Dual spaces-Existence of linear functional

[neomasterc]

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 T^t(f)=cf.
T^t denotes T transpose.

Homework Equations

T^t(f)=f°T. E_v(f)=f(v).
Let V*=Hom_F(V,F)
V*=(V*)*
E:V→V** i.e. E(v)=E_v
Theorem: E is an isomorphism iff V is finite dimensional

The Attempt at a Solution

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

 

[micromass]

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??

 

[neomasterc]

f is just a linear transformation in V*=Hom_F(V,F), so you cannot have f(v)=v, f:V→F.

 

[micromass]

f(v)=1 I meant.

 

[neomasterc]

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 won't be a linear transformation.

 

[micromass]

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 won't 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??