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 $\dim S.$

• $x + y + z= 0.$
• 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 $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$ be two elements of $S.$ That is, $A$ and $B$ are such that $a_1 + a_2 + a_3 = 0$ and $b_1 + b_2 + b_3 = 0.$ To satisfy closure under addition, $A + B$ must also be in $S.$ But $A + B$ is given by \begin{align*} A + B &= (a_1 + b_1, a_2 + b_2, a_3 + b_3), \end{align*} For $A + B$ to be an element of $S,$ it must satisfy $x + y + z= 0,$ where $x = a_1 + b_1,$ $y = a_2 + b_2,$ and $z = a_3 + b_3.$ But we know that \begin{align*} x + y + z &= (a_1 + b_1) + (a_2 + b_2) + (a_3 + b_3) \\ &= (a_1 + a_2 + a_3) + (b_1 + b_2 + b_3) \\ &= (0) + (0) \\ &= 0 \end{align*} which implies that $A + B$ is an element of $S,$ satisfying closure under addition. Now, let $c$ be a real scalar. We then have $cA = c(a_1, a_2, a_3) = (ca_1, ca_2, ca_3).$ But this means \begin{align*} ca_1 + ca_2 + ca_3 &= c(a_1 + a_2 + a_3) \\ &= c(0) \\ &= 0 \end{align*} which implies that $cA$ is also 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 $\{(1, -1, 0), (0, 1, -1)\}$ thus giving $\dim S = 2. \quad \blacksquare$