Mathematical Immaturity

1.10 Exercises

In each of Exercises 1 through 10, let $S$ denote the set of all vectors $(x, y, z)$ in $V_3$ whose components satisfy the condition given. Determine whether $S$ is a subspace of $V_3.$ If $S$ is a subspace, compute $\dim S.$

• $x = y = z.$
• 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 $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$ be two elements of $S.$ That is, $A$ and $B$ are such that $a_1 = a_2 = a_3$ and $b_1 = b_2 = b_3.$ To satisfy closure under addition, \begin{align*} A + B &= (a_1 + b_1, a_2 + b_2, a_3 + b_3), \end{align*} must be an element of $S.$ In other words, it must satisfy $a_1 + b_1 = a_2 + b_2 = a_3 + b_3.$ But since $a_1 = a_2 = a_3$ and $b_1 = b_2 = b_3,$ we find that $a_1 + b_1 = a_2 + b_2 = a_3 + b_3,$ which means that $A + B$ is an element of $S,$ satisfying closure under addition. Now, let $c$ be a real scalar. We then have \begin{align*} cA &= c(a_1, a_2, a_3) \\ &= (ca_1, ca_2, ca_3) \end{align*} But if $a_1 = a_2 = a_3,$ then multiplying all sides of the equation by $c$ gives us $ca_1 = ca_2 = ca_3.$ But since $ca_1,$ $ca_2,$ and $ca_3$ are the respective $x,$ $y,$ and $z$ coordinates of $cA,$ this implies that $cA$ is an element of $S,$ satisfying closure under multiplication by real numbers. Following Theorem 1.4, $S$ is a subspace of $V_3.$ To find its dimension, we note that $S$ is spanned by the independent set $\{(1, 1, 1)\}$ thus giving $\dim S = 1. \quad \blacksquare$