# Is this dual pairing the same as the inner product?

If $(V, \langle \cdot$ , $\cdot\rangle)$ is an inner product space with dual $V^*$ then there is a natural dual pairing $\langle \cdot$ , $\cdot \rangle ^*: V^* \times V \rightarrow \mathbb K$ given by $\langle f,v \rangle ^*= f(v)$.

Also any inner product space can be put in a dual pairing with itself via the inner product on it.

What I want to know is if $V$ is its own dual (that is $V^* = V$) then is this natural pairing the same as the inner product on $V$? That is for all $u, v \in V$ is $\langle u, v \rangle = \langle u, v \rangle ^*$

Thanks!

• I think it depends on the isomorphism from $V$ to $V^*$ Commented Jun 9, 2015 at 13:48
• Could you expand on what you mean by the same? Commented Jun 9, 2015 at 13:51
• well if $u, v \in V$ then $\langle u,v \rangle$ gives the value of the inner product of $u$ and $v$. Now if i think of $u$ as a functional then is this value equal to $u(v)$?
– R_D
Commented Jun 9, 2015 at 13:59
• hmm, Im having a hard time unwinding the question. Maybe I would ask it as, given the natural pairing $\langle f, v\rangle^*$ does there exist an inner product on $v$ s.t. every $f$ in the dual space can be represented using this inner product. Then maybe you end up with something like en.wikipedia.org/wiki/Riesz_representation_theorem Commented Jun 9, 2015 at 14:12
• Ok. I'm not sure yet how to make it any clearer. Thank you for your comment though.
– R_D
Commented Jun 9, 2015 at 14:33

An inner product is an isomorphism $V \overset{\sim}\rightarrow V^*,\ v \mapsto v^*$. Or in bracket notation it's $v \mapsto \langle \cdot,v\rangle$.

For this question to make sense, you need to be able to compare the inner product and the dual pairing. To make that comparison you need such an isomorphism, which by definition is a map $v \mapsto v^*$ that makes $\langle u,v\rangle = \langle u,v^* \rangle$ hold.

It follows that this question makes sense if and only if the answer is trivially yes. However, this doesn't always happen, such as when there's no $v \mapsto v^*$.

Let $V$ be a vector space. You have than nondegenerated pairing $<,>:V^*\times V\rightarrow\mathbb{R}.$ $$<f,v>=f(v)$$ Additionaly let $(,):V\times V\rightarrow\mathbb{R}$ be inner product in $V.$ And it gives you injective map $$\psi:V\rightarrow V^*;\psi(w)(v)=(w,v).$$ Hence for every $v,w\in V$ you get that $$<\psi(w),v>=\psi(w)(v)=(w,v).$$ You assume that $V$ is $V^*.$ Supose the word "is" means that $\psi$ is actually an isomorphism. Hence fixing $f\in V^*$ and $v\in V$ you get that $f=\psi(w)$ for some $w.$ And so $$<f,v>=f(v)=\psi(w)(v)=(w,v)=(\psi^{-1}(f),v).$$

Take, for example, $\mathbb{R}^2=X$ with the standard scalar product. His dual is isomorphic to it, but there isn't an unique isomorphism.

For example, call $e_1,e_2$ the canonical base of $X$, and call $e_1^*,e_2^*$ the elements of $X^*$ such that $$e_i^*(e_i)=1\qquad e_i^*(e_{3-i})=0$$ If you take the isomorphism $e_i\to e_{3-i}^*$, then $$e_i^*(e_i)=1\ne 0=<e_{3-i},e_i>$$

• Isn't isomorphic different from equal? I asked for the case when $V = V^*$. So I thought the isomorphism is just the identity map. Did I miss something?
– R_D
Commented Jun 9, 2015 at 14:07
• The problem here is what you call "identity map". You are seeing the elements of $V$ as functions $V\to \mathbb R$. That means you are associating the elements to functions, but the problem is that there are a lot of ways you can do such a thing. Maybe your question is "there exists an association that gives me the properties of the inner product?" Commented Jun 9, 2015 at 14:45

