$v_1=(1, 1, 2, 4), v_2=(2, -1, -5, 2),v_3 = (1, -1, -4, 0), v_4=(2, 1, 1, 6).$
I started with writing down the matrix and through row reduction ended with the solution $x_1=4/3 + (1/3)x_3$ and $x_2 = 1/3 – (2/3)x_3$, $x_3$ is free. I think the basis is supposed to be $\{v_1, v_2\}$, but I'm not sure if this is correct. They are linearly independent, but how do the two vectors generate $\mathbb{R}^4$?
Best Answer
The easy way to approach this problem is to write down a $4\times4$ matrix with the given vectors in the rows of the matrix. Then do elementary row operations to reduce the matrix to row echelon form. Let's do that...
$$\begin{bmatrix}1 & 1 & 2 & 4 \\ 2 & -1 & -5 & 2 \\ 1 & -1 & -4 & 0 \\ 2 & 1 & 1 & 6 \\ \end{bmatrix} \Rightarrow \begin{bmatrix}1 & 1 & 2 & 4 \\ 0 & -3 & -9 & -6 \\ 0 & -2 & -6 & -4 \\ 0 & -1 & -3 & -2 \\ \end{bmatrix} \Rightarrow \begin{bmatrix}1 & 1 & 2 & 4 \\ 0 & -3 & -9 & -6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$
Since the last two rows are all zeros, we know that the given set of four vectors is linearly dependent and the sub-space spanned by the given vectors has dimension 2. Only two of the four original vectors were linearly independent.
Since we put the four vectors into the rows of the matrix and elementary row operations do not change the row space of the matrix (the space spanned by the rows of the matrix), the two remaining non-zero row vectors span the row space of the matrix. That is, the vectors $\begin{pmatrix}1,1,2,4\end{pmatrix} \;\mathrm{and}\; \begin{pmatrix}0,-3,-9,-6\end{pmatrix}$ are linearly independent and span the sub-space $\mathrm{span}\left\{v_1, v_2,v_3,v_4 \right\}$ - so, they form a basis.