Let \(\Vect{a}_1,\dots ,\Vect{a}_p\) be linearly independent vectors in \(\RNrSpc{k}\), and let \(\Vect{x}_1=(x_{11},\dots ,x_{1p})\), ... , \(\Vect{x}_n=(x_{n1},\dots ,x_{np})\) be linearly independent vectors in \(\RNrSpc{p}\). Then
| \(\Vect{y}_1\) | \(\DefEq \) | \(x_{11}\Vect{a}_1+ \cdots + x_{1p} \Vect{a}_p\) |
| \(\) | \(\vdots\) | \(\) |
| \(\Vect{y}_n\) | \(\DefEq \) | \(x_{n1} \Vect{a}_1 + \cdots + x_{np} \Vect{a}_p\) |
are linearly independent vectors in \(\span\Set{ \Vect{a}_1,\dots ,\Vect{a}_p }\).
By design, each \(\Vect{y}_i\) belongs to \(\span\Set{ \Vect{a}_1,\dots ,\Vect{a}_p }\). So it remains to show that the vectors are linearly independent. So consider a representation of the zero vector as
| \(t_1 \Vect{y}_1 + \cdots + t_n \Vect{y}_n\) | \(=\) | \(\Vect{0}\) |
We need to show that \(t_1= \cdots =t_n=0\). Collecting scalar factors of the vectors \(\Vect{a}_i\), we find
| \((t_1 x_{11} + \cdots + t_n x_{n1}) \Vect{a}_1 + \cdots + (t_1 x_{1p} + \cdots + t_n x_{np}) \Vect{a}_p\) | \(=\) | \(\Vect{0}\) |
and conclude that
\[ \begin{array}{ccccccc} x_{11}t_1 & + & \cdots & + & x_{n1}t_n & = & 0 \\ \vdots & & & & \vdots & \vdots & \vdots \\ x_{1p}t_1 & + & \cdots & + & x_{np}t_n & = & 0 \end{array} \]This is a homogeneous system of linear equations in the \(n\) unknowns \(t_1,\dots ,t_n\). The column vectors of the coefficient matrix of this system are linearly independent. Therefore \(t_1=\cdots =t_n=0\) is its only solution. It follows that the vectors \(\Vect{y}_1,\dots ,\Vect{y}_n\) are linearly independent . – This completes the proof.