Mathematical Immaturity

1.10 Exercises

Let $P_n$ denote the linear space of all real polynomials of degree $\leq n,$ where $n$ is fixed. In each of Exercises 11 through 20, let $S$ denote the set of all polynomials $f$ in $P_n$ satisfying the condition given. Determine whether or not $S$ is a subspace of $P_n.$ If $S$ is a subspace, compute $\dim S.$

• $f(0) = f(2).$
• Recall from Section 1.6, Theorem 1.4, that a nonempty subset $S$ of a linear space $V$ is a subspace if and only if it satisfies the closure axioms:
  
  $\text{Axiom 1.}\quad$Closure under addition. $\quad$ For every pair of elements $x$ and $y$ in $V$ there corresponds a unique element in $V$ called the sum of $x$ and $y,$ denoted by $x + y.$
  $\text{Axiom 2.}\quad$Closure under multiplication by real numbers. $\quad$ For every $x$ in $V$ and every real number $a$ there corresponds an element in $V$ called the product of $a$ and $x,$ denoted by $ax.$
  
  The dimension of a basis for a linear space is the number of elements in the basis.
• Let $f$ and $g$ be two elements of $S.$ Then we know that $f(0) = f(2)$ and $g(0) = g(2).$ Then, since the sum of two real polynomials satisfies $(f + g)(x) = f(x) + g(x)$ for all $x,$ the sum $f + g$ is as follows: \begin{align*} (f + g)(0) &= f(0) + g(0) \\ &= f(2) + g(2) \\ &= (f + g)(2) \end{align*} Which means that $f + g$ is also an element of $S,$ satisfying closure under addition.
  
  Suppose that $h(x) = cf(x)$ for some real scalar $c.$ Then $h(x)$ is such that \begin{align*} h(0) &= cf(0) \\ &= cf(2) \\ &= h(2) \end{align*} But this means that $h(x) = cf(x)$ is also an element of $S,$ satisfying closure under multiplication by real numbers. Hence, $S$ is a subspace of $P_n.$ To compute the dimension of $S,$ we first revisit the initial condition. For the polynomial \begin{align*} f(x) &= \sum_{k=0}^n a_kx^k \end{align*} the condition $f(0) = f(2)$ implies that: \begin{align*} a_0 + \sum_{k=1}^n a_k 0^k &= a_0 + \sum_{k=1}^n a_k 2^k \end{align*} Subtracting $a_0$ from both sides and simplifying, we get \begin{align*} \sum_{k = 1}^n a_k 2^k &= 0. \end{align*} In other words, we can represent the $k^{th}$ coefficient as a nontrivial linear combination of the other $n - 2$ coefficients. For simplicity, we represent $a_n$ as the linear combination: \begin{align*} a_n &= -\frac{1}{2^n}\sum_{k=1}^{n-1}2^ka_k \\ &= -\sum_{k=1}^{n-1}2^{k-n}a_k \end{align*} Then, plugging this back into the equation for $f(x),$ we find that \begin{align*} f(x) &= a_0 + a_1x + \cdots + a_{n-1}x^{n-1} - \left(\sum_{k=1}^{n-1}2^{k-n}a_k\right)x^n \\ \\ &= a_0 + a_1\left(x - \frac{1}{2^{n-1}}x^n\right) + \cdots + a_{n-1}\left(x^{n-1} - \frac{1}{2}x^n\right) \\ \\ &= a_0 + \sum_{k=1}^{n-1}a_k\left(x - \frac{1}{2^{n - k}}x^n\right) \end{align*} From this equation for $f(x),$ we can see that every element in $S$ can be represented as a linear combination of the $n$ independent elements \begin{align*} \left\{1, x - \frac{1}{2^{n - 1}}x^n, x^2 - \frac{1}{2^{n - 2}}x^n, \cdots, x^{n-1} - \frac{1}{2}x^n\right\} \end{align*} from which we conclude that $\dim S = n.\quad\blacksquare$