My geometry textbook states that the vectors $(a, b, c)^T$ and $k(a, b, c)^T$ represent the same line for any non-zero $k$; in other words, two such vectors related by an overall scaling are considered equivalent. It then goes on to state that an equivalence class of vectors under this equivalence relationship is known as a homogeneous vector.
I want to prove that this relation is indeed an equivalence relation.
So let $X = \{(a, b, c)^T \mid a, b, c \in \mathbb{R} \}$, and define the relation $R = \{ ((a, b, c)^T, (x, y, z)^T) \mid (a, b, c)^T = k(x, y, z)^T, (x, y, z)^T \in X, k \in (\mathbb{R} \setminus \{ 0 \}) \}$ over $X$ ($R \subseteq X \times X$) . I will now prove that $R$ is reflexive, symmetric, and transitive.
I would greatly appreciate it if people could please take the time to review my work for correctness.
Reflexivity:
Let $x = (x_1, x_2, x_3)^T \in X$.
$\therefore (x_1, x_2, x_3)^T = k(x_1, x_2, x_3)^T$ when $k = 1$
Since $k = 1 \in \mathbb{R}$, $(x_1, x_2, x_3)^T \in X$, and $(x_1, x_2, x_3)^T = k(x_1, x_2, x_3)^T$, we can conclude that $((x_1, x_2, x_3)^T, (x_1, x_2, x_3)^T) \in R$ and $x \sim x$.
$Q.E.D.$
Symmetry:
Let $x = (x_1, x_2, x_3)^T \in X$ and $y = (y_1, y_2, y_3)^T \in X$.
- (Proof that $(x \sim y) \Rightarrow (y \sim x)$.)
Assume that $(x \sim y)$.
$\therefore (x_1, x_2, x_3)^T = k(y_1, y_2, y_3)^T$, where $k \in (\mathbb{R} \setminus \{ 0 \})$
(By the definition of $R$.)
$\Rightarrow \dfrac{1}{k} (x_1, x_2, x_3)^T = (y_1, y_2, y_3)^T$
$\Rightarrow K (x_1, x_2, x_3)^T = (y_1, y_2, y_3)^T$, where $\dfrac{1}{k} = K \in (\mathbb{R} \setminus \{ 0 \})$
$\Rightarrow (y_1, y_2, y_3)^T = K (x_1, x_2, x_3)^T$
$\therefore y \sim x$
(Since $(y_1, y_2, y_3)^T \in X$, $(x_1, x_2, x_3)^T \in X$, $K \in (\mathbb{R} \setminus \{ 0 \})$, and $(y_1, y_2, y_3)^T = K (x_1, x_2, x_3)^T$, we can conclude that $((y_1, y_2, y_3)^T, (x_1, x_2, x_3)^T) \in R$ and $y \sim x$.)
- (Proof that $(y \sim x) \Rightarrow (x \sim y)$.)
Same argument as above.
$Q.E.D.$
Transitivity:
Let $x^T, y^T, z^T \in X$.
$x^T = k_1 y^T$, where $k_1 \in (\mathbb{R} \setminus \{ 0 \})$
$y^T = k_2 z^T$, where $k_2 \in (\mathbb{R} \setminus \{ 0 \})$
(Since the hypothesis assumes that $x \sim y$ and $y \sim z$, we have that $(x^T, y^T) \in R$ and $(y^T, z^T) \in R$, which implies that $k_1, k_2 \in (\mathbb{R} \setminus \{ 0 \})$.)
$\therefore x = k_1(k _2 z)$
$\Rightarrow x = (k_1 k_2) z$
$\Rightarrow x = Kz$, where $k_1 k_2 = K \in (\mathbb{R} \setminus \{ 0 \})$
(Since $k_1, k_2 \in (\mathbb{R} \setminus \{ 0 \})$, we have that $K \in (\mathbb{R} \setminus \{ 0 \})$.)
$\therefore x \sim z$
(Since $x^T \in X$, $z^T \in X$, $K \in (\mathbb{R} \setminus \{ 0 \})$, and $x = Kz$, we can conclude that $(x^T, z^T) \in R$ and $x \sim z$.)
$Q.E.D.$