Explanation of the metric tensor Before anything, I just want to say that I'm studying mathematics/physics in a different language (Serbian) so some of the English terms that I use might be a bit wonky. Just ask if a term makes no sense.
So, this is the first time we're being introduced to the concept of a metric tensor, and they define it like this. It starts off by defining a basis $\{v_1,...,v_n\}$ in a vector space with a scalar product. Sounds good so far.
It also defines $m_{ij} = \{v_i,v_j\}$. I'm not quite sure what this is about already. Why is this thing defined as a scalar product of two arbitrary vectors from the basis, when there are $n$ vectors in the basis?
Then it defines $x$ and $y$ as $x = \sum_i \xi_i v_i$ and $y = \sum_i \eta_i v_i$. Sounds reasonable, it's just simply defining the vector as a linear combination of basis vectors.
Then it describes the scalar product of the two vectors in the following way. $(x,y) = \sum_{i,j} \xi_i^* m_{ij} \eta_{j}$. As far as I understand, this could also be written as $(x,y) = \sum_{i,j} \xi_i^* v_i \eta_{j} v_j$. This is the point at which I get really confused. To try to understand it, I rewrote things. Using the isomorphism $V_n(F) \equiv F^n$, I made the following conclusion: $(x,y) = \sum_{i,j} \xi_i^* v_i \eta_{j} v_j = \left(\array{\xi_1\\...\\\xi_n}\right) \cdot \left(\array{\eta_1\\...\\\eta_m}\right) = P$, where P is an arbitrary matrix that is the product of those two vectors portrayed using its coordinates and that theorem of isomorphism.
However, their conclusion is similar-ish but not the same. They say that "using representative columns x and y of vectors $x$ and $y$ in the given basis, and marking with $x^+$ the row given by transposing and conjugating of x, and using $M$ as the matrix whose elements are $m_{ij}$, we can write $(x,y) = \sum_{i,j} \xi_i^* m_{ij} \eta_{j}$ as $(x,y) = x^+ M y$. This is what truly gets me. When I try to figure this out myself, I just get a single matrix, but they get this equation, where they say that $M$ is the metric tensor or just the metric. And that $M$ fully defines the scalar product. However, I can't seem to figure out how to use this in a concrete example (for example, $V_3(F)$ in the standard orthonormal basis we use (1,0,0 0,1,0 0,0,1). I'm not even sure where to start, since I don't really get how they got $m_{ij}$ in the first place.
 A: What they mean is that $m_{ij}$ is the value of the scalar product of $v_i$ and $v_j$, so $\left<v_i,v_j\right> = m_{ij}$. They didn't "get" $m_{ij}$ from anywhere...it's just given to you, as the definition of the scalar product. Let's do a concrete example. Let's say for $\Bbb{R}^2$, with basis $e_1,e_2$. For example. let's say our scalar product is given by $\left<e_1,e_1\right> = 2$, $\left<e_2,e_2\right> = 3$, and $\left<e_1,e_2\right> = 5$. Then the matrix $M = (m_{ij})$ is
$$ M = \left( \begin {array}{cc} 2 & 5 \\ 5 & 3 \end{array} \right) $$
Now let's say we want to compute the scalar product of the vectors $(1,2) = e_1 + 2e_2$ and $(3,-4) = 3e_1 - 4e_2$. Since the scalar product is bilinear, we can compute as:
$$ \begin {align*}
\left<e_1+2e_2, \, 3e_1-4e_2\right> &= 3m_{11} -4m_{12} + 6m_{21} -8m_{22} \\
&= 3(2) + (6-4)(5) - 8(3) \\
&= -8
\end {align*}
$$
The claim is just that this computation could have also been done using matrix multiplication:
$$ \begin {align*}
\left<(1,2),(3,-4)\right> &= (1,2) \left(\begin{array}{cc} 2&5\\5&3 \end{array}\right) \left( \begin{array}{c} 3\\-4 \end{array}\right)
\end {align*}
$$
