(n-1)-dimensional subspace is the null space of a linear functional

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Given that N is an (n-1)-dimensional subspace of an n-dimensional vector space V, show that N is the null space of a linear functional.

My thoughts:

suppose [tex]\alpha_i[/tex]([tex]1\leq i \leq n-1[/tex]) is the basis of N, the linear functional in question has to satisfy f([tex]\alpha_i[/tex])=0.

Am I correct?

Thanks
 

Answers and Replies

  • #2
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Yes, that is already correct!
 
  • #3
HallsofIvy
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Given a basis for any vector space, any linear functional depends only upon its "action" on those basis vectors- that, is given f(v1)= x1, f(v2)= x2, ..., f(vn)= xn where the v's are the basis vectors and the a's are scalars, then any vector v can be written as v= a1v1+ a2v2+ ...+ anvn for some scalars, a1, a2, ..., an. Since f is linear, f(v)= a1f(v1)+ a2f(v2)+ ...+ anf(vn)=a1x1+ a2x2+ ... anxn.

Given any subspace U of vector space V, select a basis {u1, u2,...,um} for U and extend it to a basis for V. Define f(u1)= f(u2)= ...= f(um)= 0 and define f to be whatever nonzero values you like for the other basis vectors. Then the nullspace of f is exactly U.
 

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