I am studying mechanical engineering and we are doing LA in higher mathematics.
I was preparing myself for the exam and stumbled across the following thing:
Show that: $$v_1 =\begin{bmatrix} 1 & 2 & 3 & 4 \end{bmatrix}^T, v_2 =\begin{bmatrix} 5 & 6 & 7 & 8 \end{bmatrix}^T, v_3 =\begin{bmatrix} 1 & 6 & 3 & 8 \end{bmatrix}^T, v_4 =\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix}^T$$
are linear indipendent.
My approach was to use the gauss-algorithm:
$$ A=\begin{bmatrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8\\ 1 & 6 & 3 & 8\\ 1 & 0 & 0 & 0\\ \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 3 & 4\\ 0 & 4 & 8 & 12\\ 0 & 4 & 0 & 4\\ 0 & 2 & 3 & 4\\ \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 3 & 4\\ 0 & 4 & 8 & 12\\ 0 & 0 & 8 & 8\\ 0 & 0 & 2 & 4\\ \end{bmatrix} \to \begin{bmatrix} 1 & 2 & 3 & 4\\ 0 & 4 & 8 & 12\\ 0 & 0 & 8 & 8\\ 0 & 0 & 0 & 2\\ \end{bmatrix} $$ Therefor, I concluded that those vectors are linear independent.
Then, I saw their solution and they did the same algorithm but they applied the algorithm to $A^{-1}$
My question now is:
If n vectors are written into a matrix row by row, are the vectors, column by column, also linear independent?
Yes of course, indeed for any matrix the number of linearly independent columns is equal to the number of linearly independent rows, and that number is defined the rank of the matrix.