The Minkowski distance is defined as: $$\sum_{i=1}^n (|x_i-y_i|^p)^\frac{1}{p} $$
Also, the norm of a vector can be defined as the $ ||x|| = \sqrt{x^T\bullet g\bullet x} \text{, g is the metric matrix}$ . If $ p = 2 $, this means it is the euclidean space and for that reason the $ g_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$.
In physics, when you are dealing with the Minkowski space the metric matrix could be defined as $$ g_2 = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{pmatrix}. $$
For the Euclidean space, I know that it holds the properties of Positivity, Symmetry and Triangle Inequality.
For that reason, I'm wondering, if there is a way, knowing which matrix is the matrix metric, if it is possible to prove the properties that will be valid.
I might said that the prof of the Symmetry could be made by showing that $g^T = g^{-1}$, which is the case for $g_1$ and $g_2$. In the case of Positivity, I could prove showing that $det(g) = 1$, which is not the case for $g_2$. Could you tell me if this is the correct approach? Also, about the Triangle Inequality. How can I prove, knowing the matrix, if the property will be valid for the given space?
Note: I have no formal background in mathematics, as I'm a student in the third period of engineering. So, please don't judge too hard my prof approach.
It is correct that symmetry is equivalent to the matrix $g$ being symmetric.
Positivity is equivalent to $g$ being positive definite, which is equivalent to $g$'s eigenvalues being all positive. Notice that if $g$ is symmetric, the Spectral Theorem guarantees that $g$'s eigenvalues are real. Clearly, your $g_2$ above fails positivity, since it has an eigenvalue $-1$ $($with multiplicity $3)$.
Moreover, observe that this does not have much to do with the $\det(g)$. Since the determinant is the product of the eigenvalues, the most we can say is that if $\det(g)\leq0$ then positivity fails. In particular, we may have $\det(g)>0$ without $g$ being positive definite $($such as when $g=-I$ on a space of even dimension$)$, and $g$ may be positive definite with $\det(g)\neq 1$ $($such as when $g=\lambda I$ for some positive $\lambda \neq 1)$.
Finally, if $g$ is symmetric and positive definite, then the bilinear form $x^Tgx$ defines an inner product, and hence $\lVert x \rVert = \sqrt{x^Tgx}$ is indeed a norm. In other words, we get the triangle inequality automatically from symmetry and positive-definiteness.