Faraday tensor, antisymmetric rank two $F^{\mu \nu}$ is defined in 
http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-7/relativistic-lorentz-force/
How to show that $F^{\mu \nu}$$F_{\mu \nu}$, is a Lorentz scalar and how to find its value in terms of vectors, Electric field (E), and magnetic field(B).
This is mentioned in properties (point number 3) in wikipedia article
https://en.wikipedia.org/wiki/Electromagnetic_tensor
How to prove that?
 A: It's really easy.
First use the definition of the Faraday Tensor: $F_{\mu\nu}\equiv\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$, and then write down the same expression but in an another inertial system of reference, i.e. $F'^{\mu\nu}=\Lambda_{\ \alpha}^{\mu}\Lambda_{\ \beta}^{\nu}F^{\alpha\beta}$. And using the property of the $\Lambda$ matrix: $\eta_{\alpha\beta}=\Lambda_{\ \alpha}^{\mu}\Lambda_{\ \beta}^{\nu}\eta_{\mu\nu}$, you will get that $F'^{\mu\nu}F'_{\mu\nu}=F^{\mu\nu}F{}_{\mu\nu}$, i.e. $F^{\mu\nu}F{}_{\mu\nu}$ is an Lorentz invariant quantity. Then for expressing this in terms of the EM Fields, you should use the matrix expression of $F^{\mu\nu}$: let say $F$, in terms of the EM fields. And the quantity $F^{\mu\nu}F{}_{\mu\nu}$ is, in matrix terms, the trace of the matrix $FF^T$. i.e.
$$F^{\mu\nu}F{}_{\mu\nu}=Tr(FF^T)$$
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Now, we know that the interval $s^2:=x_{\mu}x^{\mu}$ must be Lorentz invariant, in other words if $x'^{\mu}=\Lambda_{\ \nu}^{\mu}x^{\nu}$ is the world line of a particle measured by an inertial observer that moves with velocity $\vec v=v \hat u_x$, where
$$\Lambda_{\ \nu}^{\mu}=\begin{pmatrix}\gamma & -\beta\gamma & 0 & 0\\
-\beta\gamma & \gamma & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}$$
then is must be hold that
$$x_{\mu}'x'^{\mu}=x_{\mu}x^{\mu}$$
where we know that $x_{\mu}=\eta_{\mu\nu}x^{\nu}$ and the metric is
$$\eta_{\mu\nu}:=\begin{pmatrix}1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{pmatrix}=\mathrm{diag}(1,-1,-1,-1)=:\eta^{\mu\nu}$$
then  we have that
$$x_{\mu}'x'^{\mu}=\eta_{\mu\nu}x'^{\nu}x'^{\mu}$$
$$\eta_{\mu\nu}x'^{\nu}x'^{\mu}=\eta_{\mu\nu}\Lambda_{\ \alpha}^{\nu}x^{\alpha}\Lambda_{\ \beta}^{\mu}x^{\beta} $$
$$ \eta_{\alpha\beta}x'^{\alpha}x'^{\beta}=\eta_{\mu\nu}\Lambda_{\ \alpha}^{\nu}\Lambda_{\ \beta}^{\mu}x^{\alpha}x^{\beta}$$
i.e. the matrix $\Lambda$ must satisfy
$$ \eta_{\alpha\beta}=\eta_{\mu\nu}\Lambda_{\ \alpha}^{\nu}\Lambda_{\ \beta}^{\mu} \qquad (1)$$
Then using (1),
$$F_{\mu\nu}=\eta_{\mu\alpha}\eta_{\nu\beta}F^{\alpha\beta}$$
and
$$F'^{\mu\nu}=\Lambda_{\ \sigma}^{\mu}\Lambda_{\ \rho}^{\nu}F^{\sigma\rho}$$
we are going to show that $F'^{\mu\nu}F'_{\mu\nu}=F^{\mu\nu}F_{\mu\nu}$ (here the order does not matter due to $F^{\mu\nu}$ is actually a scalar, the $\mu,\nu$ component of the matrix $F$). OK, then we have that
$$F'^{\mu\nu}F'_{\mu\nu}=F'^{\mu\nu}\eta_{\mu\alpha}\eta_{\nu\beta}F'^{\alpha\beta}$$
$$ =\Lambda_{\ \delta}^{\mu}\Lambda_{\ \varepsilon}^{\nu}F^{\delta\varepsilon}\eta_{\mu\alpha}\eta_{\nu\beta}\Lambda_{\ \sigma}^{\alpha}\Lambda_{\ \rho}^{\beta}F^{\sigma\rho}$$
$$ =\left(\Lambda_{\ \delta}^{\mu}\Lambda_{\ \sigma}^{\alpha}\eta_{\mu\alpha}\right)\left(\Lambda_{\ \varepsilon}^{\nu}\Lambda_{\ \rho}^{\beta}\eta_{\nu\beta}\right)F^{\delta\varepsilon}F^{\sigma\rho}$$
$$ \equiv\eta_{\delta\rho}\eta_{\varepsilon\sigma}F^{\delta\varepsilon}F^{\sigma\rho}\equiv F^{\delta\varepsilon}F_{\delta\varepsilon}$$
So
$$F'^{\mu\nu}F'_{\mu\nu}= F^{\delta\varepsilon}F_{\delta\varepsilon}$$
Here the indices are dummy (they are being summed over), so they don't have to match.
So, in conclusion the quantity $F^{\mu\nu}F{}_{\mu\nu}$ (or $F{}_{\mu\nu}F^{\mu\nu}$ ) if is measured in any inertial frame, it would have the same value, i.e. it is an Lorentz invariant.
Finally I'll write down the part of the calculation of $F^{\mu\nu}F_{\mu\nu}$ in terms of the EM fields, but maybe later. You know, because of the time. But! You should read this question, I asked a few months ago the same, and here is the answer. Of course, if you have questions, ask us.
Clic here:
Calculating electromagnetic invariant in matrix form
A: Geometrically, there's a reason it's also called the Faraday bivector: bivectors represent oriented planes in spacetime, and the Faraday field is just a field of these oriented planes, all with magnitudes and orientations.  $F^{\mu \nu} F_{\mu \nu}$ is just the squared magnitude of the Faraday field.  This is no more exotic than talking about the magnitude of a vector.
A: $${\displaystyle{ F^{\mu \nu }=\begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\\E_{x}/c&0&-B_{z}&B_{y}\\E_{y}/c&B_{z}&0&-B_{x}\\E_{z}/c&-B_{y}&B_{x}&0\end{bmatrix}}.}$$
$$
{\displaystyle F_{\mu \nu }=\eta _{\mu \alpha }F^{\alpha \beta }\eta _{\beta \nu }={\begin{bmatrix}0&E_{x}/c&E_{y}/c&E_{z}/c\\-E_{x}/c&0&-B_{z}&B_{y}\\-E_{y}/c&B_{z}&0&-B_{x}\\-E_{z}/c&-B_{y}&B_{x}&0\end{bmatrix}}.}
$$
$$
F^{\mu \nu }F_{\mu \nu }=2 \left[ {\vec E} {}^2 -\vec B{}^2  \right]
$$
