# Finite Dimensional Inner-Product Space Equals its Dual?

Finite Dimensional Inner-Product Space Equals its Dual!?

Let V be a finite dimensional inner-product space. Then V is 'essentially' equal to its dual space V'.

By the Reisz Representation theorem, V is isomorphic to V'. However, I've been told that V=V', which I am having a hard time believing. It seems to me that the two spaces do not contain the same elements: V' contains linear functionals, while V contains any kind of vectors. Therefore, V does not equal V'.

Could someone clear this up for me? Is V only isomorphic to V', or are the two spaces REALLY equal?

Thanks!

## Answers and Replies

Fredrik
Staff Emeritus
Gold Member

They are isomorphic, not equal. Hence the phrase "essentially equal".

You only need the Riesz representation theorem when V is infinite-dimensional, since you can easily see that any two n-dimensional vector spaces (with the same n) are isomorphic.

The isomorphism between V and V** is more interesting, because it can be defined without an inner product or a choice of bases for V and V**. The function f:V→V** defined by f(x)(ω)=ω(x) for all ω in V* and all x in V is such an isomorphism.

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