Inner products on vector spaces is usually defined to be sesquilinear. Let $$\mathcal{V}$$ be a finite dimensional inner product space on a field $$\mathbb{F}$$, for any vectors $$\mathbf{a},\mathbf{b}\in\mathcal{V}$$ and scalar $$\alpha\in\mathbb{F}$$ the inner product is defined so as, besides other axioms, $$\left(\mathbf{a}|\mathbf{b}\right)=\overline{\left(\mathbf{b}|\mathbf{a}\right)}$$ and $$\left(\mathbf{a}|\alpha\mathbf{b}\right)=\overline{\alpha}\left(\mathbf{a}|\mathbf{b}\right)$$, where the overline denotes complex conjugate.

On the other hand, the natural paring of a vector $$\mathbf{v}\in\mathcal{V}$$ and a linear functional $$w\in\mathcal{V}^*$$ is defined as a linear map $$\left\langle\cdot,\cdot\right\rangle:\mathcal{V}\times\mathcal{V}^*\rightarrow\mathbb{F},\left\langle\mathbf{v},w\right\rangle\equiv w\left(\mathbf{v}\right)$$.

Given a basis $$\left\{\mathbf{e}_i\right\}$$ of $$\mathcal{V}$$ and its dual basis $$\left\{e_i\right\}$$ in $$\mathcal{V}^*$$, the inner product of two vector $$\left(\mathbf{v}|\mathbf{w}\right)=\sum_{i,j}v_i\overline{w_j}G_{ij}$$, where $$G_{ij}\equiv\left(\mathbf{e}_i|\mathbf{e}_j\right)$$ are the Gram matrices.

By saying "$$\mathcal{V}$$ is its own dual" maybe you mean a natural embedding from $$\mathcal{V}$$ to $$\mathcal{V}^{**}$$, where the double dual space $$\mathcal{V}^{**}$$ is the set of all linear functionals on $$\mathcal{V}^*$$, such that each linear functionals in $$\mathcal{V}^{**}$$ is defined corresponding to a vector in $$\mathcal{V}$$ as $$\tilde{\mathbf{v}}:\mathcal{V}^*\rightarrow\mathbb{F},\tilde{\mathbf{v}}\left(f\right)\equiv f\left(\mathbf{v}\right)$$. The tilde for distinguishing a vector in $$\mathcal{V}$$ and its image of natural embedding in $$\mathcal{V}^{**}$$ is often omitted in other texts. Particularly, $$\left\{\tilde{\mathbf{e}}_i\right\}$$, the natural embedded images of the basis of $$\mathcal{V}$$, form a basis of $$\mathcal{V}^{**}$$ that are the dual basis of $$\left\{e_i\right\}$$. Now a natural paring of interest would be $$\left\langle\mathbf{v},w\right\rangle$$ or $$\left\langle v,\tilde{\mathbf{w}}\right\rangle$$, neither of which, under the bases given above, equals to $$\left(\mathbf{v}|\mathbf{w}\right)$$ in general.

We can define linear functionals on $$\mathcal{V}$$ as, corresponding to a vector $$\mathbf{w}$$ in $$\mathcal{V}$$, $$f_{\mathbf{w}}:\mathcal{V}\rightarrow\mathbb{F}, f_\mathbf{w}\left(\mathbf{v}\right)\equiv\left(\mathbf{v}|\mathbf{w}\right),\forall \mathbf{v}\in\mathcal{V}$$. It is easy to prove that for each $$\mathcal{w}\in\mathcal{V}$$ there is uniquely one $$f_\mathbf{w}\in\mathcal{V}^*$$. Now as we wish, the natural paring $$\left\langle\mathbf{v},f_\mathbf{w}\right\rangle\equiv\left(\mathbf{v}|\mathbf{w}\right)$$. But this case does not look any similar to "$$\mathcal{V}$$ is its own dual".