Is any inner product given by... Is it true that any inner product $\langle \cdot, \cdot \rangle :  V \times V \to \mathbb{C}$ is given by $$\langle \boldsymbol{x}, \boldsymbol{y} \rangle = \boldsymbol{x}^T \boldsymbol{M} \boldsymbol{\bar{y}}$$ where $\boldsymbol{M}$ is a positive definite, hermitian matrix, $\boldsymbol{x}, \boldsymbol{y} \in V$?
Is it true that any inner product is also given by $$\langle \boldsymbol{x}, \boldsymbol{y} \rangle = \boldsymbol{y}^* \boldsymbol{M} \boldsymbol{x}$$ where $\boldsymbol{x}^*$ denotes the conjugate transpose of $\boldsymbol{x}$?
 A: The answer to the first question is yes.  
Let $V$ be a finite dimensional complex vector space, and let $\langle\cdot,\cdot\rangle:V\times V\rightarrow\Bbb{C}$ be an inner product.  Fix a basis $\{e_1,\ldots,e_n\}$, so that for any two vectors $u,v\in V$, we have 
$$
\langle u,v\rangle=\left\langle\sum_{j=1}^n\alpha_je_j,\sum_{k=1}^n\beta_ke_k\right\rangle=\sum_{j=1}^n\sum_{k=1}^n\alpha_j\bar{\beta}_j\langle e_j,e_k\rangle
$$
So if we define $M=(M_{ij})_{n\times n}=(\langle e_i,e_j\rangle)_{n\times n}$, we have exactly the expression 
$$
\langle u,v\rangle = u^\intercal M \bar{v}
$$
as desired.  It is easy to check that $M$ is Hermitian - this follows from conjugate symmetry of the inner product: $\overline{\langle x,y\rangle}=\langle y,x\rangle$. Positive definiteness follows from the positivity of the inner product - $\langle x,x\rangle>0$ for all $x\in V\backslash\{0\}$.
For the second claim, it suffices to show the identity $x^\intercal M \bar{y}=y^*\overline{M} x$.  This is straightforward: 
$$
x^\intercal M\bar{y}=\sum_j\alpha_j\sum_kM_{jk}\bar{\beta}_k=\sum_k\bar{\beta}_k\sum_jM_{jk}\alpha_j=y^*\overline{M}x
$$
Notice of course that we must then use the matrix $\overline{M}$, i.e. the same matrix won't work.  (Thanks to @user1551 for pointing this out)
A: It gets even better...imo, much better: if you have an orthonormal basis $\,\{u_1,...,u_n\}\,$ , then the inner product is given by the extremely simple formula
$$x=\sum_{k=1}^nx_iu_i\,\,,\,\,y=\sum_{k=1}^ny_iu_i\,\,\,,\,\,x_i,y_i\in\Bbb C\Longrightarrow\langle\,x\,,\,y\,\rangle=\sum_{i=1}^nx_i\overline y_i$$
