# I Reflexive relation question

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1. Oct 12, 2016

### Math_QED

Hello all. I have a question about a reflexive relation.

Consider $1_V : V \rightarrow V$ with $V$ a vector space. Obviously, this is an isomorphism. My book uses this example to show that V is isomorphic with V (reflexive relationship). However, suppose I have a function $f: V\rightarrow V$ and this function is not $1_V$ but the function is an isomorphism too. Does that prove the reflexivity too? I do understand that using the identityis the easiest way, though.

2. Oct 12, 2016

### andrewkirk

Isomorphism of two vector spaces is an equivalence relation, which means it is reflexive, as well as transitive and symmetric. So yes, if a vector space has an isomorphism to itself (which is called an automorphism) that is not the identity, that isomorphism is a reflexive relation.

An example of a non-identity isomorphism of a vector space is the map from $\mathbb R^2$ to itself that switches coordinates, mapping (x,y) to (y,x).

3. Oct 12, 2016

### Staff: Mentor

Isomorphisms define an equivalence relation, so especially they are reflexive.

4. Oct 12, 2016

### Math_QED

Define? My book shows that it must be an equivalence relation by showing that

1) $1_V$ is an isomorphism
2) $f^{-1}$ is an isomorphism
3) $f: V \rightarrow W, g: W \rightarrow X$ isomorphisms, then $g \circ f: V \rightarrow X$ isomorphism

5. Oct 12, 2016

### Math_QED

Yes but to show that it is an equivalence relation, I have to show that it is transitive, reflexive and symmetric first, right?

6. Oct 12, 2016

### andrewkirk

Yes. Have you tried doing that? Which of the three have you managed?

EDIT: Actually, on reflection, we need to be a bit careful about what the equivalence relation is on. Isomorphism is an equivalence relation on vector spaces so, writing 'is isomorphic to' by $\cong$ we have

1. $V\cong V$ (reflexivity)
2. $U\cong V\Rightarrow V\cong U$ (symmetry)
3. $U\cong V\wedge V\cong W\Rightarrow U\cong W$ (transitivity).

This all becomes a bit trivial when we are talking about maps from a vector space to itself.

7. Oct 12, 2016

### Math_QED

All of them. But my question is:

Is it nessecary to use $1_V$ for reflexivity, or can I use an other automorphism to show the reflexitivity "V is isomorphic with V" for all vector spaces V?

Using the identity, is obviously the easiest way.

8. Oct 12, 2016

### Staff: Mentor

"Define", because "is isomorphic" establishes an equivalence relation, but you could define other relations. So in the end, the question is, how the relation is defined. E.g. the relation between two vector spaces doesn't have to be a linear map.
Yes.

9. Oct 12, 2016

### andrewkirk

Yes, you can use any automorphism, not just the identity. However, for some vector spaces there may be no automorphisms other than the identity. So it's safest to use the identity because it's the one automorphism that we know every vector space, of any kind whatsoever, must have.

10. Oct 12, 2016

### Math_QED

Okay. I get it now. Thank you both.

11. Oct 12, 2016

### Stephen Tashi

That's an interesting question. When we say two vectors spaces are isomorphic, we mean (by definition) that there exists an at least one isomorphism between them. This does not rule out the possibility that there could be more than one isomorphism between them.

The property of "reflexivity" can be interpreted as a property of the relation "V1 is isomorphic to V2" or it could be interpreted (differently) as a property of a particular isomorphism - i.e. " f is an isomorphism between V1 and V2" can be viewed as the relation "V1 is isomorphic to V2 with respect to the mapping f" ( - similar to attaching "mod K" to the relation "=" in the congruence of integers A=B (mod K) ).

There is a minor technical distinction between the proofs of reflexive property for those two interpretations of "is isomorphic to".

If you want to prove the relation "V1 is isomorphic to V2" is a reflexive, the argument needs to begin with the step like: "By (that) definition of isomorphic there exists at least one mapping g that is an isomorphism between V1 and V2".

If you want to prove the relation "V1 is isomorphic to V2 with respect to the isomorphism f" is reflexive, by (that) definition of isomorphic, the isomorphism "f" is handed to you.

12. Oct 13, 2016

### Math_QED

My book says that V is isomorphic to W if $f: V \rightarrow W$ is an isomorphism.

13. Oct 13, 2016

### Stephen Tashi

Is that a direct quote of the definition? Or does the books say "V is isomorphic to W if and only there exists an isomorphism $f:\rightarrow W$" ?

If it doesn't mention "there exists" then what "$f$" is it talking about? - one that was mentioned in a preceeding sentence ?

14. Oct 13, 2016

### Math_QED

No, the definition from my book says:

A bijective linear function is an isomorphism. If there exists an isomorphism between 2 vector spaces V and W, we say that these two vector spaces are isomorphic.

15. Oct 13, 2016

### Staff: Mentor

This defines isomorphisms in general.
And this defines a special relation between vector spaces. Namely the relation given / established / defined by the existence of some isomorphism.

So $R(V,W) \Longleftrightarrow \exists \; f\, : \, V \rightarrow W \text{ such that } f \text{ is an isomorphism }$.

$R$ is also an equivalence relation (reflexive, symmetric, transitive) and $f = id_V$ an example (that surely always exists, regardless of the nature of $V$) to prove reflexivity.

If a certain $f_R$ had been part of the definition of $R$, then things would have been more complicated. E.g. $f_R \, : \, V \rightarrow W$ with $V \neq W$ would have removed all of the three properties from $R$. In the case of $V = W$, then $R$ would have been a relation among the elements of (this single) $V$, which could have any (or none, or all) of the properties. In this case $f_R$ even doesn't need to be a function.

This explains, that the essential part is the definition of $R$ and the quantifier at $f$ in it.
Also important is the domain of the relation $R$. In your case it is a relation on the set of vector spaces.
A relation on a single vector space (a certain set $V$) is something else.

16. Oct 13, 2016

### zinq

A rather different interpretation of the original question can be given in light of the fact that any subset of V x V defines a relation on the set underlying V.

Any mapping

f: V → V​

gives rise to its graph:

G(f): {(x, y) ∈ V×V | y = f(x)},​

which in turn defines a relation ~

given by

x ~ y ⇔ (x, y) ∈ G(f),​

which is therefore equivalent to saying y = f(x).

For which f is this relation reflexive? Left as an exercise.