(r\) such that some \(r \times r\) submatrix of \(A\) has a non-zero determinant. Prove that the determinant rank of \(A\) is equal to the row rank of \(A\) (= column rank \(A\)).
Solution
Let \(\rho(A)\) denote the ordinary rank of \(A\) and let \(d(A)\) denote its determinant rank.
If an \(s\times s\) submatrix \(M\) has nonzero determinant, then the \(s\) selected columns of \(M\) are linearly independent. Any linear relation among the corresponding columns of \(A\) would restrict to the same relation among the columns of \(M\), so those columns of \(A\) are linearly independent. Hence \[ s\le \rho(A). \] Taking the largest such \(s\) gives \[ d(A)\le \rho(A). \]
Conversely, put \(r=\rho(A)\). Choose \(r\) linearly independent columns of \(A\) and form the resulting \(n\times r\) matrix \(C\). Then \(C\) has rank \(r\). Since row rank equals column rank, \(C\) has \(r\) linearly independent rows. Restricting \(C\) to those \(r\) rows gives an \(r\times r\) matrix \(M\) whose rows are linearly independent; therefore \(\det M\ne0\). This \(M\) is an \(r\times r\) submatrix of \(A\), so \[ d(A)\ge r=\rho(A). \] Therefore \[ d(A)=\rho(A), \] and this equals both the row rank and the column rank.