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.
27. All vectors $(x, y, z)$ in $V_3$ whose components satisfy a system of three linear equations of the form:
\begin{align*}
a_{11}x + a_{12}y + a_{13}z = 0, \quad a_{21}x + a_{22}y + a_{23}z = 0, \quad a_{31}x + a_{32}y + a_{33}z = 0.
\end{align*}
Suppose we have two vectors $X = (x_1, x_2, x_3)$ and $Y = (y_1, y_2, y_3)$ that satisfy the system of linear equations defined above. Then, we have
\begin{align*}
& a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0, \quad a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0, \quad a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0.
\\
& a_{11}y_1 + a_{12}y_2 + a_{13}y_3 = 0, \quad a_{21}y_1 + a_{22}y_2 + a_{23}y_3 = 0, \quad a_{31}y_1 + a_{32}y_2 + a_{33}y_3 = 0.
\end{align*}
But, if we add the top equations to the bottom, we find that
\begin{align*}
a_{11}(x_1 + y_1) + a_{12}(x_2 + y_2) + a_{13}(x_3 + y_3) &= 0
\\
a_{21}(x_1 + y_1) + a_{22}(x_2 + y_2) + a_{23}(x_3 + y_3) &= 0
\\
a_{31}(x_1 + y_1) + a_{32}(x_2 + y_2) + a_{33}(x_3 + y_3) &= 0
\end{align*}which shows that the set of all vectors $(x, y, z)$ satisfying this system of linear equations satisfies closure under addition. Now, let $b$ be a real scalar. If we multiply both sides of
\begin{align*}
a_{11}x_1 + a_{12}x_2 + a_{13}x_3 &= 0
\\
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 &= 0
\\
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 &= 0
\end{align*}by $b,$ we get:
\begin{align*}
b(a_{11}x_1 + a_{12}x_2 + a_{13}x_3) &= 0
\\
b(a_{21}x_1 + a_{22}x_2 + a_{23}x_3) &= 0
\\
b(a_{31}x_1 + a_{32}x_2 + a_{33}x_3) &= 0
\end{align*}
But since the components of $X$ and all of the coefficients $a_{ij}$ are real-valued, we can use the distributive law of addition to get
\begin{align*}
ba_{11}x_1 + ba_{12}x_2 + ba_{13}x_3 &= 0
\\
ba_{21}x_1 + ba_{22}x_2 + ba_{23}x_3 &= 0
\\
ba_{31}x_1 + ba_{32}x_2 + ba_{33}x_3 &= 0
\end{align*}
Then, we can use the commutative law of multiplication to get
\begin{align*}
a_{11}bx_1 + a_{12}bx_2 + a_{13}bx_3 &= 0
\\
a_{21}bx_1 + a_{22}bx_2 + a_{23}bx_3 &= 0
\\
a_{31}bx_1 + a_{32}bx_2 + a_{33}bx_3 &= 0
\end{align*}
As we can see, $bX$ satisfies the system of linear equations, which means the set of vectors satisfying the system of equations satisfies closure under multiplication by real numbers.
The vector $O = (0, 0, 0)$ trivially satisfies this system of equations, and for any $X$ in $V_3,$ we have $X + O = X,$ thus confirming the existence of a zero element.
For the remaining axioms, since the set of $X$ satisfying this system of linear equations is a subset of $V_3,$ we know from previous exercises that any $X$ and $Y$ in this set satisfy the axioms on addition and multiplication by real numbers. Thus, the set of vectors $(x, y, z)$ in $V_3$ satisfying the system of linear equations
\begin{align*}
a_{11}x_1 + a_{12}x_2 + a_{13}x_3 &= 0
\\
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 &= 0
\\
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 &= 0
\end{align*}is a real linear space. $\,\blacksquare$
