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 $\dim S.$

• $x + y + z = 0$ and $x - y - z = 0.$
• 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.
• If $x + y + z = 0$ and $x - y - z = 0,$ then we have $x = y + z$ and $x = -y - z,$ or $y + z = -(y + z).$ But, adding $(y+z)$ to both sides, we get $2(y + z) = 0,$ which implies that $y + z = - y - z = 0.$ But this in turn implies that $x = 0,$ with $y = -z.$ Now, let $A = (0, a, -a)$ and $B = (0, b, -b)$ be two elements of $S,$ where $a$ and $b$ are real scalars. Then, \begin{align*} A + B &= (0, a + b, -a - b) \\ \end{align*} from which we can see that $S$ satisfies closure under addition, with \begin{align*} x + y + z &= 0 + (a + b) + (-a - b) \\ &= a + b - a - b \\ &= 0 \\ \\ x - y - z &= -(a + b) - (-a - b) \\ &= -a - b + a + b \\ &= 0 \end{align*} Then, if $c$ is some real scalar, we have $cA = (0, ca, -ca),$ with \begin{align*} x + y + z &= 0 + ca - ca \\ &= 0 \\ \\ x - y - z &= 0 -ca + ca \\ &= 0 \end{align*} satisfying closure under multiplication. As such, $S$ is a subspace of $V_3.$ And since we can see that all vectors in $S$ are scalar multiples of $(0, 1, -1),$ we find that the $S$ has dimension $1. \quad \blacksquare$