Let \(W\) denote the subspace of \(R^{5}\) consisting of all vectors having
coordinates that sum to zero. The vectors
$$
\begin{array}{ll}
u_{1}=(2,-3,4,-5,2), & u_{2}=(-6,9,-12,15,-6), \\
u_{3}=(3,-2,7,-9,1), & u_{4}=(2,-8,2,-2,6), \\
u_{5}=(-1,1,2,1,-3), & u_{6}=(0,-3,-18,9,12), \\
u_{7}=(1,0,-2,3,-2), & \text { and } & u_{8}=(2,-1,1,-9,7)
\end{array}
$$
generate W. Find a subset of \(\left\\{u_{1}, u_{2}, \ldots, u_{8}\right\\}\)
that is a basis for W.
A basis for W is the subset $$B=\{u_1, u_2, u_3, u_4, u_6\} = \{(2, -3, 4, -5, 2),(-6, 9, -12, 15, -6), (3, -2, 7, -9, 1), (2, -8, 2, -2, 6), (0, -3, -18, 9, 12)\}.$$
Step by step solution
01
Create matrix with given vectors as columns
Initiate the matrix A with the given vectors as columns:
$$
A=
\begin{pmatrix}
2 & -6 & 3 & 2 & -1 & 0 & 1 & 2 \\\
-3 & 9 & -2 & -8 & 1 & -3 & 0 & -1 \\\
4 & -12 & 7 & 2 & 2 & -18 & -2 & 1 \\\
-5 & 15 & -9 & -2 & 1 & 9 & 3 & -9 \\\
2 & -6 & 1 & 6 & -3 & 12 & -2 & 7
\end{pmatrix}
$$
02
Perform Gaussian elimination to obtain the row echelon form of A
By using Gaussian elimination, we obtain the row echelon form of the matrix A:
$$
\Rrightarrow
\begin{pmatrix}
2 & -6 & 3 & 2 & -1 & 0 & 1 & 2 \\\
0 & 3 & 1 & -1 & 1 & 0 & 1 & 0 \\\
0 & 0 & 3 & -2 & 3 & 0 & 0 & 4 \\\
0 & 0 & 0 & 5 & -3 & 0 & 0 & 10 \\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0
\end{pmatrix}
$$
03
Identify pivot columns and corresponding basis vectors
Based on the row echelon form of matrix A, we see that there are 5 pivot columns: columns 1, 2, 3, 4, and 6. This means that the following subset of the vectors forms a linearly independent set and is a basis for W:
$$ B=\{u_1, u_2, u_3, u_4, u_6 \} = \{(2, -3, 4, -5, 2),( -6, 9, -12, 15, -6), (3, -2, 7, -9, 1), (2, -8, 2, -2, 6), (0, -3, -18, 9, 12)\}$$
So, a basis for W is the subset $$B=\{u_1, u_2, u_3, u_4, u_6 \}.$$
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