What does it mean for a vector space to be self-dual? The Euclidean space is sometimes said to be "self-dual". I was wondering what this term actually means, because I couldn't find definitions of it when this "self-duality" is mentioned.
I find it a bit confusing because the elements of the dual vector space are supposed to be linear functionals, so I don't quite see how a vector space can be "the same" as its dual. For example, in the Euclidean space $\mathbb{R}^n$ with the inner product $\langle x, y \rangle = x^T y$, does it make sense to refer to $x^T$ as an element of the dual space? In what way is the Euclidean space "self-dual"? Is every inner product space self-dual in this way?
I'd appreciate any elaboration on these ideas!
 A: "Self-dual" is not commonly used as a term with a precise definition in this context.  When someone says that a vector space $V$ is self-dual, that normally means (at a minimum) that there exists an isomorphism $V\to V^*$ from $V$ to its dual space.  Depending on context, it may also mean that a specific such isomorphism has been chosen, or that there is a specific canonical such isomorphism which can be defined in terms of some extra structure that $V$ has.
So in the first, weakest sense, where you just say there exists an isomorphism, every finite-dimensional vector space is self-dual.  Given a vector space $V$ with a bilinear form $\langle \cdot,\cdot\rangle:V\times V\to\mathbb{R}$, there is a canonical map $f:V\to V^*$ which takes $v\in V$ to the functional $w\mapsto\langle v,w\rangle$.  If the bilinear form is nondegenerate, then $f$ is injective.  If $V$ is additionally finite-dimensional, then $f$ is automatically surjective as well.  So any finite-dimensional vector space with a nondegenerate bilinear form (e.g., an inner product) is self-dual in the stronger sense of having a canonical isomorphism to its dual determined by its extra structure.
(In fact, conversely, an isomorphism $V\to V^*$ determines a nondegenerate bilinear form by reversing the construction above, so fixing such an isomorphism on a finite-dimensional vector space is equivalent to choosing a nondegenerate bilinear form.)
