Equivalence relation on Vector Space

[ych22]

Let W be a subspace of a vector space V. We define a relation v~w if v-w is an element of W.

It can be shown that ~ is an equivalence relation on V.

Suppose that V is R^2. Say W1 is a representative of the equivalence class that includes (1,0). Say W2 is a representative of the equivalence class that includes (0,1). Obviously the zero vector is related to (1,0) and (0,1).

But either two equivalence classes are similar, or they are disjoint. Am I missing something out?

 

[Fredrik]

When you specified V=ℝ², you didn't specify the subspace. Suppose that W={(x,0)|x in ℝ}. Now the equivalence class that contains (1,0) (which can be written as [(1,0)] but is more commonly written as (1,0)+W) is =W. (0,1) on the other hand, is a member of (0,1)+W, which is disjoint from W.

In this case, the subspace W is the x axis, which is a horizontal line in a diagram of the xy-plane. All of the equivalence classes are horizonal lines.

 

[HallsofIvy]

Let W be a subspace of a vector space V. We define a relation v~w if v-w is an element of W.

It can be shown that ~ is an equivalence relation on V.

Suppose that V is R^2. Say W1 is a representative of the equivalence class that includes (1,0). Say W2 is a representative of the equivalence class that includes (0,1). Obviously the zero vector is related to (1,0) and (0,1).

Let W be the subspace spanned by (1, 1) (that is all (x, x)). The 0 vector (0, 0) is NOT equivalent to either (1, 0) or (0,1). If fact, it is easy to show that, no matter what W is, a vector is equivalent to (0, 0) if and only if it is in W.

But either two equivalence classes are similar, or they are disjoint. Am I missing something out?

Are you under the impression that these equivalence classes are themselves subspaces- and so all contain the 0 vector? That's not true. Of all the equivalence classes defined by W, only W itself is a subspace.

For example, in R², all (non-trivial) subspaces are lines through the origin. The various equivalence classes defined by such a subspace are lines parallel to that line. So that all of them except the subspace itself are NOT through the origin, do NOT include (0, 0), and so are not subspaces.

 

[Landau]

Obviously the zero vector is related to (1,0) and (0,1).

No, this is not obviously so. 0 being related to (1,0) means that their difference is an element of W. But that difference is precisely (up to sign) (1,0) itself. In other words, 0 is related to v if and only if $v\in W$. So you are in fact asserting that 'obviously (1,0) and (0,1) are in W'.

 

[blue_raver22]

I can confirm that the definition of ~ as an equivalence relation on a vector space is accurate. In this case, the relation ~ is reflexive, symmetric, and transitive, which are the three properties required for an equivalence relation.

As for your question about the equivalence classes, it is important to note that the equivalence classes are not necessarily similar to each other. In this case, the equivalence class containing (1,0) is related to the equivalence class containing (0,1) through the zero vector. This does not mean that these two equivalence classes are the same or similar, but rather they are related through the subspace W.

In general, equivalence classes are used to group together elements that share a common property or relation, but they do not necessarily have to be identical or similar to each other. In the context of vector spaces, the equivalence classes are used to group together elements that are related through the subspace W, but they may have different values or coordinates.

Therefore, you are not missing anything out, but it is important to understand that equivalence classes in this context are not necessarily similar or identical. They are simply a way to group together elements that are related through the subspace W.