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Abstract Linear Algebra: Dual Basis

  • Thread starter Fringhe
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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 ym+1, ..., yn.
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, ..., yn
=> y = [tex]\Sigma[/tex]j=1n njyj
Since by assumption y is in N we have for every i=1,...,m
[xi,y] =0

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

Answers and Replies

  • #2
Dick
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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.
 
  • #3
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Ok, so let x=([tex]\xi[/tex]1,[tex]\xi[/tex]2,[tex]\xi[/tex]3) (where [tex]\xi[/tex]1=[tex]\xi[/tex]2) and let y be the functional such that y = [tex]\xi[/tex]1+[tex]\xi[/tex]2+0*[tex]\xi[/tex]3
So for xi (i from 0 to n) y(xi) would equal 0.
 
Last edited:
  • #4
Dick
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Ok, so let x=([tex]\xi[/tex]1,[tex]\xi[/tex]2,[tex]\xi[/tex]3) (where [tex]\xi[/tex]1=[tex]\xi[/tex]2) and let y be the functional such that y = [tex]\xi[/tex]1+[tex]\xi[/tex]2+0*[tex]\xi[/tex]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?
 

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