Abstract Linear Algebra: Dual Basis

[Fringhe]

Homework Statement

Define a non-zero linear functional y on C^2 such that if x₁=(1,1,1) and x₂=(1,1,-1), then [x₁,y]=[x₂,y]=0.

Homework Equations

N/A

The Attempt at a Solution

Le X = {x1,x2,...,xn} be a basis in C³ whose first m elements are in M (and form a basis in M). Let X' be the dual basis in C³'. Let N be the subspace of V' spanned by y_m+1, ..., y_n.
Let's assume that y is any element in N.
1) y is in V'
2) y is a linear combination of the basis vectors y₁, ..., y_n
=> y = $$\Sigma$$_j=1ⁿ n_jy_j
Since by assumption y is in N we have for every i=1,...,m
[x_i,y] =0

P.S: I am new to the abstract linear algebra world.

 

[Dick]

Apparently you mean a linear functional on C^3. What is C? Is it the complex numbers? And I think you are taking this 'abstract' thing a little far. You just want a linear functional y such that y((1,1,1))=y((1,1,-1))=0. Try thinking of it as a 'not abstract' problem. You want to write down a concrete linear functional that maps x1 and x2 to zero.

 

[Fringhe]

Ok, so let x=($$\xi$$1,$$\xi$$2,$$\xi$$3) (where $$\xi$$1=$$\xi$$2) and let y be the functional such that y = $$\xi$$1+$$\xi$$2+0*$$\xi$$3
So for xi (i from 0 to n) y(xi) would equal 0.

 

[Dick]

Ok, so let x=($$\xi$$1,$$\xi$$2,$$\xi$$3) (where $$\xi$$1=$$\xi$$2) and let y be the functional such that y = $$\xi$$1+$$\xi$$2+0*$$\xi$$3
So for xi (i from 0 to n) y(xi) would equal 0.

i from 0 to n? There's only x1 and x2. How about if x=(a,b,c) define y(x)=1*a+(-1)*b+0*c? Then y(x1)=0 and y(x2)=0. That's really all you need. Your notation [x1,y] has got to mean y(x1), right?