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.
28. All vectors in $V_n$ that are linear combinations of two given vectors $A$ and $B$.
Let $a, b, c, d, x,$ and $y$ be real scalars and let $X = aA + bB,$ $Y = cA + dB.$ Then, we have
\begin{align*}
xX + yY &= x(aA + bB) + y(cA + dB)
\\
&= (xa + yc)A + (xb + yd)B
\end{align*}
As we can see, this is another linear combination of $A$ and $B.$ Hence, the set of all linear combinations of $A$ and $B$ satisfies closure under addition and closure under multiplication by real numbers. We can also confirm the associative law of multiplication by noting that in the above equation, $x(aA) = (ax)A.$
The zero vector can be found trivially by setting $a = b = 0,$ with $X + O = X$ for all $X$ in the set. If we set $x = 1$ we find that $xX = 1(aA + bB) = aA + bB = X,$ confirming the distributive law of addition and the existence of identity. Setting $x = -1,$ we find
\begin{align*}
xX + X &= -(aA + bB) + aA + bB
\\
&= (a - a)A + (b - b)B
\\
&= O
\end{align*}confirming the commutative law, distributive law for addition of numbers, and existence of negatives. And if $X,$ $Y,$ and $Z$ are vectors in $V_n,$ we know that $X + (Y + Z) = (X + Y) + Z$
As such, we have shown that the set of vectors in $V_n$ that are linear combinations of given vectors $A$ and $B$ is a rea linear space.$\,\blacksquare$
