Is a bra the adjoint of a ket?

The instructor in my quantum computation course sometimes uses the equivalence
I understand that this is true for the typical matrix implementation of Dirac’s bra-ket notation, and taking $x^\dagger\equiv\bar{x}^T$; but does it follow from the definition of an inner product space and the adjoint? Should I be able to derive this as a fact from definition of an inner product space and the adjoint of linear operators on that space or is it simply a notational convention?

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$\newcommand{\ip}[1]{\left\langle{#1}\right\rangle}$Actually, you can view a bra vector as the honest-to-goodness adjoint of the corresponding ket vector, but it requires a trick.

So, let $V$ be a finite-dimensional complex inner product space. The trick is that you have a canonical isomorphism $\Phi : V \to L(\mathbb{C},V)$ given by $\Phi(v) : \lambda \mapsto \lambda v$ for all $v \in V$; indeed, one has that $\Phi^{-1}(s) = s(1)$. Now, since $\mathbb{C} = \mathbb{C}^1$ is also an inner product space, for any $v \in V$ we can form the adjoint $\Phi(v)^\ast \in L(V,\mathbb{C}) = V^\ast$ of $\Phi(v)$, and lo and behold, for any $w \in V$,
\Phi(v)^\ast(w) = \ip{1,\Phi(v)^\ast(w)}_{\mathbb{C}} = \ip{\Phi(v)(1),w}_V = \ip{v,w}_V,
so that $\Phi(v)^\ast : w \mapsto \ip{v,w}_V$, as required. Thus, up to application of a canonical isomorphism, a bra vector really is the adjoint of the corresponding ket vector.

$ADDENDUM: In more physics-friendly notation, here’s what’s going on. Let $H$ be your (finite-dimensional) Hilbert space, and let $\ket{a} \in H$. You can interpret $\ket{a}$, in a completely natural way, as defining a linear transformation $\Phi[\ket{a}] : \mathbb{C} \to H$ by $\Phi[\ket{a}](\lambda) := \lambda \ket{a}$. The Hermitian conjugate of $\Phi[\ket{a}]$, then, is a linear transformation $\Phi[\ket{a}]^\dagger : H \to \mathbb{C}$, so that $\Phi[\ket{a}]^\dagger$ is simply a bra vector. The computation above then shows that $\Phi[\ket{a}]^\dagger = \bra{a}$. Thus, as long as you’re fine with identifying $\ket{a}$ with $\Phi[\ket{a}]$ (which is actually completely rigorous), you do indeed have that $\bra{a} = \ket{a}^\dagger$.

If $L:V\to W$ is linear then the adjoint of $L$ is a linear map from the dual of $W$ to the dual of $V$. So I would argue that linear maps have adjoints, but vectors do not. I do not think either bra or ket is a linear map.

The correspondence that sends $|a\rangle$ to $\langle a|$ is an isomorphism
from a vector space into its dual space. In projective geometry this might be called a correlation (but I have not seen the term used outside this context).
Any inner product on a vector space $V$ determines an isomorphism for $V$ to its dual.

It is important to bear in mind that physicists have their own way of thinking about linear algebra, and this can be quite different from what you will meet in mathematics courses. (For example, one physics text writes: “a vector in $\mathbb{R}^3$ is not just a triple of numbers, it has a meaning”.)