# Question on a proof.

1. May 26, 2010

### 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!

Last edited: May 27, 2010
2. May 27, 2010

### 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.

Last edited: May 27, 2010
3. May 27, 2010

### 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.

4. May 27, 2010

### 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.