# linear algebra basic proof

F is a linear functional in $V'$ a linear vector space which operates on $\phi\in V$. Show that there is a one-to one correspondence between F and $f\in V$ such that

$F(\phi)=(f,\phi)$

where $V$ is also a linear vector space. $(,)$ represents inner product. Any hint will be helpful.

Attempt: I don't even know if this is right but...

If $F(\phi)=c$ , $c\in \mathbb{C}$ then let $(f_1, \phi)=c$ and $(f_2, \phi)=c$ $\rightarrow$ $(f_1-f_2, \phi)=0$

But this only means orthogonality.

EDIT: so, from hint

let $\{e_i\}$ be a linearly independent basis for vectorspace $V$

$F(\Phi) = F (\sum_n \phi_n e_n) =\sum_n \phi_n F(e_n)$

$(f,\Phi) = \sum_n f_n^* \phi_n$

Therefore if $F(\phi)=c$ , $c\in \mathbb{C}$ and $(f,\Phi)=c$ then

$\sum_n (f_n^* -F(e_n)) \phi_n =0$

implies (?) $f_n^* =F(e_n)$ $\forall n$?

EDIT Ok. So the above implies that if some $(e_i,\Phi)$ vanishes then we cannot have a unique $f$. So is this solved by specifying at the beginning that $\{e_i\}$ is a basis where $\phi_i=(e_i,\Phi)$ is non vanishing for all i?

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If by one-to-one you mean bijection, then you probably need to suppose finite dimension. In that case, this is standard textbook material. – lhf Mar 2 '11 at 0:33
yes, indeed i wish to ignore convergence questions of infinite dimenstions. infact, i am resisting looking it up on the book on my table where i encountered this as I dont even have an idea how to begin. (book: quantum mechanics by l ballentine) – Please Delete Account Mar 2 '11 at 0:37

Make sure it's clear what you want to prove: For each linear functional $F$, there is a vector $f$ such that $F(\phi) = (f,\phi)$.
You have computed $F(\phi)$ after expressing $\phi$ in a basis. Which vector $f$ do you need to take so that $F(\phi) = (f,\phi)$? Will choosing a special kind of basis help?
edited my answer. If i choose a basis such that $\phi_i$ are all non zero, i think there would be a unique f in V for every F in V'. – Please Delete Account Mar 2 '11 at 1:24