Proof regarding linear functionals

[Adgorn]

Homework Statement

Let V be a vector space over R. let Φ₁, Φ₂ ∈ V* (the duel space) and suppose σ:V→R, defined by σ(v)=Φ₁(v)Φ₂(v), also belongs to V*. Show that either Φ₁ = 0 or Φ₂ = 0.

Homework Equations

N/A

The Attempt at a Solution

Since σ is also an element of the duel space, it is linear, so σ(v+u)=σ(v)+σ(u). Translating both sides into terms of Φ₁ and Φ[SUB}2[/SUB], I came up with the equation
(1) Φ₁(v)Φ₂(u)+Φ₁(u)Φ₂(v)=0.

Doing the same with σ(av+bu) and plugging in equation (1) produced the equation
(2) a²Φ₁(v)Φ₂(v)+b²Φ₁(u)Φ₂(u)=aΦ₁(v)Φ₂(v)+bΦ₁(u)Φ₂

This is where I got stuck, perhaps there is a trick I am not seeing to this equation or perhaps I approached this the wrong way. Any help would be appreciated.

 

[PeroK]

Homework Statement

Let V be a vector space over R. let Φ₁, Φ₂ ∈ V* (the duel space) and suppose σ:V→R, defined by σ(v)=Φ₁(v)Φ₂(v), also belongs to V*. Show that either Φ₁ = 0 or Φ₂ = 0.

Homework Equations

N/A

The Attempt at a Solution

Since σ is also an element of the duel space, it is linear, so σ(v+u)=σ(v)+σ(u). Translating both sides into terms of Φ₁ and Φ[SUB}2[/SUB], I came up with the equation
(1) Φ₁(v)Φ₂(u)+Φ₁(u)Φ₂(v)=0.

Doing the same with σ(av+bu) and plugging in equation (1) produced the equation
(2) a²Φ₁(v)Φ₂(v)+b²Φ₁(u)Φ₂(u)=aΦ₁(v)Φ₂(v)+bΦ₁(u)Φ₂

This is where I got stuck, perhaps there is a trick I am not seeing to this equation or perhaps I approached this the wrong way. Any help would be appreciated.

You're on the right track. Go back to equation (1) and try $u = v$.

 

[Adgorn]

You're on the right track. Go back to equation (1) and try $u = v$.

I see, thank you for the help, I solved it.

 

[PeroK]

I see, thank you for the help, I solved it.

Are you sure?

 

[Adgorn]

Are you sure?

I think so. Plugging in u=v into Φ₁(v)Φ₂(u)+Φ₁(u)Φ₂(v)=0 resulted in Φ₁(v)Φ₂(v)+Φ₁(v)Φ₂(v)=0, meaning Φ₁(v)Φ₂(v)=0. Since these are elements of a field, it means either Φ₁(v)=0 or Φ₂(v)=0, and since v is arbitrary, this means either Φ₁=0 or Φ₂=0.

 

[PeroK]

I think so. Plugging in u=v into Φ₁(v)Φ₂(u)+Φ₁(u)Φ₂(v)=0 resulted in Φ₁(v)Φ₂(v)+Φ₁(v)Φ₂(v)=0, meaning Φ₁(v)Φ₂(v)=0. Since these are elements of a field, it means either Φ₁(v)=0 or Φ₂(v)=0, and since v is arbitrary, this means either Φ₁=0 or Φ₂=0.

I guessed that is what you had done! What you have shown there is:

$\forall \ v: \ \Phi_1(v) = 0$ or $\Phi_2(v) = 0$ (Equation 3)

That means that for some $v$ it could be $\Phi_1(v) = 0$ and for some other $v$ it could be $\Phi_2(v) = 0$. But, you haven't shown that either $\Phi_1 = 0$ or $\Phi_2 = 0$.

So, there is still work to do. Let me help with the next step. Suppose $\Phi_1 \ne 0$. We have to show that then $\Phi_2 = 0$.

Can you use equations (1) and (3) to do this?

