Existence of isomorphism ϕ:V→V s.t. ϕ(ϕ(v))=−v for all v∈V

[CGandC]

Problem: Let $V$ be a vector space over $\mathbb{F}$ and suppose its dimension is even, $dimV=2k$. Show there exists an isomorphism $\phi:V→V$ s.t. $\phi(\phi(v))=−v$ for all $v \in V$

Generally that way to solve this is to define a basis for the vector space $V$ and on that basis to define the isomorphism. However I thought about 2 ways to solve the problem, similar at-first sight but different in logic and wondered if both could be accepted as a proof to the problem:

Solution 1:
Since $V$ is a vector space, then there exists a basis $\{ v_1,...,v_{2k} \}$ of $V$. Define $\phi:V→V$ as follows,
$\begin{array}{l} \phi(v_{2i-1})= v_{2i}\\ \phi(v_{2i})= -v_{2i-1} \end{array} \qquad i=1,\ldots, k.$
We then get $\phi(\phi(v_{2i-1}))=\phi(v_{2i})=-v_{2i-1}$ and $\phi(\phi(v_{2i}))=\phi(-v_{2i-1})=-\phi(v_{2i-1})= -v_{2i}$ . Therefore $\phi \circ \phi= -id_V$

Solution 2:
Let $\{ v_1,...,v_{2k} \}$ be an arbitrary basis of $V$. Define $\phi:V→V$ as follows,
$\begin{array}{l} \phi(v_{2i-1})= v_{2i}\\ \phi(v_{2i})= -v_{2i-1} \end{array} \qquad i=1,\ldots, k.$
We then get $\phi(\phi(v_{2i-1}))=\phi(v_{2i})=-v_{2i-1}$ and $\phi(\phi(v_{2i}))=\phi(-v_{2i-1})=-\phi(v_{2i-1})= -v_{2i}$ . Since $\{ v_1,...,v_{2k} \}$ was an arbitrary basis of $V$, therefore $\phi \circ \phi= -id_V$

Solution 1 follows the logic ( I didn't write it fully formally, but it's mostly for intuition ):
$\exists \text{basis of } V . \exists \phi:V→V . \forall v \in V ( \phi(\phi(v))=−v )$
Solution 2 follows the logic:
$\forall \text{basis of } V . \exists \phi:V→V . \forall v \in V ( \phi(\phi(v))=−v )$

So the proofs are not identical ( as seen by the logic ): 1st solution shows there exists a basis for the vector space and then I'm defining the isomorphism $\phi$ on that specific vector space and not on an arbitrary one. In 2st solution I'm proving the existence of $\phi$ for an arbitrary basis.

Question: I wanted to know which solution is the correct one and why? ( I couldn't really discern the correct solution from the question and it feels ambiguous as to which one I should choose )

 

[Office_Shredder]

The solutions are really the same, since your basis is arbitrary in solution 1, despite your ability to make it more confusing by introducing formal logical. And they are both correct.

 

[CGandC]

In solution 1 my basis is arbitrary since the vector space $V$ is arbitrary?

 

[mfb]

No, it's an arbitrary basis within the vector space V you look at. You do not set any requirements on it, so you automatically show that you can choose any basis - that's what proof 2 does explicitly.

 

[CGandC]

Is my original question analogous to the following problem ( In terms of what I asked )?:
It is given that there exists $x \in \mathbb{R}$ s.t. 2|x. Prove the statement P(x).
So suppose we proved P(x). Then logical format of the problem is: $\exists x \in \mathbb{R} ( 2|x ) . P(x)$
In the given of the problem, since we didn't make any other requirements on $x$ besides being real number and being divided by 2, Then problem is equivalent to : $\forall x \in \mathbb{R} ( 2|x ) . P(x)$

So is it true in this problem to say $\exists x \in \mathbb{R} ( 2|x ) . P(x)$ is equivalent to $\forall x \in \mathbb{R} ( 2|x ) . P(x)$ ?

 

[Office_Shredder]

I'm not really sure what P(x) is, but the things you wrote at the bottom are not true. For example, if P(x) is the statement that x=2, then there exists and x such that P(x), but it's not true that for all x, P(x). And this would be apparent in the proof, because you wouldn't select an arbitrary x at the start of it.

