Definition of adjoint of a linear map I am having a tough time understanding adjoint of a linear map.
Consider a linear map between two vector spaces $\, f:V\rightarrow W,$ let us denote $f^*$ to denote its adjoint.


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*Accroding to this video https://www.youtube.com/watch?v=SjCs_HyYtSo (around time 5:50) the author explains that adjoint of a linear map is a function from dual of $\,W$ (denoted by $\,W^*$) to the dual of $\,V$ (denoted by $\,V^*$). So this implies $\,f^*:W^*\rightarrow V^*.$

*On the other hand in the pdf http://math.mit.edu/~trasched/18.700.f10/lect17-article.pdf , the adjoint of the linear map is defined as another linear map from $\,W$ to $\,V.$ So this implies  $\,f^*:W\rightarrow V.$


Can some body clarify this discrepancy?
 A: If the vector spaces V and W have respective nondegenerate bilinear forms $B_V$ and $B_W$, a concept closely related to the transpose – the adjoint – may be defined:
If $f : V → W$ is a linear map between vector spaces $V$ and $W$, we define $g$ as the adjoint of f if $g : W → V$ satisfies :
$ B_V(υ,g(w))=B_W(f(υ),w)$ $ \forall υ \in V, w\in W $.
These bilinear forms define an isomorphism between $V $and $V^∗$, and between W and $W^∗$, resulting in an isomorphism between the transpose and adjoint of $f$. The matrix of the adjoint of a map is the transposed matrix only if the bases are orthonormal with respect to their bilinear forms. In this context, many authors use the term transpose to refer to the adjoint as defined here.
The adjoint allows us to consider whether $g : W → V$ is equal to $f^{−1} : W → V$. In particular, this allows the orthogonal group over a vector space $V$ with a quadratic form to be defined without reference to matrices (nor the components thereof) as the set of all linear maps $V → V$ for which the adjoint equals the inverse
A: The adjoint of a linear map $f: \Bbb V \to \Bbb W$ between two vector spaces is given by the definition in the first source: It is the map $f^* : \Bbb W^* \to \Bbb V^*$ defined by
$$(f^*(\phi))(v) := \phi(f(v))$$ for all $\phi \in \Bbb W^*$ and $v \in \Bbb V$.
For ease of exposition I'll henceforth restrict to the case that $\Bbb V$ and $\Bbb W$ are finite dimensional, though the notion of adjoint makes sense in the infinite-dimensional setting, too. In the second source, $\Bbb V$ and $\Bbb W$ are inner product spaces, that is, $\Bbb V$ and $\Bbb W$ come equipped with inner products, say, $\langle \,\cdot\, , \,\cdot\, \rangle$ and $\langle\!\langle \,\cdot\, , \,\cdot\, \rangle\!\rangle$, respectively. Now, an inner product $\langle \,\cdot\, , \,\cdot\, \rangle$ on a vector space $\Bbb U$ defines an isomorphism $\Phi : \Bbb U \stackrel{\cong}{\to} \Bbb U^*$ by
$$(\Phi(u))(u') := \langle u, u' \rangle .$$
Thus, for any linear map $f: \Bbb V \to \Bbb W$ we can identify $\Bbb W^*$ with $\Bbb W$ and $\Bbb V^*$ with $\Bbb V$, and hence $f^*$ with a map $\Bbb W \to \Bbb V$. Unwinding the definitions shows that this map satisfies the identity $$\langle\!\langle w, f(v) \rangle\!\rangle = \langle f^*(w), v \rangle$$ given in the second source.
It is an instructive exercise to write out all of these objects in terms of their matrix representations with respect to some bases of $\Bbb V, \Bbb W$. In particular, if $\Bbb V$ is a finite-dimensional real inner product space, one can show that, with respect to an orthogonal basis, the matrix representations $[f]$ and $[f^*]$ of a map $f: \Bbb V \to \Bbb V$ and its adjoint $f^*$, respectively, are related by the transpose operation: $[f^*] = [f]^{\top}$.
