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 $\dim S.$

• $f(0) = 0.$
• 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 $A$ and $B$ be polynomials of degree $\leq n$ defined as follows: \begin{align*} A &= \sum_{k = 0}^n a_k x^k \qquad B = \sum_{k = 0}^n b_k x^k \end{align*} where $a_1, ..., a_n$ and $b_1, ..., b_n$ are real coefficients. Then, for $A$ and $B$ to be elements of $S,$ $a_0$ and $b_0$ must be zero. The sum of $A$ and $B$ is then \begin{align*} A + B &= \sum_{k = 1}^n a_k x^k + \sum_{k = 1}^n b_k x^k \\ &= \sum_{k = 1}^{n} (a_k + b_k)x^k \end{align*} Setting $x = 0,$ we find that $(A + B)(0)$ is \begin{align*} (A + B)(0) &= \sum_{k = 1}^{n} (a_k + b_k)0 \\ &= 0, \end{align*} satisfying closure under addition. If we let $c$ be some real scalar, then the product $cA$ is given by \begin{align*} cA &= c\sum_{k = 1}^n a_k x^k \\ &= \sum_{k = 1}^n ca_k x^k \end{align*} then, setting $x = 0,$ we get \begin{align*} cA(0) &= \sum_{k = 1}^n (ca_k)(0) \\ &= 0, \end{align*} satisfying closure under multiplication. As such, $S$ is a subspace of the linear space of real polynomials of degree $\leq n.$ To find the dimension of $S,$ we note that since $S$ is spanned by the independent set of $n$ elements $\{x, ..., x^n\},$ thus $\dim S = n. \quad \blacksquare$