In general, which is the difference between this tensors: $T_{ij}$ and $T^{ij}$? For the metric, $g_{ij}=\vec{e}_{i}\cdot \vec{e}_{j}$, why $g^{ij}$ is as simple as its inverse matrix?
So, given a general tensor $T_{ij}$ in its matricial form, how can I find the tensor $T^{ij}$ also in its matricial form?
An example would be the following. Given this metric, $g_{ij}(\mu,\nu)$ Metric ant this tensor $R_{ij}(\mu, \nu)$, the Ricci tensor, Ricci tensor how can I find the tensor $R^{ij}$ using matrix notation?
First of all, $T_{ij}$ and $T^{ij}$ are not tensors, strictly speaking. They are respectively the covariant coordinates and contravariant coordinates of the same second-order tensor $\boldsymbol{T}$ on the chosen basis $\lbrace\boldsymbol{e}_i\rbrace$ of the vector space (e.g. $\mathbb{R}^n$). Let us introduce $\lbrace\boldsymbol{e}^j\rbrace$, the dual basis of $\lbrace\boldsymbol{e}_i\rbrace$. On the one hand, using the Einstein summation convention [1] and the rule of raising and lowering indices [2], we write \begin{aligned} \boldsymbol{T} &= T_{i j} \left(\boldsymbol{e}^i \otimes \boldsymbol{e}^j\right) \\ &= T_{i j} \left( (g^{ik}\, \boldsymbol{e}_k) \otimes (g^{j\ell }\, \boldsymbol{e}_\ell )\right) , \end{aligned} where $g^{ij}$ are the contravariant coordinates of the metric tensor. On the other hand, we write $\boldsymbol{T} = T^{k \ell} \left(\boldsymbol{e}_k \otimes \boldsymbol{e}_\ell\right)$. By unicity of the contravariant coordinates of $\boldsymbol{T}$, we obtain the relationship $$ T^{k \ell} = g^{ik}\, g^{j\ell}\, T_{i j} \, . $$ A similar question was asked here.
Since $v_i = g_{ij}\, v^j$ and $v^j = g^{ji}\, v_i$ for every vector $\boldsymbol{v}$, one has the matrix relation $[g_{ij}] = [g^{ij}]^{-1}$.