Prove: y=ξz if [x,y]=0 & [x,z]=0

[Matthollyw00d]

Q: Prove that if y and z are linear functionals (on the same vector space) such that [x,y]=0 whenever [x,z]=0, then there exists a scalar ξ such that y=ξz.
(Hint: if [x₀,z]≠0, write ξ=[x₀,y]/[x₀,z].)

I'm fairly certain there's an obvious proof using the dual basis, but this is in the section before that, so I'm trying to do it without that, and can't seemed to get the proper result. Any help would be great, thanks!

 

[eok20]

What does [x,y] mean? Is x a vector and [x,y] the pairing between x and the linear functional y?

If so, here's a hint: Let n be the dimension of the vector space. The dimensions of the kernels of linear functionals are either n (if it is the 0 functional) or n-1. This follows from the rank nullity theorem. Assuming that the functionals are non-zero (if one is zero the problem is trivial), this means that they have the same kernels. If [x_0,z]\ne 0 then anything in the vector space can be written as x_0 + k where k is in the kernel. Now use the hint.

 

[Matthollyw00d]

[x,y]=y(x)

Your hint would be adequate if we can assume finite dimensionality, but I don't think we can in this problem.

 

[eok20]

I think you can still modify the argument to show it works in the infinite dimensional case: you can still show that everything in V can be written as a multiple of x_0 + something in the kernel of y.

 

[blue_raver22]

First, let's define the bracket notation [x,y] as the inner product between two vectors x and y. This means that [x,y]= x^T y, where x^T is the transpose of x.

Now, let's start with the given information that [x,y]=0 whenever [x,z]=0. This means that for any vector x, if [x,z]=0, then [x,y]=0.

We want to prove that there exists a scalar ξ such that y=ξz. To do this, we will use the hint provided and write ξ=[x0,y]/[x0,z] where [x0,z]≠0.

Since [x0,z]≠0, we can choose a vector x0 such that [x0,z]=1. This means that [x0,y]=0, since [x0,y]=0 whenever [x0,z]=0.

Now, let's consider a general vector x. We can write x as a linear combination of x0 and z, i.e. x=αx0+βz for some scalars α and β.

Using the given information, we know that [x,z]=0, so we have [x,y]= [αx0+βz, y]= α[x0,y]+ β[z,y]= α*0+ β[z,y]= β[z,y].

But we also know that [x,y]= x^T y, so we have x^T y= β[z,y]. Rearranging this, we get y= (x^T y)/β*z.

Now, using the definition of ξ=[x0,y]/[x0,z], we can rewrite this as y= (x^T y)/β*z= (x^T y)/[x0,y]*[x0,z]*z= (x^T y)/[x0,y]*[x0,z]*z= (x^T y)/[x0,y]*z= ξz, where ξ=[x0,y]/[x0,z].

Therefore, we have shown that for any vector x, y=ξz, which means that y and z are linearly related by a scalar ξ. This proves the statement that if y and z are linear functionals such that [x,y]=0 whenever [x,z]=0,