# Abstract Nonsense

## Canonical Isomorphism Between a Finite Dimensional Inner Product Space and its Dual

Point of Post: In this post we prove that every finite dimensional inner product space is isomorphic to its dual space.

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Motivation

We have seen in the past the proof that every finite dimensional vector space is isomorphic to its double dual. We know of course since dimension is preserved under taking duals for finite dimensional vector spaces (this is, in fact, a characterization of finite dimensionality) but there was no canonical (free of basis choice) way of defining the mapping. In this post we prove the scene is different if the vector space is supplied with an inner product (or more generally a non-degenerate bilinear form).

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The Identification

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We recall that a bilinear form $B:\mathscr{V}\times\mathscr{V}\to F$ for some finite dimensional $F$-space $\mathscr{V}$ is one such that $B(-,v)$ and $B(u,-)$ are never the zero map unless $u$ or $v$ are themselves $\bold{0}$. With this in mind we prove that:

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Theorem: Let $\mathscr{V}$ be a finite dimensional $F$-space and $B$ a distinguished non-degenerate bilinear form on $\mathscr{V}\times\mathscr{V}$. Then, the map

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$\phi:\mathscr{V}\to\text{Hom}\left(\mathscr{V},F\right):u\mapsto B(u,-)$

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,where $B(u,-)(v)=B(u,v)$, is an isomorphism.

Proof: Clearly this map is linear since for every $v\in\mathscr{V}$ and $\alpha,\beta\in F$ one has that

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$\phi(\alpha u_1+\beta u_2)(v)=B(\alpha u_1+\beta u_,v)=\alpha B(u_1,v)+\beta B(u_2,v)=\left(\alpha\phi(u_1)+\beta\phi(u_2)\right)(v)$

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and so $\phi(\alpha u_1+\beta u_2)=\alpha\phi(u_1)+\beta\phi(u_2)$. Moreover, by definition of non-degeneracy we have that $\ker\phi=\{\bold{0}\}$ and thus by a common theorem regarding linear transformations and so $\phi$ is injective. Thus, we have that $\text{im}(\phi)\leqslant \text{Hom}\left(\mathscr{V},F\right)$ and so (recalling our previous comment about finite dimensional vector spaces being isomorphic to their duals) $\dim_F\text{im}(\phi)=\dim_F\mathscr{V}=\text{dim}_F\text{Hom}\left(\mathscr{V},F\right)$ from where it follows that $\text{im}(\phi)=\text{Hom}\left(\mathscr{V},F\right)$. The conclusion follows. $\blacksquare$

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Remark: It’s clear that $v\mapsto B(-,v)$ is also an isomorphism.

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Where this most comes up is the following:

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Corollary: Let $\mathscr{V}$ be a finite dimensional inner product space over $F$ (where $F=\mathbb{R},\mathbb{C}$) then every linear functional $f\in\text{Hom}\left(\mathscr{V},F\right)$ is of the form $f(v)=\langle u,v\rangle$ for some fixed $u\in\mathscr{V}$, and moreover this $u$ is unique.

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References:

1. Roman, Steven. Advanced Linear Algebra. New York: Springer-Verlag, 1992. Print.

June 4, 2011 -

## 3 Comments »

1. […] often comes up (this is actually taken as the usual multidimensional mean value theorem, but if we recall the canonical isomorphism between an inner product space and its dual we realize that these two […]

Pingback by The Mean Value Theorem for Multivariable Maps « Abstract Nonsense | June 11, 2011 | Reply

2. The inverse to this isomorphism seems confusing (using an inner-product for simplicity/definiteness) : if we’re given v* in V* , it seems we would first have to find the specific w that “represents” v* , i.e., the w such that:

v* (z) = to find the inverse. Or is there a simpler way?

Comment by Archibald. | June 20, 2012 | Reply

• No, you are absolutely correct. The tricky part of this isomorphism is that we use one of the most powerful (non-constructive!) theorems in finite dimensional linear algebra: that if $V\leqslant W$ and $\dim V=\dim W$ then $V=W$. Using this we circumvented having to prove surjectivity via “this element of $V^\ast$ is hit by…..” Of course, as you pointed out, this makes the construction of the inverse non-trivial.

Best,
Alex

Comment by Alex Youcis | October 4, 2012 | Reply