In Exercises 1 through 28, determine whether each of the given sets is a real linear space, if addition and multiplication by real scalars are defined in the usual way. For those that are not, tell which axioms fail to hold. The functions in Exercises 1 through 17 are real-valued. In Exercises 3, 4, and 5, each function has domain containing 0 and 1. In Exercises 7 through 12, each domain contains all real numbers.
26. All vectors $(x, y, z)$ in $V_3$ which are scalar multiples of $(1, 2, 3)$.
Let $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$ be vectors $(x, y, z)$ in $V_3$ which are scalar multiples of $(1, 2, 3).$ In other words, for real scalars $a$ and $b,$ we can write $A = (a, 2a, 3a)$ and $B = (b, 2b, 3b).$ Then, we have
\begin{align*}
A + B &= (a + b, 2a + 2b, 3a + 3b)
\\
&= [(a + b), 2(a + b), 3(a + b)]
\\
&= (a + b)(1, 2, 3)
\end{align*}which we can see is another scalar multiple of $(1, 2, 3)$ in $V_3,$ satisfying closure under addition. Now, let $c$ be another real scalar, and let $C = (c, 2c, 3c).$ By component-wise addition, we can see that $A + B = B + A$ and $(A + B) + C = A + (B + C).$ Setting $c = 0,$ we can see that the zero vector $O = (0, 0, 0)$ is trivially a member of the set, with $A + O = A.$ If we set $c = 1$ we can see that $cA = A,$ confirming the existence of identity, and if we set $c = -1,$ we find that $cA + A = O,$ confirming the existence of negatives.
Now, with our vectors $A$ and $B$ and scalars $a,$ $b,$ and $c$ defined as before, we can easily verify the axioms for multiplication by real numbers:
\begin{align*}
c(aB) &= c(ab, 2ab, 3ab)
\\
&= (cab, 2cab, 3cab)
\\
&= (ca)B
\\
\\
c(A + B) &= c[(a + b), 2(a + b), 3(a + b)]
\\
&= (ca + cb, 2ca + 2cb, 3ca + 3cb)
\\
&= cA + cB
\\
\\
(b + c)A &= (b + c)(a, 2a, 3a)
\\
&= [(b + c)a, (b + c)2a, (b + c)3a]
\\
&=(ba + ca, 2ba + 2ca, 3ba + 3ca)
\\
&= bA + cA
\end{align*}
Thus, we have shown that the set of all scalar multiples of $(1, 2, 3)$ in $V_3$ is a linear space. $\,\blacksquare$
