Blogs, Linear Algebra

A Vector Space of Polynomials

As we continue our study of linear algebra, we begin to focus on sets with properties we’ve seen displayed by our matrices. Looking at other sets that have similar properties, we arrive at the topic of vector spaces.

Definition

Let \(V\) be a set on which two operations (vector addition and scalar multiplication) are defined. \(V\) is a vector space over \(F\), if for every \(\mathbf{u},\mathbf{v},\mathbf{w} \in V\) and scalars \(c,d \in F\) we have

  1. Addition:
    1. \(\mathbf{u}+\mathbf{v} \in V\),
    2. \(\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}\),
    3. \(\mathbf{u}+(\mathbf{v}+\mathbf{w})=(\mathbf{u}+\mathbf{v})+\mathbf{w}\),
    4. \(V\) has a zero vector, \(0\), such that, for every \(\mathbf{u} \in V\), \(\mathbf{u}+0=\mathbf{u}\).
    5. For every \(\mathbf{u} \in V\), there exists a \(-\mathbf{u}\) such that \(\mathbf{u}+(-\mathbf{u})=0\).
  2. Scalar Multiplication:
    1. \(c\mathbf{u} \in V\),
    2. \(c(\mathbf{u}+\mathbf{v})=c\mathbf{u}+c\mathbf{v}\),
    3. \((c+d)\mathbf{u}=c\mathbf{u}+d\mathbf{u}\),
    4. \(c(d\mathbf{u})=(cd)\mathbf{u}\),
    5. \(1\mathbf{u}=\mathbf{u}\).

The set of polynomials of degree 3 or less

Now that we have the formal definition of a vector space, we will need to be able to show that a set is a vector space. Therefore, we will work through showing the following.

Let \(P_{3}\) be the set of all polynomials of degree 3 or less. Then, with the addition and multiplication defined in the usual way, \(P_{3}\) forms a vector space over \(\mathbb{R}\).

In order to show that \(P_{3}\) is indeed a vector space, we will need to show all of the properties given above. As we work through this, we will let \(p(x)=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}\), \(q(x)=q_{3}x^{3}+q_{2}x^{2}+q_{1}x+q_{0}\) and \(r(x)=r_{3}x^{3}+r_{2}x^{2}+r_{1}x+r_{0}\) be in \(P_{3}\) with each \(p_{i},q_{j}\) and \(r_{k}\) being a real numbers. Furthermore, let \(c,d\) be real numbers. We then have the following.

Addition Properties

Closure

Note that
\begin{align*}
&p(x)+q(x) \\
&=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}+q_{3}x^{3}+q_{2}x^{2}+q_{1}x+q_{0} \\
&=(p_{3}+q_{3})x^{3}+(p_{2}+q_{2})x^{2}+(p_{1}+q_{1})x+(p_{0}+q_{0}).
\end{align*}
Note that \(p_{i}+q_{i} \in \mathbb{R}\) since \(p_{i}, q_{i} \in \mathbb{R}\) for \(i \in \mathbb{Z}\) and \(0 \leq i \leq 3\). Therefore, the coefficients before each of the powers of \(x\) are real numbers. Hence, \(p(x)+q(x)\) is again a polynomial of degree at most 3.

As a further note, the degree would be less than 3 if \(p_{3}+q_{3}=0\); however, such a polynomial would still be in \(P_{3}\).

Commutative

Using the same notation, we find that
\begin{align*}
&p(x)+q(x)\\
&=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}+q_{3}x^{3}+q_{2}x^{2}+q_{1}x+q_{0} \\
&=(p_{3}+q_{3})x^{3}+(p_{2}+q_{2})x^{2}+(p_{1}+q_{1})x+(p_{0}+q_{0}) \\
&=(q_{3}+p_{3})x^{3}+(q_{2}+p_{2})x^{2}+(q_{1}+p_{1})x+(q_{0}+p_{0}) \\
&=q_{3}x^{3}+q_{2}x^{2}+q_{1}x+q_{0}+p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0} \\
&=q(x)+p(x).
\end{align*}
Note the third and fourth lines follow due to commutativity of addition and the distributive property of real numbers. Hence, addition of polynomials is commutative.

Associative

We now note that
\begin{align*}
&(p(x)+q(x))+r(x) \\
&=(p_{3}+q_{3})x^{3}+(p_{2}+q_{2})x^{2}+(p_{1}+q_{1})x+(p_{0}+q_{0})+r_{3}x^{3}+r_{2}x^{2}+r_{1}x+r_{0} \\
&=((p_{3}+q_{3})+r_{3})x^{3}+((p_{2}+q_{2})+r_{2})x^{2}+((p_{1}+q_{1})+r_{1})x+(p_{0}+q_{0})+r_{0} \\
&=(p_{3}+(q_{3}+r_{3}))x^{3}+(p_{2}+(q_{2}+r_{2}))x^{2}+(p_{1}+(q_{1}+r_{1}))x+p_{0}+(q_{0}+r_{0}) \\
&=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}+(q_{3}+r_{3})x^{3}+(q_{2}+r_{2})x^{2}+((q_{1}+r_{1})x+q_{0}+r_{0} \\
&=p(x)+(q(x)+r(x)).
\end{align*}

Note that third line follows from associativity of real number addition. Hence, the addition of polynomials is associative.

