# Expressing $E$ as a linear combination of the dual basis.

Here is the question I am trying to solve:

Let $$V$$ be the collection of polynomials with coefficients in $$\mathbb Q$$ in the variable $$x$$ of degree at most 5 with $$1, x, x^2, \dots , x^5$$ as basis. Prove that the following are elements of the dual space of $$V$$ and express them as a linear combination of the dual basis.

$$E : V \to \mathbb Q$$ defined by $$E(p(x)) = p(3).$$

My thoughts:

I was able to prove that it is linear and hence in $$V^*.$$ but I do not know how to express it as a linear combination of the dual basis? How can I find the dual basis in this case, what are the vectors corresponding to the basis of $$V$$ that I should use to determine the dual basis? If I answered this question I can complete from there. Could someone clarify this to me please?

Edit:

My question is: How can I write the basis element $$x$$ for example as a vector, could anyone explain this to me please?

• Do you substitute a specific $x$? Because $p(x)$ is a polynomial, not a rational number.
– Mark
Dec 5, 2023 at 0:15
• @Mark I did not get your point. Dec 5, 2023 at 0:30
• you mean that for example $f_0(1) = 1$?@Mark Dec 5, 2023 at 0:31
• You defined $E:V\to\mathbb{Q}$ as $E(p(x))=p(x)$. I'm asking how exactly is that a map from $V$ to $\mathbb{Q}$. $p(x)$ is a polynomial, not an element of $\mathbb{Q}$.
– Mark
Dec 5, 2023 at 0:34
• oh sorry I will edit my question @Mark Dec 5, 2023 at 0:57

By definition, if your basis is $$1,x,x^2,...,x^5$$ then its dual basis of $$V^*$$ consists of the linear functionals $$\varphi_0,\varphi_1,...,\varphi_5$$ that satisfy:

$$\varphi_i(x^j)=\begin{cases} 1 & i=j\\ 0 & i\ne j\\ \end{cases}$$

In more detail, the map $$\varphi_i:V\to\mathbb{Q}$$ (for $$0\leq i\leq 5$$) is defined by $$\varphi_i(\sum\limits_{j=0}^5 a_jx^j)=a_i$$, i.e it gives us the $$i$$-th coefficient in the expansion of the vector as a linear combination of our basis of $$V$$.

Note: To this point it was the explanation of what is the dual basis. The next part is about solving the problem. If you want to try alone, stop reading here.

Let $$f(x)=\sum\limits_{j=0}^5 a_jx^j$$ be an element of $$V$$. Then $$\varphi_i(f)=a_i$$ for each $$i$$, and $$E(f)=\sum\limits_{j=0}^5a_j3^j$$. Therefore:

$$E(f)=\sum\limits_{j=0}^5 3^j\varphi_j(f).$$

This is true for any $$f\in V$$, and hence as functions $$V\to\mathbb{Q}$$ we have $$E=\sum\limits_{j=0}^53^j\varphi_j.$$ This is the required linear combination.

• Downvoter, care to explain the reasons?
– Mark
Dec 10, 2023 at 22:21