Problem with proving the invariance of dot product of two four vectors I am having a spot of trouble with index manipulation (its not that I am very unfamiliar with this, but I keep losing touch). This is from an electrodynamics course - we're just getting started with 4 vectors, the metric tensor and so on. So, wherever I use $\eta$ or $\Lambda$, it is in the context of special relativity.
So, what I want to show is that $\mathbf{A \cdot B}$ is a scalar where both $\mathbf{A}$ and $\mathbf{B}$ are 4 vectors and the inner product is with respect to the metric, $\eta = diag(1, -1, -1, -1)$.
I can show that $\mathbf{A'} \cdot \mathbf{B'} = \mathbf{A \cdot B}$ if I go like this,
$$ \mathbf{A'} \cdot \mathbf{B'} = A'^{\ \mu}B'_{\ \mu} = \Lambda^{\mu}_{\nu} \Lambda^{\sigma}_{\mu}\ A^{\nu}B_{\sigma} = \delta^{\sigma}_{\nu}\ A^{\nu}B_{\sigma} = \mathbf{A \cdot B}$$
The problem is when I try it this way,
$$ \mathbf{A'} \cdot \mathbf{B'} = A'^{\ \mu}B'_{\ \mu} = \eta_{\mu \nu} \ A'^{\ \mu}B'^{\ \nu} = \eta_{\mu \nu} \ \Lambda^{\mu}_{\rho} \Lambda^{\nu}_{\sigma}\ A^{\rho}B^{\sigma}$$
This is basically it; I cannot proceed further without running into problems. I know that $\Lambda^{T}\eta \ \Lambda = \eta$, which implies that $\eta_{\alpha \beta} = (\Lambda^{T})^{\alpha}_{\mu}\ \eta_{\mu \nu} \ \Lambda^{\nu}_{\beta} $, but this seems very problematic since the indices on either side do not match ($\alpha$ is on the bottom on the left but on top on the left) and both the $\mu$-s on the left are down below which seems to fall outside the purview of the summation convention. On the other hand, I don't think I have seen an $\eta$ with mixed indices in the couple of books I have gone through earlier. I am certain I am making a very elementary, possibly stupid, mistake but I have gone through this calculation for nearly an hour and without avail.
Any help would be appreciated. Thanks!
Edit: On a further note, if I write $\eta_{\alpha \beta} = (\Lambda^{T})^{\mu}_{\alpha}\ \eta_{\mu \nu} \ \Lambda^{\nu}_{\beta} $, I have a different issue. As far as I remember, upper indices are "like row indices" which seems to say that this can't be written (as, on the left, $\alpha$ is like the second index in a normal matrix $A_{ab}$). Second, I don't think this is the form that I have for the original inner product thing either. Thanks!
 A: $\newcommand{\bl}[1]{\boldsymbol{#1}} 
\newcommand{\e}{\bl=}
\newcommand{\mb}[1]{\mathbf {#1}}
\newcommand{\mc}[1]{\mathcal {#1}}
\newcommand{\tl}[1]{\tag{#1}\label{#1}}$
\begin{equation}
\mb A\bl\cdot \mb B \e \mc A^{\bl\top}\eta\; \mc B
\tl{01}
\end{equation}
\begin{equation}
\mb A'\bl\cdot \mb B' \e  {\mc A'}^{\bl\top}\eta\; \mc B'\e \mc A^{\bl\top}\Lambda^{\bl\top}\eta\; \Lambda\,\mc B\e \mc A^{\bl\top}\eta\;\mc B \e \mb A\bl\cdot \mb B
\tl{02}
\end{equation}
A: You made a mistake in the second equation, $\mathbf B$ is a covariant vector not a contravariant one. So instead of $\Lambda^\nu_\sigma$, it should be $\Lambda^\sigma_\nu $.
After you fix that, this is a rather simple question if you understand what the metric tensor really is. I'm sure you know that in the case of Minkowski metric, you can use the Kronecker delta to represent the spatial coordinates and put a negative sign for the "time dimension". However the best definition to do so would  be to use the partial derivative definition.
$$\frac {\partial{x^\mu }}{\partial x^\nu}=g^{\mu \nu}=\Lambda^\mu_{\nu}$$
So in the second equation, you have to work out the partial derivatives and you can resolve three of them to get $g_{\rho \sigma}$. After that the four-product is trivial.
