Find a basis for $\Span(S)$, where $S$ is a Set of Four Vectors

Vector Space Problems and Solutions

Problem 710

Find a basis for $\Span(S)$ where $S=
\left\{
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
-1 \\ -2 \\ -1
\end{bmatrix}
,
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
,
\begin{bmatrix}
1 \\ 1 \\ 3
\end{bmatrix}
\right\}$.

 
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Solution.

We will first use the leading $1$ method. Consider the system
\[
x_{1}
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
+x_{2}
\begin{bmatrix}
-1 \\ -2 \\ -1
\end{bmatrix}
+x_{3}
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
+x_{4}
\begin{bmatrix}
1 \\ 1 \\ 3
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}
.\] The augmented matrix for this system is
\[
\left[\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 0 \\
2 & -2 & 6 & 1 & 0 \\
1 & -1 & -2 & 3 & 0
\end{array}\right] \xrightarrow{\substack{R_{2}-2R_{1} \\ R_{3}-R_{1}}}
\left[\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 0 \\
0 & 0 & 2 & -1 & 0 \\
0 & 0 & -4 & 2 & 0
\end{array}\right] \] \[
\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 0 \\
0 & 0 & 1 & -1/2 & 0 \\
0 & 0 & 0 & 0
\end{array}\right].
\] Since the above matrix has leading $1$’s in the first and third columns, we can conclude that the first and third vectors of $S$ form a basis of $\Span(S)$. Thus
\[
\left\{
\begin{bmatrix}
1 \\ 2 \\ 1
\end{bmatrix}
,
\begin{bmatrix}
2 \\ 6 \\ -2
\end{bmatrix}
\right\}
\] is a basis for $\Span(S)$.


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