# How to find basis for subspace of $P_n(R)$

I'm working on an exercise in Friedberg's Linear Algebra. The exercise is as follows

For a fixed $$a \in R$$, determine the dimension of the subspace of $$P_n(R)$$ defined by $$S = \{f \in P_n(R) \ |\ f(a) = 0 \}$$.

(The answer in the book is $$n-1$$)

This is my work so far:

We begin by considering an arbitrary element $$f \in S$$. We express $$f$$ as \begin{align} f(x) = b_0 + b_1x + b_2x^2 + \cdots + b_nx^n. \end{align} Since $$f(a) = 0$$, it follows that \begin{align*} f(a) &= b_0 + b_1a + b_2a^2 + \cdots + b_na^n \\ 0 &= b_0 + \sum_{k=1}^n b_ka^k\\ b_0 &= -(\sum_{k=1}^n b_ka^k) \\ b_0 &= -b_1a - b_2a^2 - \cdots - b_na^n \end{align*} Reinserting this expression into our expression for $$f(x)$$, we have

\begin{align*} f(x) &= b_0 + b_1x + b_2x^2 + \cdots + b_nx^n \\ &= (-b_1a - b_2a^2 - \cdots - b_na^n) + b_1x + b_2x^2 + \cdots + b_nx^n \\ &= b_1(x-a) + b_2(x^2 - a^2) + b_3(x^3 - a^3) + \cdots + b_n(x^n - a^n)\\ \end{align*}

I'm terribly confused for the following reasons:

1. If we let $$b_1 = -(x+a)$$ , $$b_2 = 1$$, and $$b_3, \cdots , b_n = 0$$, then we have a nontrivial solution to the homogeneous equation.

2. I'm having difficulty understanding how the value of $$x$$ affects answers to the homogeneous equation. For instance, if we let $$x = a$$ in this equation, then clearly the equation equals $$0$$ (by hypothesis).

3. It seems to me that we could do the same thing as in (1.) for $$x^2 - a^2$$ and $$x^4 - a^4$$. That is, let $$b_2 = -(x^2+a^2)$$, let $$b_4 = 1$$, and the remaining coefficients equal $$0$$. Could we not do this for many of the terms of the form $$b_{2k}(x^{2k} - a^{2k})$$, producing many nontrivial solutions to the equation?

I feel as though I'm misunderstanding something fundamental about working with polynomial spaces. How do I determine the dimension of this subspace, and what role does the value of $$x$$ have to do with respect to the dimension of the subspace?

• Since $\dim S=n$, not $n-1$, there is something wrong here. Are you sure that your definition of $P_n(\mathbb R)$ is the correct one? Commented Mar 1, 2020 at 20:15
• Perhaps there is a typo in the textbook. The answer in the back of the book states $n-1$. My definition of $P_n(R)$ is "The vector space of polynomials of degree less than or equal to $n$." Commented Mar 1, 2020 at 20:39

Consider the following linear map $$\phi$$ from $$P_n(x)$$ to $$R$$: $$$$\phi(f)=f(a).$$$$
Then $$\text{ker}(\phi)=S$$ and $$\dim(\text{im}(\phi))=1$$. By Rank-Nullity Theorem, $$$$\dim(\text{ker}(\phi))=\dim(P_n(x))-\dim(\text{im}(\phi))=n+1-1=n.$$$$
$$S$$ is the subspace of polynomials $$f \in P_n(\mathbb R)$$ such that $$x-a$$ divides $$f$$. To obtain this result consider the Euclidean division of $$f$$ by $$x-a$$.
Therefore $$\dim S= n$$. And a basis of $$P_n(\mathbb R)$$ is $$x-a, (x-a)x, \dots, (x-a)x^{n-1}$$.
• I'm sorry, I'm not understanding how one goes from $b_1(x-a) + b_2(x^2 - a^2) + b_3(x^3 - a^3) + \cdots + b_n(x^n - a^n)$ to the statement that "$S$ is the subspace of polynomials $f \in P_n(R)$ such that $x-a$ divides $f$. Or is my work incorrect/unnecessary in coming to that conclusion? Commented Mar 1, 2020 at 20:48
• You can prove that $x^k-x^k$ can be divided by $x-a$ for all $p$. But an easier way is to consider the Euclidean division of $f$ by $x-a$. Commented Mar 1, 2020 at 20:52