Blogs, Linear Algebra

A Linearly Independent Set

Now that we have looked at vector spaces and subspaces, we turn our attention to finding sets of vectors that generate a given vector space. Additionally, we will want this generating set to be as small as possible. In order to generate a the vector space, we will need spanning set, whereas, it will be as small as possible if it is linearly independent. As such, we look at how to prove a set does or does not have these properties. We would call such a smallest generating set a basis of the vector space.

Let \(S=\left\{(5,2)\right\}\)

For our first example, we will look at whether or not the set \(S\) which consists of the single vector \((5,2)\) is a basis of \(\mathbb{R}^{2}\). In this case, we will need to check whether this set spans \(\mathbb{R}^{2}\) and if it is linearly independent.

Linear Independence

As we look at whether or not this set is linearly independent, we first need to recall the definition of linearly independent. Here, we have that a set of vectors \(\left\{\mathbf{v}_{1}, \ldots \mathbf{v}_{k}\right\}\) is linearly independent if the only solution to the equation
\begin{align*}
c_{1}\mathbf{v}_{1}+c_{2}\mathbf{v}_{2}+ \ldots c_{k}\mathbf{v}_{k}=0
\end{align*}
is the solution that \(C_{1}=c_{2}= \ldots =c_{k}=0\). Therefore, we need to check whether or not this is true for the set of vectors in \(S\).

Doing this, we see that if
\begin{align*}
c_{1}(5,2)=(0,0),
\end{align*}
then \(5c_{1}=0\) and \(2c_{1}=0\), hence \(c_{1}=0\). We, therefore, get that \(S\) is linearly independent. We are now ready to give a formal proof of this.

Proof

Let \(S=\left\{(5,2)\right\}\). Now suppose that
\begin{align*}
c_{1}(5,2)=0.
\end{align*}
We then have that \(5c_{1}=0\) and \(2c_{1}=0\). Hence \(c_{1}=0\). Therefore, \(S\) is linearly independent.

Spanning

In order to determine if \(S\) spans \(\mathbb{R}^{2}\), we must determine if every vector in \(\mathbb{R}^{2}\) can be written as a linear combination of the vectors of \(S\). That is, we need to show that if \((x,y) \in \mathbb{R}^{2}\), then there exists a \(c\) such that
\begin{align*}
c(5,2)=(x,y).
\end{align*}
Again, this gives us a system of equations, but we get \(5c=x\) and \(2c=y\). Therefore, we can write the vector \((x,y)\) as a linear combination of \((5,2)\) if and only if the system of equations
\begin{align*}
5c&=x \\
2c&=y
\end{align*}
has a solution. We note here, however, that by letting \(x=y=1\), we get that \(5c=1=2c\), so \(c=\frac{1}{5}\) and \(c=\frac{1}{2}\), which is a contradiction. Hence, \((1,1)\) is not in the span of \((5,2)\). We are now ready to prove that \(S\) does not span \(\mathbb{R}^{2}\).

Proof

Let \(S=(5,2)\). Then suppose, by way of contradiction, that \(S\) spans \(\mathbb{R}^{2}\). Since \((1,1) \in \mathbb{R}^{2}\), we must have that
\begin{align*}
c(5,2)=(1,1).
\end{align*}

However, this implies that \(\frac{1}{5}=c=\frac{1}{2}\), which is a contradiction. Hence, \(S\) does not span \(\mathbb{R}^{2}\).

Basis

In order to determine if \(S\) is a basis, we now just look at whether or not it was a linearly independent spanning set. Since we showed that \(S\) did not span \(\mathbb{R}^{2}\), we know that \(S\) is not a basis of \(\mathbb{R}^{2}\).

Conclusion

Here we had a set that was linearly independent but not spanning. The set was therefore not a basis. We could as well had a set that spanned and was linearly independent, spanned but was not linearly independent or a set that was linearly dependent and did not span the vector space. In the next posts, we will look at other examples of vector spaces where some of these options occur.

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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