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 $\Span(S)$.

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 $\Span(S)$.