We write \(\Mtrx{A}\) in terms of its row vectors \(R_1,\dots ,R_m\), and its column vectors \(C_1,\dots ,C_n\):

\[ \Mtrx{A} = \left[ \begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{mn} & \cdots & a_{mn} \end{array} \right] = \left[ \begin{array}{c} R_1 \\ \vdots \\ R_m \end{array} \right] = [ C_1\ \ \dots\ \ C_n] \] i.

To see the relationship between the linear independence of the columns of \(\Mtrx{A}\) and the determinant of a suitable submatrix of \(\Mtrx{A}\), recall that \(C_1,\dots , C_n\) are linearly independent if and only if the RREF of \(\Mtrx{A}\) is

\[ \left[ \begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & 0 \end{array} \right] \]

This happens exactly when \(\Mtrx{A}\) has \(n\) rows, forming an \((n,n)\)-matrix \(\Vect{B}\), whose RREF is the identity matrix; i.e. exactly when \(\det(\Mtrx{B})\neq 0\).

ii.

This follows from the previous part: The rows \(R_1,\dots ,R_m\)are the columns of \(\Mtrx{A}^T\), and the claim follows. – This completes the proof