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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?

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  • $\begingroup$ 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? $\endgroup$ Commented Mar 1, 2020 at 20:15
  • $\begingroup$ 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 $n$." $\endgroup$ Commented Mar 1, 2020 at 20:39

2 Answers 2

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Consider the following linear map $\phi$ from $P_n(x)$ to $R$: \begin{equation} \phi(f)=f(a). \end{equation}

Then $\text{ker}(\phi)=S$ and $\dim(\text{im}(\phi))=1$. By Rank-Nullity Theorem, \begin{equation} \dim(\text{ker}(\phi))=\dim(P_n(x))-\dim(\text{im}(\phi))=n+1-1=n. \end{equation}

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Hint

$S$ 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 $x-a, (x-a)x, \dots, (x-a)x^{n-1}$.

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  • $\begingroup$ 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? $\endgroup$ Commented Mar 1, 2020 at 20:48
  • $\begingroup$ 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 $x-a$. $\endgroup$ Commented Mar 1, 2020 at 20:52

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