Is this theorem true? Relationship between linear functionals and inner products

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Is this "theorem" true? Relationship between linear functionals and inner products

Suppose we have a finite dimensional inner product space V over the field F. We can define a map from V to F associated with every vector v as follows:
[tex]\underline{v}:V\rightarrow \mathbb{F}, \ w \mapsto \langle w,v\rangle[/tex]
Clearly this is a linear functional.

My question is whether all linear functionals from V to F are of this form. That is, is it true that for every f in V*, there exists a unique v such that f = v?

I have a felling that it is, but I can't prove it.
 

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  • #2
pwsnafu
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It is true for any Hilbert space (including infinite dimensional). The important thing is completeness. This is called the Riesz representation theorem.
 
  • #3
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It is quite easy to see for finite dimensional spaces. If [itex]\{e_1,...,e_n\}[/itex] are a basis for V, then we can define

[tex]\varepsilon_i:V\rightarrow \mathbb{F}:v\rightarrow <e_i,v>[/tex]

The [itex]\varepsilon_i[/itex] are easily seen to be linearly independent. Indeed, if [itex]\alpha_i[/itex] are such that

[tex]\sum_{i=1}^n \alpha_i\varepsilon_i=0[/tex]

then for all v in V holds that

[tex]0=\sum_{i=1}^n \alpha_i<e_i,v>=<\sum_i \alpha_ie_i,v>.[/tex]

Since this is true for all v, it is in particular true for [itex]\sum_i\alpha_ie_i[/itex]. And thus
[itex]\sum_i \alpha_ie_i=0[/itex]. Since [itex]\{e_1,...,e_n\}[/itex] is a basis, it follows that [itex]\alpha_1=...=\alpha_n=0[/itex]. Thus linear independence holds.

The [itex]\{\varepsilon_1,...,\varepsilon_n\}[/itex] also span [itex]V^*[/itex]. Indeed, if [itex]\varphi:V\rightarrow \mathbb{F}[/itex] is an arbitrary functional, then we define

[tex]\alpha_i=\varphi(e_i)[/tex]

For an arbitrary v holds that we can write [itex]v=\sum_i <e_i,v>e_i[/itex]. Thus

[tex]\varphi(v)=\varphi(\sum_i <e_i,v> e_i)=\sum_i\varphi(e_i) <e_i,v>[/tex]

Since this holds for all v, we have

[tex]\varphi=\sum_i \alpha_i \varepsilon_i[/tex]

So this proves the result for finite dimensional spaces. The result in infinite dimensions is false, since [itex]V^*[/itex] can really be huge.

However, if we restrict our attention to complete inner-product spaces and to continuous functionals, then the result is true. The proof is not as easy as the one I just gave though.

This Riesz representation theorem forms the justification for bra-ket notation (if you're familiar with that).
 
  • #4
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Thank you pwsnafu and micro mass.
 

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