I'm having a bit of trouble finding a source of study to metric matrices (I'm not even sure whether this is how we call them in English). It is introduced along with vector inner products. I can't find anything on them in Portuguese, and I've had no luck with English either.
I have the following exercise:
Consider an oriented euclidean vector space V and its direct (what is direct?) base $B=(v_1,v_2,v_3)$, whose vectors are such that:
$||v_1|| = 2√2$,
$||v_2||=2$,
$||v_3||=1$
$∡(v_1,v_2) = π/2; ∡(v_1,v_3) = π/4; ∡(v_2,v_3) = π/3$
I have no idea what to do with the angles. Should I use the inner product formula?
Then, the exercise is: Prove the metric matrix of V in respect to the basis B is
$$\pmatrix{8 &0 &2\\0 &4& 1\\ 2& 1 &1}$$
It would be of great help if someone could tell me where to find more info on this. I'm not sure the correct terminology is "metric matrices", but that's the only translation I can come up with from Portuguese.
Thanks in advance! 
$\endgroup$3 Answers
$\begingroup$I'm just going to use a dot for the inner product. The matrix you want is symmetric positive definite, let us call it $M.$ The rule is $$ m_{ij} = v_i \cdot v_j $$ This is often called the Gram matrix. However, sometimes half of this is called the Gram matrix, so you need some context usually.
Probably worth this note: often enough, the Gram matrix is taken as half the Hessian matrix of a given quadratic form. So, given the familiar form $x^2 + y^2 + z^2,$ the Hessian matrix of second partial derivatives is $2I.$ Many people would take the Gram matrix as $I,$ many others would use $2I.$ Caution is required. Oh: if you are given the Hessian matrix $H,$ the form (metric) applied to a column vector $x$ is $(1/2) x^T H x.$
Details: note
$$ v_i \cdot v_i = \|v_i \|^2 $$
The Law of Cosines says
$$ v_i \cdot v_j = \|v_i \| \|v_j \| \cos \theta. $$
Compare a different basis, for a root lattice called $D_3,$ at which specifies the Gram matrix. On this website, the basis vectors are the rows under BASIS, and the GRAM matrix is as I said, not divided by 2 at all.
Meanwhile, if someone specifies a basis for you as rows of a matrix, call that $R,$ from Nebe $$ R = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{array} \right) $$ we get Gram matrix $G = R R^T,$ in $D_3$ $$ G = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 0 \\ 1 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 1 \\ 0 & 2 & -1 \\ 1 & -1 & 2 \end{array} \right) $$
$\endgroup$ 1 $\begingroup$Given the base $B = (v_1,v_2,v_3)$, the "metric matrix" of $V$ with respect to $B$ is $$ \newcommand{\<}{\langle} \newcommand{\>}{\rangle} M = \pmatrix{\<v_1,v_1\> & \<v_1,v_2\> & \< v_1,v_3 \>\\ \<v_2,v_1\> & \<v_2,v_2\> & \< v_2,v_3 \>\\ \<v_3,v_1\> & \<v_3,v_2\> & \< v_3,v_3 \>} $$ You can figure out what these entries are using the inner product formula.
$\endgroup$ $\begingroup$Metric Matrix is also called a Metric Tensor. Unlike the variable change theorem which requires differentiability. Metric Tensor change could be made on any integrable function via tangent vectors. On WikiPedia:
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