Find a Basis for the Subspace spanned by Five Vectors

Problem 709

Let $S=\{\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}\}$ where
\[
\mathbf{v}_{1}=
\begin{bmatrix}
1 \\ 2 \\ 2 \\ -1
\end{bmatrix}
,\;\mathbf{v}_{2}=
\begin{bmatrix}
1 \\ 3 \\ 1 \\ 1
\end{bmatrix}
,\;\mathbf{v}_{3}=
\begin{bmatrix}
1 \\ 5 \\ -1 \\ 5
\end{bmatrix}
,\;\mathbf{v}_{4}=
\begin{bmatrix}
1 \\ 1 \\ 4 \\ -1
\end{bmatrix}
,\;\mathbf{v}_{5}=
\begin{bmatrix}
2 \\ 7 \\ 0 \\ 2
\end{bmatrix}
.\] Find a basis for the span $\Span(S)$.

 
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We will give two solutions.

Solution 1.

We apply the leading 1 method.
Let $A$ be the matrix whose column vectors are vectors in the set $S$:
\[
A=
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
.\] Applying the elementary row operations to $A$, we obtain
\begin{align*}
A=\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
\xrightarrow[R_4+R_1]{\substack{R_2-2R_1 \\ R_3-2R_1}}
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
0 & 1 & 3 & -1 & 3 \\
0 & -1 & -3 & 2 & -4 \\
0 & 2 & 6 & 0 & 4
\end{bmatrix}\\[6pt] \xrightarrow[R_4-2R_2]{\substack{R_1-R_2 \\ R_3+R_2}}
\begin{bmatrix}
1 & 0 & -2 & 2 & -1 \\
0 & 1 & 3 & -1 & 3 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 2 & -2
\end{bmatrix}
\xrightarrow[R_4-2R_3]{\substack{R_1-2R_3 \\ R_2+R_3}}
\begin{bmatrix}
1 & 0 & -2 & 0 & 1 \\
0 & 1 & 3 & 0 & 2 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}=\rref(A).
\end{align*}
Observe that the first, second, and fourth column vectors of $\rref(A)$ contain the leading 1 entries.
Hence, the first, second, and fourth column vectors of $A$ form a basis of $\Span(S)$.
Namely,
\[\left\{ \begin{bmatrix}
1 \\
2 \\
2 \\
-1
\end{bmatrix}, \begin{bmatrix}
1 \\
3 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
4 \\
-1
\end{bmatrix}\right \}\] is a basis for $\Span(S)$.

Solution 2.

Let
\[
A=
\begin{bmatrix}
1 & 1 & 1 & 1 & 2 \\
2 & 3 & 5 & 1 & 7 \\
2 & 1 & -1 & 4 & 0 \\
-1 & 1 & 5 & -1 & 2
\end{bmatrix}
.\] Then $\Span(S)$ is the column space of $A$, which is the row space of $A^{T}$. Using row operations, we have
\[
A^{T}=
\begin{bmatrix}
1 & 2 & 2 & -1 \\
1 & 3 & 1 & 1 \\
1 & 5 & -1 & 5 \\
1 & 1 & 4 & -1 \\
2 & 7 & 0 & 2
\end{bmatrix}
\to
\begin{bmatrix}
1 & 2 & 2 & -1 \\
0 & 1 & -1 & 2 \\
0 & 3 & -3 & 6 \\
0 & -1 & 2 & 0 \\
0 & 3 & -4 & 4
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 4 & 5 \\
0 & 1 & -1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 0 & -1 & -2
\end{bmatrix}
\] \[
\to
\begin{bmatrix}
1 & 0 & 0 & -13 \\
0 & 1 & 0 & 4 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
.
\] Therefore, the set of nonzero rows
\[
\left\{
\begin{bmatrix}
1 \\ 0 \\ 0 \\ -13
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 1 \\ 0 \\ 4
\end{bmatrix}
,
\begin{bmatrix}
0 \\ 0 \\ 1 \\ 2
\end{bmatrix}
\right\}
\] is a basis for the row space of $A^{T}$, which equals $\Span(S)$.

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