I the post, A Linearly Independent Set, we were able to show that a set was linearly independent, but not a spanning set of a given vector space. In this example, we will look at a set of vectors which does span a given vector space.

**Our Set**

Here we let \(S=\left\{(3,5,2),(1,7,8),(9,3,3),(4,2,7)\right\}\). We now want to determine is \(S\) is linearly independent, a spanning set or a basis of \(\mathbb{R}^{3}\).

**Linearly Independent**

In order to determine if the given set is linearly independent, we need to determine if there are any solutions to

\begin{align*}

c_{1}(3,5,2)+c_{2}(1,7,8)+c_{3}(9,3,3)+c_{4}(4,2,7)=0,

\end{align*}

other than the solution \(c_{1}=c_{2}=c_{3}=c_{4}=0\). In order to determine this, we will solve the system of equations given by this matrix equation. Note that we will have

\begin{align*}

3c_{1}+c_{2}+9c_{3}+4c_{4}&=0 \\

5c_{1}+7c_{2}+3c_{3}+2c_{4}&=0 \\

2c_{1}+8c_{2}+3c_{3}+7c{4}&=0

\end{align*}

Now, recall that in Solving a System of Equations, we found a solution to such a system by creating an augmented matrix and row reducing. If we do the same here, we get the matrix

\begin{align*}

\begin{bmatrix}

3 & 1 & 9 & 4 & 0 \\

5 & 7 & 3 & 2 & 0 \\

2 & 8 & 3 & 7 & 0

\end{bmatrix}

\end{align*}

which we can row reduce to the matrix

\begin{align*}

\begin{bmatrix}

1 & 0 & 0 & \frac{-49}{36} & 0 \\

0 & 1 & 0 & \frac{11}{12} & 0 \\

0 & 0 & 1 & \frac{43}{54} & 0

\end{bmatrix}.

\end{align*}

Now, we can see that there is a free variable, so we would arrive at the general solution

\begin{align*}

c_{4}&=t \\

c_{1}&=\frac{49}{36}t \\

c_{2}&=-\frac{11}{12}t \\

c_{3}&=-\frac{43}{54}t

\end{align*}

Since we only need one nontrivial solution, we could let \(t\) be anything and show that the set of vectors is linearly dependent. Therefore, we will let it be \(108\). We then have the solution \(c_{1}=147, c_{2}=-99, c_{3}=-86, c_{4}=108\). This gives us what we need to provide a formal proof that \(S\) is linearly dependent.

**Proof**

Let \(S=\left\{(3,5,2),(1,7,8),(9,3,3),(4,2,7)\right\}\). Then note that

\begin{align*}

147(3,5,2)-99(1,7,8)-86(9,3,3)+108(4,2,7)=0.

\end{align*}

Hence, this set of vectors is linearly dependent.

**Spanning**

Now that we want to determine if the given set of vectors is spanning, we need to determine if the set any vector in \(\mathbb{R}^{3}\) can be written as a linear combination of the vectors in \(S\). Therefore, \(S\) will be spanning if every equation of the form

\begin{align*}

c_{1}(3,5,2)+c_{2}(1,7,8) +c_{3}(9,3,3)+c_{4}(4,2,7)=(x,y,z),

\end{align*}

has a solution.

Again, we can employ the techniques used in Solving a System of Equations. This will be similar to what we did for linearly independence, however, the right side will be not consist of all 0s. We instead get the matrix

\begin{align*}

\begin{bmatrix}

3 & 1 & 9 & 4 & x \\

5 & 7 & 3 & 2 & y \\

2 & 8 & 3 & 7 & z

\end{bmatrix}

\end{align*}

which we can row reduce to the matrix

\begin{align*}

\begin{bmatrix}

1 & 0 & 0 & \frac{-49}{36} & \frac{1}{72}(-x+23y-20z) \\

0 & 1 & 0 & \frac{11}{12} & \frac{1}{24}(-x-y+4z) \\

0 & 0 & 1 & \frac{43}{54} & \frac{1}{108}(13x-11y+8z)

\end{bmatrix}.

\end{align*}

Since we only need one solution, it will be easiest to let \(c_{4}=0\), then \(c_{1}=\frac{1}{72}(-x+23y-20z), c_{2}=\frac{1}{24}(-x-y+4z), \) and \(c_{3}=\frac{1}{108}(13x-11y+8z)\). We are now ready for the proof.

**Proof**

Let \(S=\left\{(3,5,2),(1,7,8),(9,3,3),(4,2,7)\right\}\) and \((x,y,z) \in \mathbb{R}^{3}\). Then note that

\begin{align*}

&\frac{1}{72}(-x+23y-20z)(3,5,2)+\frac{1}{24}(-x-y+4z)(1,7,8) \\

&+\frac{1}{24}(-x-y+4z)(9,3,3)+0(4,2,7)=(x,y,z).

\end{align*}

Hence, \(S\) spans \(\mathbb{R}^{3}\).

**Basis**

Here, we can just look at the work we have already done. Since \(S\) is not linearly independent, we know that \(S\) is not a basis of \(\mathbb{R}^{3}\).

**Conclusion**

Here we had a set that was spanning but not linearly independent. The set was therefore not a basis. In the next post, we will look at other examples of vector spaces to see which properties it has.

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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