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 its rank, recall that \(C_1,\dots , C_n\) are linearly independent if and only if the vector equation

\(t_1 C_1 + \cdots + t_n C_n\)

\(=\)

\(\Vect{0}\)

has \((t_1,\dots ,t_n)=(0,\dots ,0)\) as its only solution. We know that this happens if and only if the reduced row echelon form 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 is so if and only if \(\Mtrx{A}\)has rank \(n\).

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