 

[CGandC]

You're right.
I don't yet fully understand why solutions 1 & 2 in my original question are the same. $\{ v_1,...,v_{2k} \}$ is not an arbitrary basis in solution 1, it is just a specific basis that exists because $V$ is a vector space ( there's a theorem that says if you have a vector space then there exists a basis for it , assume in the question that the vector space is finite) , but in solution 2 $\{ v_1,...,v_{2k} \}$ is arbitrary and not specific one.

 

[Office_Shredder]

How is it specific in solution 1? If I picked an arbitrary basis of V, what would make the proof not work?

Your are obviously using the same theorem about the existence of a basis in solution 2 - if a basis doesn't exist, then you can't actually construct an isomorphism.

 

[wrobel]

In physics there is a good operator $J$. It helps to write the Hamilton equations
$$\dot z=J\frac{\partial H}{\partial z},\quad J^2=-E$$

 

[Office_Shredder]

In physics there is a good operator $J$. It helps to write the Hamilton equations
$$\dot z=J\frac{\partial H}{\partial z},\quad J^2=-E$$

Did you post this in the wrong thread?

 

[wrobel]

Did you post this in the wrong thread?

No. I intentionally posted it here. And I explained why

 

[mfb]

I don't see the connection either.

 

[wrobel]

Direct connection. The operator $J$ provides an example of the isomorphism required in the initial post.
Moreover this operator prompts a way to produce such examples .

Indeed, consider a cross product of two isomorphic vector spaces $Z=X\times Y$ and let $A:X\to Y$ be an isomorphism. Define an operator $F:Z\to Z$ as follows $F(x,y)=(-A^{-1}y,Ax)$ then $F^2=-id_Z$

 

[mfb]

Indeed, consider a cross product of two isomorphic vector spaces $Z=X\times Y$ and let $A:X\to Y$ be an isomorphism. Define an operator $F:Z\to Z$ as follows $F(x,y)=(-A^{-1}y,Ax)$ then $F^2=-id_Z$

That's similar to what we discussed before. But I still don't see the connection to your other posts.

 

[wrobel]

That's similar to what we discussed before.

Indeed? My construction does not rely on finite dimension.

But I still don't see the connection to your other posts.

I'm desperate.
The operator J is a special case of my construction with A and F. Invite other mathematicians here. Perhaps they can explain better than me.

 

[Office_Shredder]

I think the point is no one else here knows what the Hamilton equations are, or how J is defined.

Anyway, CGandC, do you have any more questions?

 

[CGandC]

No, it's an arbitrary basis within the vector space V you look at. You do not set any requirements on it, so you automatically show that you can choose any basis - that's what proof 2 does explicitly.

I think I've got it. So in solution 1, if for example I would've imposed a requirement/constraint on the basis somewhere during the proof then solution 2 would be different ( since I'm not making a constraint on the basis in that solution ).
Also, because I know that there exists a basis for every vector space and I'm not setting any requirement on the basis during the proof of solution 1, then the proof would also work for every basis ( hence solution 2 is the same as solution 1 because in solution 2 I'm not making any requirement on the basis other than assuming it exists ).

I didn't really understand this at first because looking at these logical statements:
$\exists \text{basis of } V . \exists \phi:V→V . \forall v \in V ( \phi(\phi(v))=−v )$ (proof 1 )
$\forall \text{basis of } V . \exists \phi:V→V . \forall v \in V ( \phi(\phi(v))=−v )$ (proof 2 )
I was mislead by thinking that proof 1 wouldn't be true if there wasn't a basis, but proof 2 would be vacuously true even if there wasn't a basis , Hence these two proofs are different from each other.

But I erred since I know there must exist a basis for every vector space ( even for the null space ) so the situation in which a basis does not exist never occurs.

 

[Office_Shredder]

I agree the thing you proved in 2 would be vacuously true if there was no basis, but if there was no basis then you wouldn't have proven the original thing you wanted to prove, that there actually exists an isomorphism. So proof 2 is actually incomplete until you observe that there always exists a basis. Most people just know there is a basis so will fill that detail in in their head without asking for it.