Find a basis for the subspace of \(\mathbb{R}^4\) spanned by the four vectors \[ \alpha_1 = (1, 1, 2, 4), \quad \alpha_2 = (2, -1, -5, 2), \quad \alpha_3 = (1, -1, -4, 0), \quad \alpha_4 = (2, 1, 1, 6) \] of Exercise 2.
Solution
1. The four vectors are \(\alpha_1 = (1,1,2,4)\), \(\alpha_2 = (2,-1,-5,2)\), \(\alpha_3 = (1,-1,-4,0)\), \(\alpha_4 = (2,1,1,6)\).
from Exercise 2.
2. Row-reduce the matrix with these vectors as rows:\[\begin{pmatrix} 1 & 1 & 2 & 4 \\ 2 & -1 & -5 & 2 \\ 1 & -1 & -4 & 0 \\ 2 & 1 & 1 & 6 \end{pmatrix}.\]
set up the matrix.
3. Row-reducing gives the row echelon form\[\begin{pmatrix} 1 & 1 & 2 & 4 \\ 0 & -3 & -9 & -6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix},\]so the rank is \(2\).
\(R_2 \leftarrow R_2 - 2R_1\) gives \((0, -3, -9, -6)\); \(R_3 \leftarrow R_3 - R_1\) gives \((0, -2, -6, -4) = \frac{2}{3}(0,-3,-9,-6)\); \(R_4 \leftarrow R_4 - 2R_1\) gives \((0, -1, -3, -2) = \frac{1}{3}(0,-3,-9,-6)\).
4. Hence the span has dimension \(2\), and a basis is \(\{\alpha_1, \alpha_2\}\).
3 (the first two rows are linearly independent and span the row space).
5. Q.E.D.
4.