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.


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 $x \sim x$.



Let $x = (x_1, x_2, x_3)^T \in X$ and $y = (y_1, y_2, y_3)^T \in X$.

  1. (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 $y \sim x$.)

  1. (Proof that $(y \sim x) \Rightarrow (x \sim y)$.)

Same argument as above.



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 $x \sim z$.)


  • $\begingroup$ Reflexivity is just wrong. What you want to show is that for every $(x_1,x_2,x_3)\in X$, you have $(x_1,x_2,x_3)\sim (x_1,x_2,x_3)$, by producing an explicit $k$ (or showing such a $k$ exists). But you claim that the fact that $(x_1,x_2,x_3)$ is in $X$ somehow implies $(x_1,x_2,x_3)=k(x,y,z)$. You never say who this mysterious $(x,y,z)$ is, you never say who $k$ is. That’s wrong. $\endgroup$ – Arturo Magidin May 10 at 3:59
  • $\begingroup$ For symmetry, you should explicitly note that $K$ is a real number because $k\neq 0$, and you also need to note that $K\neq 0$. Similarly, for Transitivity, you need to show that $K$ is not zero. $\endgroup$ – Arturo Magidin May 10 at 4:00
  • $\begingroup$ @ArturoMagidin Thank you for the review. I have attempted to fix the errors that you described. Is it correct now? $\endgroup$ – The Pointer May 10 at 4:11
  • $\begingroup$ No; you still don’t say who $k$ is in the reflexivity; you need to exhibit a scalar that works, but you don’t exhibit one. You just introduce one without saying where it came from or what it is. And you still do not explain why your $K$s in the second and third part are nonzero, you just assert that this is the case. You need to justify the assertion, not just make it. (Also, the “therefore” in the second line of your Reflexivity proof does not make any sense; don’t think that putting in symbols substitutes for making sense) $\endgroup$ – Arturo Magidin May 10 at 4:15
  • $\begingroup$ @ArturoMagidin Thanks again. I have gone through every step and made changes and added justifications (especially for $K$). What do you think? $\endgroup$ – The Pointer May 10 at 5:13

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