Relationship between transpose/symmetric/hermitian/unitary operators and their respective matrices. I am studying linear algebra by myself, using Serge Lang's "Linear Algebra", and I'm having some difficuties undersending the relationship between transpose, symmetric, Hermitian and unitary operators and their matrices. I have a few questions but, seeing as they all seem very closely related, I decided to post them together. Here they are:
1) Let $V$ be an n-dimensional vector space over $\mathbb{R}$, let $\langle\cdot,\cdot\rangle$ be a scalar product defined on $V$, and let $A:V\rightarrow V$ be a linear operator$^{[1]}$. Moreover, let $\mathcal{B} = \{v_1,v_2,\dots,v_n\}$ be a base of $V$, and $[L]^{\mathcal{B}}_{\mathcal{B}}$ be the matrix representing a linear application $L$ with respect to $\mathcal{B}$.
What conditions (non-degenerate, definite positive, etc) are needed on $\langle\cdot,\cdot\rangle$ and/or on $\mathcal{B}$, to make sure that, for each possible choice of $A$, the following hold:


*

*$[^tA]^{\mathcal{B}}_{\mathcal{B}}=([A]^{\mathcal{B}}_{\mathcal{B}})^T$, where $^tA$ is the only operator such that $\forall v,w \in V, \langle A(v),w\rangle = \langle v,^tA(w)\rangle$

*$^tA=A \iff [^tA]^{\mathcal{B}}_{\mathcal{B}}=([A]^{\mathcal{B}}_{\mathcal{B}})^T$

*$A$ is unitary (that is, $\langle A(v),A(w)\rangle = \langle v,w\rangle$) $\iff [A]^{\mathcal{B}}_{\mathcal{B}}*([A]^{\mathcal{B}}_{\mathcal{B}})^T=I$


2) What would change if, instead of on $\mathbb{R}$, we defined $V$ on any other field $\mathbb{K}$?
3)Let $V$ now be defined on $\mathbb{C}$ and $\langle\cdot,\cdot\rangle$ be an Hermitian product. Same as before, what conditions (non-degenerate, definite positive, etc) are needed on $\langle\cdot,\cdot\rangle$, or possibly on $\mathcal{B}$, to make sure that, for each possible choice of $A$, the following hold:


*

*$[A^*]^{\mathcal{B}}_{\mathcal{B}}=\overline{([A]^{\mathcal{B}}_{\mathcal{B}})}^T$, where $A^*$ is the only operator such that $\forall v,w \in V, \langle A(v),w\rangle = \langle v,A^*(w)\rangle$

*$A^*=A \iff [^tA]^{\mathcal{B}}_{\mathcal{B}}=\overline{([A]^{\mathcal{B}}_{\mathcal{B}})}^T$

*$A$ is complex unitary (that is, $\langle A(v),A(w)\rangle = \langle v,w\rangle$) $\iff [A]^{\mathcal{B}}_{\mathcal{B}}\ \overline{([A]^{\mathcal{B}}_{\mathcal{B}})}^T=I$


4) What would change if $V$ were not finite dimensional?
I admittedly found some similar questions here on stackexchange, but most of them seem to limit themselves to the standard dot product on $\mathbb{R^n}$ or the standard Hermitian product on $\mathbb{C^n}$, whereas I'm interested in the generic version. Working by myself I found that most of these seem to hold if the matrix $M$ representing the scalar or Hermitian product with respect to $\mathcal{B}$ is either $I$ or $-I$ (which, in a way, makes them isomorphic to the standard dot and Hermitian products), and the implications from matrix equalities to operator equalities also seem to hold if $M$ is something akin to:
$$\left(\begin{matrix}
\pm I & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 0\\
\end{matrix}\right).$$
but I'm really not sure if there are other cases which work.
I'd be grateful even for a partial answer, seeing as my post is quite long.
$^{[1]}$: The definition of operator I've learned up until now is simply "linear funcion from a space into itself", so there might be something I'm missing
 A: *

*The necessary and sufficient condition is that the basis you use is orthonormal up to scale, that is, $\langle v_i, v_j \rangle = c \delta_{ij}$ for some $c \neq 0$. (Note that we need nondegeneracy in order for adjoints to be unique: for example adjoints clearly can't be unique if the scalar product is zero. This orthonormality condition then implies that the scalar product is symmetric.)

*Nothing except that "positive definite" no longer makes sense.

*The same conditions, but $c$ must be (positive and) real. 

*In this case the interpretation of transposes is more complicated. Things don't behave much like the finite-dimensional case unless $V$ is assumed to be a Hilbert space. 
Here is the relevant computation. Write 
$$A v_i = \sum_j A_{ji} v_j$$
so that
$$\langle A v_i, v_j \rangle = \sum_k A_{ki} \langle v_k, v_j \rangle.$$
Similarly, writing $B$ for the adjoint of $A$ with respect to the scalar product (this is to make referring to the components of $B$ slightly less confusing), we have
$$B v_i = \sum_j B_{ji} v_j$$
so that
$$\langle v_i, B v_j \rangle = \sum_k B_{kj} \langle v_i, v_k \rangle.$$
We want the matrix of $B$ to be the transpose of the matrix of $A$, meaning that $B_{kj} = A_{jk}$. This means that we want
$$\sum_k A_{ki} \langle v_k, v_j \rangle = \sum_k A_{jk} \langle v_i, v_k \rangle$$
for all $i, j$ and all $A$. We can treat the entries of $A$ as variables. The two sides of this linear equation have no variables in common except $A_{ji}$, and equating their coefficients gives
$$\langle v_j, v_j \rangle = \langle v_i, v_i \rangle.$$
All of the other coefficients vanish, hence $\langle v_i, v_j \rangle = 0$ for $i \neq j$. Now nondegeneracy / the uniqueness of adjoints requires that the common value of $\langle v_i, v_i \rangle = c$ not vanish. 
