Vector Space Notation question

linear algebranotationvector-spacesvectors

I have a question on notation used for vectors and a vector spaces. I'm using the Wiki page for Euclidean Space and Wiki for vectors. Definitions are provided and questions are identified in bold.

A bound vector or pair of points, with initial point $P$ and terminal point $Q$ is:
$$
\vec{PQ}_b = (P, Q)
$$
where $P,Q \in E$.

A free vector from the origin:
$$
\mathbf v = \vec v = \vec {0\mathbf v} = \vec {O\mathbf v} = a_0\hat {\mathbf b_0} + …+ a_n\hat {\mathbf b_1}
$$
where the ordered basis is $B = (\mathbf b_0, …,\mathbf b_n)$, $\mathbf b_n = (p_0,…,p_n) \in E$ and Euclidean Space $E = \mathbb{R}^n$.

A free vector is also defined on Wiki as an equivalence class of bound vectors, where $\bumpeq$ is equipollence (have the same length and direction), defined as true when a parallelogram can be drawn between $a,b,P,Q$:
$$
w = [\vec {PQ}_b]= \{(a, b) : a \in E \land b \in E \land \vec{ab}_b \bumpeq \vec{PQ}_b\}
$$
or using the difference operator:
$$
w = [\vec {PQ}_b] = \{(a, b) : a \in E \land b \in E \land ( b – a = Q – P)\}
$$

It's my understanding when using the equivalence relation above, the vector addition operation is defined as another equivalence relation:
$$
\vec{E} = \{[\vec {PQ}_b] : P \in E \land Q \in E\}
$$
$$
+_v : \vec{E} \times \vec{E} \rightarrow \vec{E}
$$
$$ v = +_v(w_0, w_1) = \{(a,c) : (a,b) \in w_0 \land (b,c) \in w_1\}
$$

Question do the above definitions look correct?

The definitions below define $\vec{PQ}_b$ as a free vector and not a equivalence class of bound vectors as above. The following definition agrees with Hamilton – Elements of Quaternions Page 1:
$$
\vec{PQ} = Q – P = \mathbf v
$$

which can be put into correspondence with bound vectors:

From a free vector to equivalence class:
$$
f(\vec{PQ}) = [\vec{PQ}_b]
$$

From equivalence class to a bound vector with $\vec 0$ origin:
$$
g([\vec{PQ}_b]) = \cup \{(\vec 0, y) \in [\vec{PQ}_b] : y \in E\}
$$
From a bound vector to a free vector:
$$
\vec{PQ} = f^-1([\vec{PQ}_b]) = \cup \{b – a : (a, b) = g(\vec{PQ}_b) \}
$$

Vector space of free vectors:
$$
\vec E = \{\vec{PQ} : P \in E \land Q \in E\}
$$

$P$ is a point of $E$ then:
$$
E = \{P + \mathbf v : \mathbf v \in \vec{E}\}
$$
$$
P + \vec{E} = \{P + \mathbf v : \mathbf v \in \vec{E}\}
$$
A line:
$$
L = \{P + \lambda \vec{PQ} : \lambda \in \mathbb{R}\}
$$
A line segment:
$$
PQ = QP = \{P + \lambda \vec{PQ} : 0 \le \lambda \le 1\}
$$

Finally, distance:
$$
d(P, Q) = ||\vec{PQ}||
$$

Best Answer

The primary mistake in the definitions I see is that there is no "vector space of bound vectors". Only the free vectors form a vector space.

You can add the bound vectors $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ to get the bound vector $\overrightarrow{PR}$. However, two bound vectors cannot be added if the endpoint of the first is not the start of the second. In a vector space, we should be able to add any two vectors, so this doesn't satisfy those conditions.

For any bound vector $\overrightarrow{PQ}$, we have a corresponding free vector: the free vector represented by the ordered pair $Q-P$. Most people don't talk about bound vectors at all, and so for them $\overrightarrow{PQ}$ simply is the free vector $Q-P$. In particular:

  • In the Wikipedia article on Euclidean vectors, $\overrightarrow{PQ}$ denotes a bound vector.
  • In the Wikipedia article on Euclidean spaces, bound vectors do not appear, and $\overrightarrow{PQ}$ denotes the free vector $Q-P$.

Another way to define free vectors, instead of the ordered pair $Q-P$, is as the equivalence class of all bound vectors with the same value of $Q-P$. Since such an equivalence class goes hand in hand with the value of $Q-P$, it doesn't matter too much which we use. Some considerations:

  • You might want to use the equivalence class definition if bound vectors are in the picture. The nice property it has is that the definition of $\vec v + \vec w$ for free vectors appears directly out of its definition for bound vectors: you choose any vector $\overrightarrow{PQ}$ in the equivalence class $\vec v$, choose specifically the vector $\overrightarrow{QR}$ in the equivalence class $\vec w$ that starts at $Q$, then add them together to get a representative of $\vec v + \vec w$.
  • On the other hand, just defining a free vector as $Q-P$ means it's a more concrete object rather than an equivalence class, and doesn't require us to talk about bound vectors at all.

As a side note, I would not say that a bound vector is an ordered pair $(P,Q)$, but rather that there is a one-to-one correspondence between bound vectors and such ordered pairs. They contain all the same data, but are different types of things. But this is a very minor thing.

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