 

[Adgorn]

I see. So let us assume Φ₁≠0, Then it means there exists w∈V such that Φ₁(w)≠0, let us denote the set of all such vectors with S. since for every vector v∈V Φ₁(v) is either 0 or not 0, V=S∪Ker(Φ₁).We plug one of these (arbitrary) w into equation 3 and get Φ₁(w)Φ₂(w)=0, since Φ₁(w)≠0, it means Φ₂(w)=0. And since this applies to every w∈S, it means Φ₂(S)=0.

Now we must show that Φ₂(u)=0 ∀u∈Ker(Φ₁). Let us take an arbitrary vector u from Ker(Φ₁) and plug it in equation 1 along with another vector w∈S: Φ₁(u)Φ₂(w)+Φ₁(w)Φ₂(u)=0. Since u∈Ker(Φ₁), Φ₁(u)Φ₂(w)=0. Thus, Φ₁(w)Φ₂(u)=0, and since Φ₁(w)≠0 by definition, this means Φ₂(u)=0 ∀u∈Ker(Φ₁).

Combining these two results give us Φ₂(v)=0 ∀v∈V, and so Φ₂=0. Doing the same process by assuming Φ₂≠0 produces the result Φ₁=0.

 

[PeroK]

I see. So let us assume Φ₁≠0, Then it means there exists w∈V such that Φ₁(w)≠0, let us denote the set of all such vectors with S. since for every vector v∈V Φ₁(v) is either 0 or not 0, V=S∪Ker(Φ₁).We plug one of these (arbitrary) w into equation 3 and get Φ₁(w)Φ₂(w)=0, since Φ₁(w)≠0, it means Φ₂(w)=0. And since this applies to every w∈S, it means Φ₂(S)=0.

Now we must show that Φ₂(u)=0 ∀u∈Ker(Φ₁). Let us take an arbitrary vector u from Ker(Φ₁) and plug it in equation 1 along with another vector w∈S: Φ₁(u)Φ₂(w)+Φ₁(w)Φ₂(u)=0. Since u∈Ker(Φ₁), Φ₁(u)Φ₂(w)=0. Thus, Φ₁(w)Φ₂(u)=0, and since Φ₁(w)≠0 by definition, this means Φ₂(u)=0 ∀u∈Ker(Φ₁).

Combining these two results give us Φ₂(v)=0 ∀v∈V, and so Φ₂=0. Doing the same process by assuming Φ₂≠0 produces the result Φ₁=0.

I can't see any mistakes there, although your solution is more complicated than I was expecting. If I pick up your proof from here:

I see. So let us assume Φ₁≠0, Then it means there exists w∈V such that Φ₁(w)≠0

Then, I would use equation 3 to note that $\Phi_2(w) = 0$ and hence (from equation 1) we have:

$\forall v \in V: \ \Phi_1(w)\Phi_2(v) = 0$

Hence $\forall v \in V: \ \Phi_2(v) = 0$

And the result follows.

Note also that there is no need to repeat the process for $\Phi_2$. If $\Phi_1 = 0$, then we are done. And, if $\Phi_1 \ne 0$ we have shown that $\Phi_2 = 0$. In either case, one of the functionals must be $0$. And, there is no need to do any more.

 

[zwierz]

Let V be a vector space over R. let Φ1, Φ2 ∈ V* (the duel space) and suppose σ:V→R, defined by σ(v)=Φ1(v)Φ2(v), also belongs to V*. Show that either Φ1 = 0 or Φ2 = 0.

Assume that both functions $\Phi_1,\Phi_2$ are not equal to zero identically. Then there exist a vector $v$ such that $\Phi_1(v)\Phi_2(v)\ne 0$.
So that $\sigma(\lambda v)=\lambda^2\Phi_1(v)\Phi_2(v)\Longrightarrow \sigma( v)=\lambda\Phi_1(v)\Phi_2(v),\quad \forall\lambda\ne 0$
This is possible if only $\Phi_1(v)\Phi_2(v)=0$ Contradiction.