The following vectors are linearly independent -
$v1 = (1, 2, 0, 2)$
$v2 = (1,1,1,0)$
$v3 = (2,0,1,3)$
Find a fourth vector v4 so that the set { v1, v2, v3, v4 } is a basis fpr $\mathbb{R}^4$?
I asked this question before here - Show vectors are linearly independent and finding a basis - and someone suggested a way of doing it. However I am wondering if there is a simpler way. I put the vector
$\begin{bmatrix} 0 \\ 0 \\ 0 \\ x \end{bmatrix}$
as the fourth column in a matrix of these vectors, then row reduce.
$\begin{bmatrix} 1 & 1 & 2 & 0 & | & 0 \\ 2 & 1 & 0 & 0 & | & 0 \\ 0 & 1 & 1 & 0 & | & 0 \\ 2 & 0 & 3 & x & | & 0 \end{bmatrix}$
By doing this I will be left with the fourth column looking like, for example, $(0, 0, 0, x-2)$. Then as long as x is not equal to 2, there will be pivots in each column and the vectors will be linearly independent?
In the matrix above the fourth column ends up as $(0, 0, 0, x)$. So as long as x is not equal to 0 the vectors will be linearly independent?
Edit: David Mitra's answer here - Show vectors are linearly independent and finding a basis - is the best way to do this imo.