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

Problems and solutions in Linear Algebra

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

 
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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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