# A Basis

In A Spanning Set we found a set which spanned a given vector space, but that was linearly dependent. While as in A Linear Independent Set we found a set that was linearly independent, but did not span the given vector space. In this post, we will look at a set which is both linearly independent and spans a given vector space. That is, we will show that the given set is a basis.

## Our Set

Let $$S=\left\{(2,0,0),(4,0,-2),(6,1,-1)\right\}$$. Show that $$S$$ is a basis of $$\mathbb{R}^{3}$$.

### Basis

In the last two examples we saw, we had to go through both cases of linear independence and spanning as separate cases. However, we should point out that this set of vectors is linearly independent if
\begin{align*}
c_{1}(2,0,0)+c_{2}(4,0,-2)+c_{3}(6,1,-1)=(0,0,0)
\end{align*}
only has the solution $$c_{1}=c_{2}=c_{3}=0$$. Whereas, if we want to show that this set is a spanning set, we must show that
\begin{align*}
c_{1}(2,0,0)+c_{2}(4,0,-2)+c_{3}(6,1,-1)=(x,y,z)
\end{align*}
has at least one solution for every $$x,y,z \in \mathbb{R}$$.

Now, if we instead show that
\begin{align*}
c_{1}(2,0,0)+c_{2}(4,0,-2)+c_{3}(6,1,-1)=(x,y,z)
\end{align*}
has a unique solution for each $$x,y,z \in \mathbb{R}$$ we get both of these at the same time. Therefore, we need to show that this equation has a unique solution.

The first thing we need to note is that
\begin{align*}
c_{1}(2,0,0)+c_{2}(4,0,-2)+c_{3}(6,1,-1)=(x,y,z)
\end{align*}
is equivalent to the matrix equation
\begin{align*}
\begin{bmatrix}
2 & 4 & 6 \\
0 & 0 & 1 \\
0 & -2 & -1
\end{bmatrix}
\begin{bmatrix}
c_{1} \\ c_{2} \\ c_{3} \end{bmatrix}
=\begin{bmatrix} x \\ y \\ z \end{bmatrix}.
\end{align*}

Now, we know that this equation will have a unique solution if and only if the matrix
\begin{align*}
\begin{bmatrix}
2 & 4 & 6 \\
0 & 0 & 1 \\
0 & -2 & -1
\end{bmatrix}
\end{align*}
is invertible.

Furthermore, we know that this matrix has an inverse if and only if the determinant of this matrix is nonzero. Using the techniques in Finding a Determinant (Part 1) or Finding a Determinant (Part 2) we find that the determinant of this matrix is 4. Hence, this matrix equation will have a unique solution for all $$x,y,z \in \mathbb{R}$$. Therefore, $$S$$ will be a basis. We are now ready for a proof.

### Proof

Let $$S=\left\{(2,0,0),(4,0,-2),(6,1,-1)\right\}$$. Then note that $$S$$ is a basis if and only the equation
\begin{align*}
c_{1}(2,0,0)+c_{2}(4,0,-2)+c_{3}(6,1,-1)=(x,y,z)
\end{align*}
has a unique solution for each $$x,y,z \in \mathbb{R}$$. Furthermore, this is equivalent to the matrix equation
\begin{align*}
\begin{bmatrix}
2 & 4 & 6 \\
0 & 0 & 1 \\
0 & -2 & -1
\end{bmatrix}
\begin{bmatrix}
c_{1} \\ c_{2} \\ c_{3} \end{bmatrix}
=\begin{bmatrix} x \\ y \\ z \end{bmatrix}.
\end{align*}

We now have that this equation has a unique solution if and only if the determinant of
\begin{align*}
\begin{bmatrix}
2 & 4 & 6 \\
0 & 0 & 1 \\
0 & -2 & -1
\end{bmatrix}
\end{align*}
is nonzero. Since
\begin{align*}
|\begin{bmatrix}
2 & 4 & 6 \\
0 & 0 & 1 \\
0 & -2 & -1
\end{bmatrix}|&=2|\begin{bmatrix} 0 & 1 \\ -2 * -1 \end{bmatrix}| \\
&=2(0-(-2))=4,
\end{align*}
we have that there is a unique solution to the system of equations. Hence, $$S$$ is a basis of $$\mathbb{R}^{3}$$. Furthermore, this implies that $$S$$ is both linearly independent and a spanning set.

## Conclusion

Here we had a set that was a basis of the given vector space. For this case, we could directly show that we had a basis without having to look individually at spanning and linear independence.

I hope this helped you with determining whether or not a set spanned a given vector space or was linearly independent and how to give a proof thereof. If you learned something and enjoyed the process, make sure to share the post with other people that would find it useful on Social Media. Also, make sure to see our videos and subscribe to our channel on YouTube.

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