I am trying to understand the following product:
$$\eta_{\mu\lambda}\eta^{\nu\rho}\Lambda^{\lambda}_\rho.$$
I understand that the first metric lowers the $\lambda$ and changes it for a $\mu$, while the second one raises $\rho$ and changes it for $\nu$, giving as a result $\Lambda_\mu^\nu$.
I also know that that matrix is the inverse of $\Lambda_\mu^\nu$.
However, I want to do the multiplication to actually arrive at the result.
For example, starting with $$\Lambda^{\mu}_\nu=\left(\begin{matrix}
\cosh\omega & -\sinh\omega & 0 & 0 \\
-\sinh\omega & \cosh\omega & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 &1
\end{matrix}\right)$$
And applying the metric twice:
$$\eta_{\lambda\sigma}=\eta^{\lambda\sigma}=\left(\begin{matrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{matrix}\right)$$
But multiplying that matrix twice gives the identity matrix, which would leave the Lorentz transformation intact, Clearly I must be understanding something about the metric wrong, probably regarding the different indexes, but I don't know what it is.
Best Answer
Side note: You need to be careful with index notation. Instead of writing $\Lambda^{\mu}_\nu$ it is better to write $\Lambda^{\mu}{}_\nu$ to make clear what is the first index (i.e. the row index) and what is the second index (i.e. the column index).
Below we will use the definition of matrix multiplication several times: $$\begin{align} & (AB)^i{}_j=A^i{}_kB^k{}_j \\ \text{or}\quad & (AB)^{ij}=A^i{}_kB^{kj} \\ \text{or}\quad & (AB)_{ij}=A_{ik}B^k{}_j \\ \text{or}\quad & (AB)_i{}^j=A_{ik}B^{kj} \\ \text{or}\quad & ... \end{align}$$
(Mnemonic trick: The summing indices (here $k$) need to be side by side.)
We start with the Lorentz matrix: $$\begin{align} \Lambda^{\mu}{}_\nu=\begin{pmatrix} \cosh\omega & -\sinh\omega & 0 & 0 \\ -\sinh\omega & \cosh\omega & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &1 \end{pmatrix}\end{align}$$
Now let us calculate the actual product. $$\begin{align} \Lambda_\mu{}^\nu =&\ \eta_{\mu\lambda}\ \eta^{\nu\rho}\Lambda^\lambda{}_\rho \quad\quad\quad \text{(reorder to get the $\lambda$ indices side by side)} \\ =&\ \eta_{\mu\lambda}\ \Lambda^\lambda{}_\rho\ \eta^{\nu\rho} \quad\quad\quad \text{(use the transposed matrix $\eta^T$ to get the $\rho$ indices side by side)} \\ =&\ \eta_{\mu\lambda}\ \Lambda^\lambda{}_\rho\ (\eta^T)^{\rho\nu} \quad\quad \text{(now we can interpret this as matrix products)} \\ =&\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \cosh\omega&-\sinh\omega&0&0\\ -\sinh\omega&\cosh\omega&0&0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \\ =&\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} \cosh\omega & \sinh\omega & 0 & 0 \\ -\sinh\omega & -\cosh\omega & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \\ =&\begin{pmatrix} \cosh\omega & \sinh\omega & 0 & 0 \\ \sinh\omega & \cosh\omega & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \\ =&\begin{pmatrix} \cosh\omega & -\sinh\omega & 0 & 0 \\ -\sinh\omega & \cosh\omega & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}^{-1} \end{align}$$ So we finally got the inverse matrix of $\Lambda^\mu{}_\nu$, as it should be.