I was reading about Pontryagin's duality and came accross the following quote
a finite-dimensional vector space $V$ and its dual vector space $V^*$ are not naturally isomorphic, but the endomorphism algebra (matrix algebra) of one is isomorphic to the opposite of the endomorphism algebra of the other: $\text{End}(V)\cong \text{End}(V^{*})^\text{op}$, via the transpose.
which confused me, so I have two questions:
$V^*$ is of the same dimension as $V$, hence they are isomorphic. What does it mean that they are not naturally isomorphic? I am not an expert in category theory, where clicking on the link of natural led me to, so I'd prefer a bit less abstract explanation - perhaps with some basic examples. Also, why this naturality is of any importance, how can it be more useful than some other isomorphisms?
It is said that their endomorpshism algebras are also isomorphic. Why is "but" there? Would
the endomorphism algebra (matrix algebra) of one is naturally isomorphic to the opposite of the endomorphism algebra of the other
a better way to convey the idea there?
I will try to answer the first question, at least. The point is that the isomorphism isn't "natural" because you have to make a choice to fully specify the isomorphism. See the section "Bilinear products and dual spaces" at Wikipedia.
Long story short, when $V$ is finite-dimensional, and you have a bilinear form $\langle \cdot, \cdot \rangle$ on $V$, then the mapping $v \mapsto \langle v , \cdot \rangle$, where the right side is a "functional" on $V$ (a map that takes a vector and gives a scalar, i.e. an element of the dual $V^*$), is an isomorphism between $V$ and $V^*$.
But notice that you had to choose $v$! That's why and $V$ and $V^*$ are not "naturally" isomorphic: there is no one most natural isomorphism.