I am stuck with some exercise and hope that someone can give me a hint:
Let $V$ and $W$ be vector spaces and let $B_V = \big\{ v_1, \ldots, v_n \big\}$ and $B_w = \big\{ w_1, \ldots, w_n \big\}$ be bases of $V$ and of $W$.
Let furthermore be $\big\{ \alpha^1, \ldots, \alpha^n \big\}$ be the dual basis of $B_V$ and let $\big\{ \beta^1, \ldots, \beta^n \big\}$ be the dual basis of $B_W$.
Let $\varphi : V \longrightarrow W$ be a linear map and let $A^i{}_j$ be its matrix representation with respect to the bases $B_V$ and $B_W$. Show that the matrix representation of the adjoint map $\varphi^{\ast} : W^{\ast} \longrightarrow V^{\ast}$ is the transpose of $(A^i{}_j)^T$.
I cannot quite close the proof for this. My attempt is as follows:
1. Let $\varphi(X) = Y$, then
$$ \beta^i(Y) = \beta^i(Y^j w_j ) = Y^j \beta^i(w_j) = Y^j \delta^i_j = Y^i $$
and therefore
$$Y^i = \beta^i(Y) = \beta^i \big( \varphi(X) \big) = X^j \beta^i \big( \varphi(v_j) \big)$$
Consequently, the matrix representation of $\varphi$ in said bases is
$$ A^i{}_j = \beta^i \big( \varphi( v_j ) \big)$$
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2. Now, doing an analogous calculation yields the following: Let $\Psi = \varphi^{\ast}(\Omega)$, then
$$ \Psi(v_i) = \big( \Psi_j \alpha^j \big)(v_i) = \Psi_j \alpha^j(v_i) = \Psi_j \delta^j_i = \Psi_i$$
and therefore
$$\Psi_i = \Psi(v_i) = \varphi^{\ast}(\Omega)(v_i) = \varphi^{\ast} \big( \Omega_j \beta^j \big) (v_i) = \Omega_j \, \varphi^{\ast}(\beta^j)(v_i) $$
and consequently the matrix representation $C^i{}_j$ of $\varphi^{\ast}$ with respect to the considered dual bases is
$$C^j{}_i = \varphi^{\ast}(\beta^j)(v_i)$$.
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3. The adjoint map is defined as
$$\varphi^{\ast}(\Omega)(X) = \Omega \big( \varphi(X) \big) \quad , \qquad \text{for some} \enspace \Omega \in W^{\ast} , X \in V$$
and so
$$C^j{}_i = \varphi^{\ast}(\beta^j)(v_i) = \beta^j \big( \varphi(v_i) \big) = A^j{}_i$$
What am I missing here? The transposed matrix of $A^i{}_j$ is supposed to be $A_j{}^i$. This is however not my result here. Can anyone help me find and correct my error?
I think you've just made a mistake in part 2. I'm not used to the notation you're using, but I'll try to keep it the same (so if something is off in my notation let me know). I'm going to use lower case letters for the matrix entries though.
In your part 1 you get $a^i_j=\beta^i(\varphi(v_j))=\varphi^*(\beta^i)(v_j)$, which is correct. A way to think about this is the following. The map $\varphi$ sends $v_j$ to $\sum_l a^l_j w_l$, i.e. $a^i_j$ is the $i$-th coordinate of $\varphi(v_j)$ with respect to the $B_W$ basis. But, by definition of dual basis, $\beta^i$ picks out the $i$-th coordinate of a vector in $W$ with respect to the $B_W$ basis. So $$\beta^i(\varphi(v_j))=\text{$i$-th coordinate of $\varphi(v_j)$}=a^i_j.$$
Now let's apply this approach to $\varphi^*$. If $C$ is the matrix, then $\varphi^*$ sends $\beta^i$ to $\sum_l c^l_i \alpha^l$, i.e. $c^j_i$ is the $j$-th coordinate of $\varphi^*(\beta^i)$ with respect to the dual basis $B_V^*$. That is, $$\varphi^*(\beta^i)=\sum_l c^l_i\alpha^l.$$
Let's evaluate both sides of the above at $v_j$: $$(LHS):\quad \varphi^*(\beta^i)(v_j)=a^i_j$$ $$(RHS):\quad \sum_l c^l_i\alpha^l(v_j) = \sum_l c^l_i \delta^l_j = c^j_i.$$
Comparing the above we get $a^i_j=c^j_i$, as desired.