# Determine Whether Each Set is a Basis for $\R^3$

## Problem 579

Determine whether each of the following sets is a basis for $\R^3$.

(a) $S=\left\{\, \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 4 \end{bmatrix} \,\right\}$

(b) $S=\left\{\, \begin{bmatrix} 1 \\ 4 \\ 7 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix}, \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} \,\right\}$

(c) $S=\left\{\, \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 7 \end{bmatrix} \,\right\}$

(d) $S=\left\{\, \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix}, \begin{bmatrix} 7 \\ 4 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 8 \\ 6 \end{bmatrix}, \begin{bmatrix} -1 \\ 9 \\ 10 \end{bmatrix} \,\right\}$

## Definition (A Basis of a Subspace).

A subset $S$ of a vector space $V$ is called a basis if

1. $S$ is linearly independent, and
2. $S$ is a spanning set.

## Solution.

Recall that any three linearly independent vectors form a basis of $\R^3$.
(See the post “Three Linearly Independent Vectors in $\R^3$ Form a Basis. Three Vectors Spanning $\R^3$ Form a Basis.” for the proof of this fact.)

### (a) $S=\left\{\, \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 4 \end{bmatrix} \,\right\}$

Let us check that whether $S$ is a linearly independent set.
Consider the linear combination
$x_1\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}+x_2 \begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}+x_3\begin{bmatrix} -2 \\ 1 \\ 4 \end{bmatrix} =\mathbf{0}.$ This is equivalent to the matrix equation
$\begin{bmatrix} 1 & 2 & -2 \\ 0 &1 &1 \\ -1 & -1 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\mathbf{0}.$ To find the solution, consider the augmented matrix.
Applying elementary row operations, we obtain
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 2 & -2 & 0 \\
0 &1 & 1 & 0 \\
-1 & -1 & 4 & 0
\end{array} \right] \xrightarrow{R_3+R_1}
\left[\begin{array}{rrr|r}
1 & 2 & -2 & 0 \\
0 &1 & 1 & 0 \\
0 & 1 & 2 & 0
\end{array} \right]\6pt] \xrightarrow[R_3-R_2]{R_1-2R_2} \left[\begin{array}{rrr|r} 1 & 0 & -4 & 0 \\ 0 &1 & 1 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] \xrightarrow[R_2-R_3]{R_1+4R_3} \left[\begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 &1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right]. \end{align*} It follows that the solution is x_1=x_2=x_3=0. Hence S is linearly independent. As S consists of three linearly independent vectors in \R^3, it must be a basis of \R^3. ### (b) S=\left\{\, \begin{bmatrix} 1 \\ 4 \\ 7 \end{bmatrix}, \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix}, \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} \,\right\} As in part (a), we determine whether the set S is linearly independent or not by considering the following augmented matrix: \begin{align*} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 0 \\ 4 &5 & 6 & 0 \\ 7 & 8 & 9 & 0 \end{array} \right] \xrightarrow[R_3-7R_1]{R_2-4R_1} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 0 \\ 0 &-3 & -6 & 0 \\ 0 & -6 & -12 & 0 \end{array} \right]\\[6pt] \xrightarrow{-\frac{1}{3}R_2} \left[\begin{array}{rrr|r} 1 & 2 & 3 & 0 \\ 0 & 1 & 2& 0 \\ 0 & -6 & -12 & 0 \end{array} \right] \xrightarrow[R_3+6R_2]{R_1-2R_2} \left[\begin{array}{rrr|r} 1 & 0 & -1 & 0 \\ 0 &1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]. \end{align*} Thus, the general solution is x_1=x_3, x_2=-2x_3, where x_3 is a free variable. Hence, in particular, there is a nonzero solution. So S is linearly dependent, and hence S cannot be a basis for \R^3. ### (c) S=\left\{\, \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 7 \end{bmatrix} \,\right\} A quick solution is to note that any basis of \R^3 must consist of three vectors. Thus S cannot be a basis as S contains only two vectors. Another solution is to describe the span \Span(S). Note that a vector \mathbf{v}=\begin{bmatrix} a \\ b \\ c \end{bmatrix} is in \Span(S) if and only if \mathbf{v} is a linear combination of vectors in S. Equivalently, the vector \mathbf{v} is in \Span(S) if and only if the system \[\begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 2 &7 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} is consistent.

Let us consider the augmented matrix and reduce it by elementary row operations.
\begin{align*}
\left[\begin{array}{rr|r}
1 & 0 & a \\
1 & 1 &b \\
2 & 7 & c
\end{array}\right] \xrightarrow[R_3-2R_1]{R_2-R_1}
\left[\begin{array}{rr|r}
1 & 0 & a \\
0 & 1 &b-a \\
0 & 7 & c-2a
\end{array}\right]\6pt] \xrightarrow{R_3-7R_2} \left[\begin{array}{rr|r} 1 & 0 & a \\ 0 & 1 &b-a \\ 0 & 0 & 5a-7b+c \end{array}\right]. \end{align*} Note that we obtained the (3,3)-entry by c-2a-7(b-a)=5a-7b+c. It follows that the system is consistent if and only if \[5a-7b+c=0. Thus, for example, the vector $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ is not in $\Span(S)$ as $5\cdot 1-7\cdot 0+0\neq 0$.
Hence $\Span(S)$ is not $\R^3$, and we conclude that $S$ is not a basis.

### (d) $S=\left\{\, \begin{bmatrix} 1 \\ 2 \\ 5 \end{bmatrix}, \begin{bmatrix} 7 \\ 4 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 8 \\ 6 \end{bmatrix}, \begin{bmatrix} -1 \\ 9 \\ 10 \end{bmatrix} \,\right\}$

The set $S$ contains four $3$-dimensional vectors. Hence $S$ is linearly dependent, and thus $S$ is not a basis.

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