# Equivalence relation on Vector Space

## Main Question or Discussion Point

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?

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Fredrik
Staff Emeritus
Gold Member
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
Homework Helper
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'.