Zero Vector

Note that \(0\) (the real number) is a constant function, as is therefore a polynomial of degree 0. Hence \(0 \in P_{3}\). Furhtermore,
\begin{align*}
&p(x)+0 \\
&=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}+0 \\
&=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+(p_{0}+0) \\
&=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0} \\
&=p(x).
\end{align*}
Therefore, there is a additive identity for polynomials (the constant 0) of degree 3 or less.

Additive Inverse

Let \(s(x)=(-p_{3})x^{3}+(-p_{2})x^{2}+(-p_{1})x+(-p_{0})\). Note that \(-p_{i} \in \mathbb{R}\) since \(p_{i} \in \mathbb{R}\) for \(i \in \mathbb{Z}\) and \(0 \leq i \leq 3\). Therefore, \(s(x) \in P_{3}\). Furthermore,
\begin{align*}
&p(x)+s(x)\\
&=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}+(-p_{3})x^{3}+(-p_{2})x^{2}+(-p_{1})x+(-p_{0}) \\
&=(p_{3}-p_{3})x^{3}+(p_{2}-p_{2})x^{2}+(p_{1}-p_{1})x+(p_{0}-p_{0}) \\
&=0x^{3}+0x^{2}+0x+0 \\
&=0.
\end{align*}
Hence, \(s(x)\) is the additive inverse of \(p(x)\). Therefore, every polynomial has a \(-p(x)\) such that \(p(x)+(-p(x))=0\).

Scalar Multiplication

Closure

Here we note that
\begin{align*}
cp(x)&=c(p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}) \\
&=cp_{3}x^{3}+cp_{2}x^{2}+cp_{1}x+cp_{0}.
\end{align*}
Furthermore, \(c p_{i} \in \mathbb{R}\) since \(c, p_{i} \in \mathbb{R}\) for all \(i \in \mathbb{K}\) with \(0 \leq i \leq 3\). Hence, \(cp(x) \in P_{3}\).

Distributive

We now have that
\begin{align*}
(c+d)p(x)&=(c+d)(p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}) \\
&=(c+d)p_{3}x_{3}+(c+d)p_{2}x^{2}+(c+d)p_{1}x+(c+d)p_{0} \\
&=(cp_{3}+dp_{3})x^{3}+(cp_{2}+dp_{2})x^{2}+(cp_{1}+dp_{1})x+cp_{0}+dp_{0} \\
&=cp_{3}x^{3}+cp_{2}x^{2}+cp_{1}x+cp_{0}+dp_{3}x^{3}+dp_{2}x^{2}+dp_{1}x+dp_{0} \\
&=cp(x)+dp(x).
\end{align*}
Hence, you can distribute across scalar addition.

Associative

Here, we see that
\begin{align*}
(cd)p(x)&=(cd)(p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}) \\
&=(cd)p_{3}x_{3}+(cd)p_{2}x^{2}+(cd)p_{1}x+(cd)p_{0} \\
&=c(dp_{3})x^{3}+c(dp_{2})x^{2}+c(dp_{1})x+c(dp_{0}) \\
&=c((dp_{3})x^{3}+(dp_{2})x^{2}+(dp_{1})x+(dp_{0})) \\
&=c(dp(x)).
\end{align*}
Therefore, scalar multiplication is associative.

Identity

Note that \(1\) is a real number. Furthermore, we have that
\begin{align*}
1*p(x)&=1*(p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0}) \\
&=1*p_{3}x^{3}+1*p_{2}x^{2}+1*p_{1}x+1*p_{0} \\
&=p_{3}x^{3}+p_{2}x^{2}+p_{1}x+p_{0} \\
&=p(x).
\end{align*}
Therefore, \(1\) acts as an identity for scalar multiplication in \(P_{3}\).

Conclusion

We have now shown that the set of polynomials of degree 3 or less satisfied all of the required properties for a vector space when using the usual addition and multiplication. Hence, we have proving that \(P_{3}\) is a vector space.

While it is helpful to know that \(P_{3}\) is a vector space, note that we would use a very similar process when trying to prove that any set forms a vector space. That is, we would have to work through each of the properties one by one in order to ensure that all of them are satisfied. For further work, we could also show that \(P_{n}\) is a vector space for any \(n \in \mathbb{N}\).

As always, I hope you learned something and enjoyed the process. If you did, make sure to like the post and share it on Social Media.

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