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 $\dim S.$

• $f(0) + f'(0) = 0.$
• 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 $P_n$ such that $f(0) + f'(0) = 0$ and $g(0) + g'(0) = 0.$ Adding these two equations, we get: \begin{align*} f(0) + f'(0) + g(0) + g'(0) &= (f + g)(0) + (f + g)'(0) \\ &= 0 \end{align*} Which means that the sum $f + g$ is also an element of $S,$ satisfying closure under addition.
  
  If $c$ is some real scalar, then $cf$ is also an element of $S,$ with \begin{align*} cf(0) + cf'(0) &= c\left[f(0) + f'(0)\right] \\ &= c(0) \end{align*} thus satisfying closure under multiplication by real numbers. Hence, $S$ is a subspace of $P_n.$ To compute the dimension of $S,$ we must find a set of independent elements that spans $S.$ To do so, we revisit the condition $f(0) + f'(0) = 0.$ If $f$ is a real polynomial of degree $\leq n,$ we can express $f$ and its derivative $f'$ as \begin{align*} f(x) &= \sum_{k = 0}^n a_k x^k, \qquad f'(x) = \sum_{k = 1}^n a_k x^{k - 1} \end{align*} where $a_k$ are real-valued coefficients. Evaluating at $x = 0,$ we get \begin{align*} f(0) &= a_0, \qquad f'(0) = a_1 \end{align*} Thus, for all $f$ in $S,$ we have $a_0 + a_1 = 0$ or $a_0 = -a_1.$ In turn, this means that for all $f$ in $S,$ the first two terms of the polynomial must be of the form $a_0(1 - x).$ As such, $S$ is spanned by the set of $n$ vectors $\{1 - x, x^2, ..., x^n\}$ which means that $\dim S = n.\quad \blacksquare$