# Three Linearly Independent Vectors in $\R^3$ Form a Basis. Three Vectors Spanning $\R^3$ Form a Basis.

## Problem 574

Let $B=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a set of three-dimensional vectors in $\R^3$.

**(a)** Prove that if the set $B$ is linearly independent, then $B$ is a basis of the vector space $\R^3$.

**(b)** Prove that if the set $B$ spans $\R^3$, then $B$ is a basis of $\R^3$.

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## Definition (A Basis of a Subspace).

A subset $B$ of a vector space $V$ is called a **basis** if $B$ is linearly independent spanning set.

## Proof.

### (a) Prove that if the set $B$ is linearly independent, then $B$ is a basis of the vector space $\R^3$.

To show that $B$ is a basis, we need only prove that $B$ is a spanning set of $\R^3$ as we know that $B$ is linearly independent.

Let $\mathbf{b}\in \R^3$ be an arbitrary vector.

We prove that there exist $x_1, x_2, x_3$ such that

\[x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{b}.\]
This is equivalent to having a solution $\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix}$ to the matrix equation

\[A\mathbf{x}=\mathbf{b}, \tag{*}\]
where

\[A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]\]
is the $3\times 3$ matrix whose column vectors are $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$.

Since the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent, the matrix $A$ is nonsingular.

It follows that the equation (*) has the unique solution $\mathbf{x}=A^{-1}\mathbf{b}$.

Hence $\mathbf{b}$ is a linear combination of the vectors in $B$.

This means that $B$ is a spanning set of $\R^3$, hence $B$ is a basis.

### (b) Prove that if the set $B$ spans $\R^3$, then $B$ is a basis of $\R^3$.

As we know that $B$ spans $\R^3$, it suffices to show that $B$ is linearly independent.

Note that the assumption that $B$ is a spanning set of $\R^3$ means that any vector $\mathbf{b}$ in $\R^3$ is a linear combination of $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$: there exist $c_1, c_2, c_3$ such that

\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{b}.\]
Equivalently, for any $\mathbf{b}\in \R^3$, the equation $A\mathbf{x}=\mathbf{b}$ has a solution.

Let $A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]$ as before.

It follows that the equation

\[A\mathbf{x}=\begin{bmatrix}

1 \\

0 \\

0

\end{bmatrix}\]
has a solution $\mathbf{x}=\mathbf{u}_1$.

Similarly the equations

\[A\mathbf{x}=\begin{bmatrix}

0 \\

1 \\

0

\end{bmatrix}, \quad A\mathbf{x}=\begin{bmatrix}

0 \\

0 \\

1

\end{bmatrix}\]
have solutions $\mathbf{u}_2, \mathbf{u}_3$, respectively.

Define the $3\times 3$ matrix $A’$ by

\[A’=[\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3].\]

Then it follows that

\begin{align*}

AA’&=A[\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3]\\

&=[A\mathbf{u}_1, A\mathbf{u}_2, A\mathbf{u}_3]\\

&=\begin{bmatrix}

1 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & 1

\end{bmatrix}.

\end{align*}

As the identity matrix is nonsingular, the product $AA’$ is nonsingular.

Thus, the matrix $A$ is nonsingular as well.

This implies that the column vectors of $A$ are linearly independent.

Hence the set $B$ is linearly independent and we conclude that $B$ is a basis of $\R^3$.

## Related Question.

Use the result of the problem, try the next problem about a basis for $\R^3$.

**Problem**.

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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## 1 Response

[…] 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 Basi…” for the proof of this […]