Raising and Lowering Indices using the Metric Tensor Given the next tensor:
$X^{\mu \nu}= \left(\begin{array}{cccc} 
2 & 0 & 1 & -1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \\
\end{array}\right)$ and the metric tensor $\eta_{\mu\nu} =\left(\begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}\right)$
I need to calculate the following tensors: $X^\mu$$_\nu$, $X_\mu$$^\nu$, $X^\alpha$$_\alpha$
So I get that for the first two i have to raise an lower the $\mu$ and $\nu$ indices using the metric tensor, and i get how, for example, $A^{\alpha\beta}=g^{\alpha\gamma}g^{\beta\delta}A_{\gamma\delta}$ and $A_{\alpha\beta}=g_{\alpha\gamma}g_{\beta\delta}A^{\gamma\delta}$, but what if I only want to lower(raise) one index ?
And for the third case ($X^\alpha$$_\alpha$) I imagine I'll need a Kronecker $\delta$ in order to have the same index, and then, due that it is a dummy index I can change the name.. right?
 A: 
I need to calculate the following tensors: $X^\mu$$_\nu$, $X_\mu$$^\nu$, $X^\alpha$$_\alpha$


So I get that for the first two i have to raise an lower the $\mu$ and $\nu$ indices using the metric tensor, and i get how, for example, $A^{\alpha\beta}=g^{\alpha\gamma}g^{\beta\delta}A_{\gamma\delta}$ and $A_{\alpha\beta}=g_{\alpha\gamma}g_{\beta\delta}A^{\gamma\delta}$, but what if I only want to lower(raise) one index?

If you only want to lower one index, use one instance of the metric:
$$
{X^\mu}_\nu = g_{\nu\rho}X^{\mu\rho}
$$
If you want to also sum like in ${X^\alpha}_\alpha$, where the sum is implied by the usual convention:
$$
{X^\alpha}_\alpha = g_{\alpha\rho}X^{\alpha\rho}\;.
$$
A: I am using explicit summation in this answer, I am not using the Einstein summation convention here. For example, the triply repeated indices in the equation below do not imply a sum.
I am also assuming the metric $\eta_{\mu\nu}$ is diagonal.
We have: $${X^\mu}_\nu=\sum_{\kappa=1}^4X^{\mu\kappa}\eta_{\kappa\nu}=X^{\mu\nu}\eta_{\nu\nu}=\begin{pmatrix}
-2 & 0 & 1 & -1 \\
1 & 0 & 3 & 2 \\
1 & 1 & 0 & 0 \\
2 & 1 & 1 & -2
\end{pmatrix}.$$
The first column was multiplied with $-1$. We have: $${X_\mu}^\nu=\sum_{\lambda=1}^4X^{\lambda\nu}\eta_{\lambda\mu}=X^{\mu\nu}\eta_{\mu\mu}=\begin{pmatrix}
-2 & 0 & -1 & 1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2
\end{pmatrix}.$$
The first row was multiplied with $-1$. The contraction: $${X^\mu}_\mu=\sum_{\mu,\nu=1}^4X^{\mu\nu}\eta_{\mu\nu}=-X^{00}+X^{11}+X^{22}+X^{33}=-2-2=-4$$ is the trace of those matrices.
A: I assume you would have been able to solve the exercise if you had been given the following equations:
\begin{align}\tag{1}
X^k{}_l&=\sum_iX^{ki}\eta_{il}\\
X_k{}^l&=\sum_iX^{il}\eta_{ik}
\end{align}
Thus, I will explain how to derive this two equations.
Derivation of $(1)$:
We consider the following general setting: Let $V$ an $n$-dimensional real vector space,
\begin{equation}
g\in V^*\otimes V^*
\end{equation}
and $e_1,\ldots,e_n$ a basis of $V$.
Firstly, we apply the natural isomorphism
\begin{equation}
V^*\otimes V^*\to L(V,V^*)
\end{equation}
to $g$ to obtain a function $\displaystyle{g\in L(V, V^*)}$ (note the abuse of notation). The function $g$ has the following property: Let
\begin{equation}
\begin{pmatrix}
g_{11}&\cdots&g_{1n}\\
\vdots&&\vdots\\
g_{n1}&\cdots&g_{nn}
\end{pmatrix}
\end{equation}
be the matrix defined by
\begin{equation}
g=\sum_{k,l}g_{kl}e^k\otimes e^l\in V^*\otimes V^*
\end{equation}
then $g(e_i)(e_j)=g_{ij}$ for all $1\leq i,j\leq n$.
Now suppose $X\in V\otimes V$ and let
\begin{pmatrix}
X^{11}&\cdots&X^{1n}\\
\vdots&&\vdots\\
X^{n1}&\cdots&X^{nn}
\end{pmatrix}
be the matrix defined by
\begin{equation}
X=\sum_{i,j}X^{ij}e_i\otimes e_j
\end{equation}
Then
\begin{pmatrix}
X^1{}_{1}&\cdots&X^1{}_{n}\\
\vdots&&\vdots\\
X^n{}_{1}&\cdots&X^n{}_{n}
\end{pmatrix}
is the matrix defined by
$$(1\otimes g)X=\sum_{k,l}X^k{}_l e_k\otimes e^l$$
where $1$ is the identity on $V$.
We now obtain the desired result (in your exercise, it is assumed that $g_{ij}=\eta_{ij}$):
\begin{align}
&(1\otimes g)X=(1\otimes g)\sum_{i,j}X^{ij}e_i\otimes e_j\\
&=\sum_{i,j}X^{ij}(1\otimes g)(e_i\otimes e_j)\\
&=\sum_{i,j}X^{ij}e_i\otimes g(e_j)\\
&=\sum_{i,j}X^{ij}e_i\otimes \sum_kg_{jk}e^k=\sum_{i,k}\sum_jX^{ij}g_{jk}e_i\otimes e^k
\end{align}
The other equation can be derived analogously by considering the function$$\displaystyle{g\otimes 1\colon V\otimes V\to V^*\otimes V.}$$
