Lorentz Transform of the Electromagnetic Field

Question

I'm trying to prove that the Transformation of the electric and magnetic field under a boost of arbitrary direction is $$ E_i^{\prime\parallel}=E_i^\parallel\qquad B_i^{\prime\parallel}=B_i^\parallel $$ $$ E_i^{\prime\perp}=\Gamma(E_i^\perp+\epsilon_{ijk}V_jB_k^\perp)\qquad B_i^\perp=\Gamma(B_i^\perp-\epsilon_{ijk}V_jE_k^\perp) $$ To prove this what I did was to start from $$ F_{\mu\nu}^\prime=\Lambda_\mu^\alpha\Lambda_\nu^\beta F_{\alpha\beta} $$ And got $$ E_i^\prime=(\Lambda^0_0\Lambda^i_j-\Lambda^i_0\Lambda^0_j)E_j+\epsilon_{jkl}\Lambda^0_j\Lambda^i_kB_l $$ $$ B_i^\prime=\epsilon_{ijk}\Lambda^j_0\Lambda^k_lE_l+\frac{1}{2}\epsilon_{ijk}\epsilon_{lmn}\Lambda^j_l\Lambda^k_mB_n $$ The Lorentz transform for a boost in an arbitrary direction is $$ \Lambda_\mu^\nu= \begin{pmatrix} \Lambda^0_0 & \Lambda^0_j\\ \Lambda^i_0 & \Lambda^i_j \end{pmatrix}= \begin{pmatrix} \Gamma & -\Gamma V_j\\ -\Gamma V_i & P_{ij}^\perp+\Gamma P_{ij}^\parallel \end{pmatrix} $$ Substituting in the expression of the electric field I get this $$ E_i^\prime=(\Gamma P_{ij}^\perp+\Gamma^2 P_{ij}^\parallel-\Gamma^2V^2P_{ij}^\parallel)E_j+\epsilon_{kjl}\Gamma V_j(P_{ik}^\perp+\Gamma P_{ik}^\parallel)B_l $$ And my question comes here. The last term of that expression should be $$ \epsilon_{kjl}\Gamma V_j(P_{ik}^\perp+\Gamma P_{ik}^\parallel)B_l=\Gamma \epsilon_{ijk}V_jB_k^\perp $$ But I don't see why, also, how is it that only the magnetic field was multiplied by the projector and not the velocity? If I write this in another way we have $$ \Gamma(P_{ik}^\perp+\Gamma P_{ik}^\parallel)(V\times B)_k $$ That cross product will yield a vector perpendicular to the velocity, so the parallel projector times that vector will be 0 and we get $$ \Gamma P_{ik}^\perp(V\times B)_k $$ But here, why is it that the projector only affects the magnetic field?

Note: I'm using the adimensional units, where C=1.