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.

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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*}

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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 $\R^3$.

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