Dual spaces-Existence of linear functional

  • Thread starter neomasterc
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  • #1

Homework Statement


Let V be a finite dimensional vector field over F. Let T:V→V
Let c be a scalar and suppose there is v in V such that T(v)=cv, then show there exists a non-zero linear functional f on V such that Tt(f)=cf.
Tt denotes T transpose.

Homework Equations


Tt(f)=f°T. Ev(f)=f(v).
Let V*=HomF(V,F)
V*=(V*)*
E:V→V** i.e. E(v)=Ev
Theorem: E is an isomorphism iff V is finite dimensional


The Attempt at a Solution


Ev(Tt(f))=f°T(v)=cf(v).
But now I need to show there is f such that Tt(f)=cf, not just for specific v that satisfies T(v)=cv.
 

Answers and Replies

  • #2
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How would you define f?? Obviously you will want to have f(v)=v. How would you define it for other vectors?? Maybe set f(w)=0 for some other vectors??
 
  • #3
f is just a linear transformation in V*=HomF(V,F), so you cannot have f(v)=v, f:V→F.
 
  • #4
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f(v)=1 I meant.
 
  • #5
We cannot define f(kv)=1 and f(w)=0 for w≠v, since if a is a multiple of v and b is a multiple of v, a+b is a multiple of v, so 1=f(a+b)≠f(a)+f(b)=2. So f wont be a linear transformation.
 
  • #6
22,089
3,296
We cannot define f(kv)=1 and f(w)=0 for w≠v, since if a is a multiple of v and b is a multiple of v, a+b is a multiple of v, so 1=f(a+b)≠f(a)+f(b)=2. So f wont be a linear transformation.
Indeed we can't. Can you solve that?? Don't put ALL the vectors equal to 0, but only some. Do you see it??
 

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