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Let $V$ and $W$ be finite-dimensional vector spaces over some arbitrary field $K$, and let $V^\ast$ and $W^\ast$ be their respective dual spaces. Let $f:V \rightarrow W$ be a linear map.

a) Define the dual of $f$ as the map $f^\ast : W^\ast \rightarrow V^\ast$, $e \mapsto e \circ f$.

b) Suppose we define two fixed non-degenerate bilinear forms $\langle \cdot,\cdot \rangle_V : V^2 \rightarrow K$, $\langle \cdot,\cdot \rangle_W : W^2 \rightarrow K$. Define the adjoint of $f$ as the map $\bar{f}: W \rightarrow V$ satisfying $\langle v,\bar{f}(w) \rangle_V = \langle f(v),w \rangle_W$.

Since the bilinear forms are non-degenerate, the linear maps $\phi_V : V \rightarrow V^\ast$ and $\phi_W : W \rightarrow W^\ast$ given by $$ \phi_V(v)(v_0) = \langle v,v_0 \rangle_V, \qquad \phi_W(w)(w_0) = \langle w,w_0 \rangle_W $$ are isomorphisms. Given this, how can we express $\bar{f}$ explicitly in terms of $f^\ast$ (and if required, $\phi_V$ and $\phi_W$)? Alternatively, how can we prove that the map $\bar{f}$ is guaranteed to exist?

EDIT: As pointed out by @levap I have fixed the definition of $\phi_W$.

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For some reason, your definition of the map $\phi_W$ is mangled (the $w$ is missing). Let me fix a convention. Given a vector space $U$, and a bilinear form $\left< \cdot, \cdot \right>_U$ on $U$, define the map $\phi^{R}_U \colon U \rightarrow U^{*}$ by the formula

$$ \phi^{R}_U(u_0)(u) = \left< u, u_0 \right>_U $$

(the letter $R$ is there to designate that we use $u_0$ as the "rightmost" argument in $\left< \cdot, \cdot \right>_U$). Another meaningful choice would be to define

$$ \phi^{L}_U(u_0)(u) = \left< u_0, u \right>_U $$

(so we use $u_0$ in the left slot instead of the right slot). Unless the bilinear form is symmetric, the maps will be different.

Your definition of the adjoint is the map $\overline{f}$ which satisfies

$$ \left< v, \bar{f}(w) \right>_{V} = \left< f(v), w \right>_W. $$

To express this adjoint, it is better to work with the maps $\phi^R$ and then you have $\overline{f} = \left( \phi_V^R \right)^{-1} \circ f^{*} \circ \phi_W^R$. To see why, note that

$$ \phi_{V}^R(v_0) = \varphi \iff \left< v, v_0 \right>_V = \varphi(v) \,\,\, \forall v \in V $$

where $v_0 \in V, \varphi \in V^{*}$ (one direction is clear from the definition and the other follows because $\varphi_V$ is an isomorphism). Now, fix $v \in V, w \in W$ and then using the above we have

$$ \left< v, \left( \phi_{V}^R \right) ^{-1}(f^{*}(\phi_W^R(w))) \right>_V = f^{*}(\phi_W^R(w))(v) = \phi_W^R(w)(f(v)) = \left< f(v), w \right>_W.$$

You could have defined an adjoint of $f$ as the map which satisfies

$$ \left< \bar{f}(w), v \right>_V = \left< w, f(v) \right>_W $$

instead and in general, this would give you a different adjoint which is given by $\left( \phi_V^{L} \right)^{-1} \circ f^{*} \circ \phi_W^L$.

In fact, there are two other reasonable definitions of the adjoint in this general context and each of them can be expressed in terms of $\phi_V^{L},\phi_V^{R},\phi_W^L,\phi_W^{R},f^{*}$. When both $\left< \cdot, \cdot \right>_{V}, \left< \cdot, \cdot \right>_{W}$ are symmetric, all the adjoints coincide.

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  • $\begingroup$ It looks like you are using $\phi_V(v)(v') = \langle v',v\rangle_V$ instead of $\phi_V(v)(v') = \langle v, v'\rangle_V$ as given by OP. $\endgroup$ – Christoph Feb 20 '18 at 13:48
  • $\begingroup$ @Christoph: Arg, you are right. The problem is that the definition of $\phi_V, \phi_W$ of the OP is not consistent (the roles of $v_0,v$ and $w_0,w$ are switched which doesn't make any sense as $V,W$ are arbitrary vector spaces so why choose to use one convention for one and another convention for the other). I'll edit my answer. $\endgroup$ – levap Feb 20 '18 at 13:51
  • $\begingroup$ I think it is consistent and just missing a "$(w)$" in "$\phi_W(w)(w_0)$", no? $\endgroup$ – Christoph Feb 20 '18 at 13:59
  • $\begingroup$ @Christoph: Well, it depends whether the OP meant to write $\phi_W(w)(w_0)$ or $\phi_W(w_0)(w)$. In any case, I've edited the answer to treat both conventions. $\endgroup$ – levap Feb 20 '18 at 14:21
  • $\begingroup$ @Christoph Indeed, I meant $\phi_W(w)(w_0)$. Thanks for pointing it out; I have corrected the definition. $\endgroup$ – EpsilonDelta Feb 21 '18 at 4:53
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Here's an approach using matrices.

By a choice of basis identify $V=K^n$ and $W=K^m$, elements are column vectors. We can identify their duals $V^*$ and $W^*$ with spaces of row vectors so that the usual matrix product serves as application of linear forms to vectors. In this setting, when $f\colon V\to W$ is given by an $m\times n$-matrix $A$ such that $f(v) = Av$, its dual $f^*\colon W^*\to V^*$ is given by $f^*(\beta) = \beta A$ since then $$ (f^*(\beta))(v) = \beta A v = \beta(f(v)). $$ Your bilinear forms are given by square matrices, namely $$ \langle v, v'\rangle_V = v^t X v' \qquad\text{and}\qquad \langle w, w'\rangle_W = w^t Y w'. $$ for invertible matrices $X_{ij}=(\langle e_i, e_j\rangle_V)_{i,j=1,\dots,n}$ and $Y_{ij}=(\langle e_i, e_j\rangle_W)_{i,j=1,\dots,m}$.

Your isomorphisms $\phi_V \colon V\to V^*$ and $\phi_W\colon W\to W^*$ are given by $\phi_V(v)(v') = v^t X v'$ and $\phi_W(w)(w') = w^t Y w'$.

Now let's find the $n\times m$-matrix $B$ describing the adjoint $\bar f\colon W\to V$ as $\bar f(w) = Bw$. We want to have $$ \langle v, \bar f(w) \rangle_V = \langle f(v), w\rangle_W, $$ which can be written as $$ v^t X B w = (Av)^t Y w = v^t A^t Y w. $$ So we have $XB=A^t Y$ and hence $B = X^{-1} A^t Y$.

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