Matrix representation of electromagnetic tensor, with 1 index covariant and 1 index contravariant Leafing through the notes of one of my professors, I found this particular relationship:
$$
\det F^\mu{}_\nu=-(\bar B\cdot\bar E)^2.
$$
I already know the matrix structure of $F^{\mu\nu}$ and $F_{\mu\nu}$. But how can this representation be proven?
 A: The transformation of co- to contravariant and back is given in special relativity by the most simple metric tensor: $$\eta^\mu {}^\nu=\eta_\mu {}_\nu=\text{diag}(1,-1,-1,-1).$$ (Sometimes one also finds the convention $ \text{diag}(-1,1,1,1)$)
The explicit transformation is: $$W^\nu=W_\mu \,\eta^\mu{}^\nu\,\,\,\,\,\text{and}\,\,\,\,\,W_\nu=W^\mu \,\eta_\mu{}_\nu$$ Thus we get $$F^\mu{}^\nu=F^\mu{}_\lambda \eta^\lambda{}^\nu(=F_\kappa{}_\lambda \eta^\kappa{}^\mu \eta^\lambda{}^\nu)\\ \Leftrightarrow F^\mu{}_\nu=F^\mu{}^\lambda \eta_\lambda{}_\nu $$ In this case the sum over $\lambda$ lets us write this contraction in ordinary matrix multiplication, because the first index is summed with the second: $$F^\mu{}_\nu=\begin{pmatrix}
0 & -E_x/c & -E_y/c & -E_z/c\\
E_x/c & 0 & -B_z &B_y\\
E_y/c&B_z&0&-B_x\\
E_z/c&-B_y&B_x&0
\end{pmatrix}\cdot \begin{pmatrix}
1 & 0 & 0 & 0\\
0 & -1 & 0 &0\\
0&0&-1&0\\
0&0&0&-1
\end{pmatrix}\\=\begin{pmatrix}
0 & E_x/c & E_y/c & E_z/c\\
E_x/c & 0 & B_z &-B_y\\
E_y/c&-B_z&0&B_x\\
E_z/c&B_y&-B_x&0
\end{pmatrix}.$$
Now use $$\text{det}F^\mu{}_\nu=\sum_i (-1)^i F^i{}_0\cdot\text{det}(\text{Min}F^\mu{}_\nu(F^i{}_0)),$$ where "$\text{Min}F^\mu{}_\nu(F^i{}_0)$" is the Minor Matrix taking out the $i$-th row and $0$-th column of $F^\mu{}_\nu$.
Example: $$\text{Min}F^\mu{}_\nu(F^0{}_0)=\begin{pmatrix}
 0 & B_z &-B_y\\
-B_z&0&B_x\\
B_y&-B_x&0
\end{pmatrix}.$$
After some work you should get $$\text{det}F^\mu{}_\nu=-\frac{1}{c^2}(\vec E\cdot\vec B)^2(=\text{det}F^\mu{}^\nu\text{det}\eta_\mu{}_\nu).$$ I am getting an extra expression $-2\frac{1}{c^2}(E_xB_xE_yB_y+E_xB_xE_zB_z+E_zB_zE_yB_y)$. Maybe my expression for $\text{det}F^\mu{}_\nu$ is wrong?
