Verification of the Poincare Algebra The generators of the Poincare group $P(1;3)$ are supposed to obey the following commutation relation to be verified:
$$\left[ M^{\mu\nu}, P^{\rho} \right] = i \left(g^{\nu\rho} P^{\mu} - g^{\mu\rho} P^{\nu} \right)$$
where $M^{\mu\nu}$ are the 6 generators of the Lorentz group and $P^\mu$ are the 4 generators of the four-dimensional translation group $T(4)$.
For $\mu = 3, \nu=1, \rho=0$ the LHS becomes: $ [M^{31},P^{0}] = M^{31}P^{0} - P^{0}M^{31}$.
Here $M^{31} = J^2 = -J_2= \begin{pmatrix}
       0 & 0 & 0 & 0   \\
       0 & 0 & 0 & -i   \\
       0 & 0 & 0 & 0   \\
       0 & i & 0 & 0   
     \end{pmatrix}$ and $ P^0 = P_0 = -i \begin{pmatrix}
       0 & 0 & 0 & 0 & 1  \\
       0 & 0 & 0 & 0 & 0  \\
       0 & 0 & 0 & 0 & 0  \\
       0 & 0 & 0 & 0 & 0  \\
       0 & 0 & 0 & 0 & 0 
     \end{pmatrix}$.
My question is that how can I multiply $M^{31}$ and $P^0$ when they are $4\times4$ and $5\times 5$ matrices respectively? 
 A: Consider $$M_{31} = \begin{pmatrix}
       0 & 0 & 0 & 0  & 0 \\
       0 & 0 & 0 & -i & 0  \\
       0 & 0 & 0 & 0  & 0  \\
       0 & i & 0 & 0  & 0  \\
       0 & 0 & 0 & 0  & 0 
     \end{pmatrix} \text{ and } P_0 = -i \begin{pmatrix}
       0 & 0 & 0 & 0 & 1  \\
       0 & 0 & 0 & 0 & 0  \\
       0 & 0 & 0 & 0 & 0  \\
       0 & 0 & 0 & 0 & 0  \\
       0 & 0 & 0 & 0 & 0 
     \end{pmatrix},$$
Then the commutator vanishes! As expected from $\left[ M_{31}, P_{0} \right] = i \left(g_{10} P_{3} - g_{30} P_{1} \right) = 0$.
If you take 
$$M_{01} = 
\begin{pmatrix}
 0 & i & 0 & 0 & 0 \\
 i & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0  \\
 0 & 0 & 0 & 0 & 0  \\
 0 & 0 & 0 & 0 & 0 
\end{pmatrix}
\text{ and }
P_0 = -i
\begin{pmatrix}
  0 & 0 & 0 & 0 & 1  \\
  0 & 0 & 0 & 0 & 0  \\
  0 & 0 & 0 & 0 & 0  \\
  0 & 0 & 0 & 0 & 0  \\
  0 & 0 & 0 & 0 & 0 
\end{pmatrix},$$
then
$$
\left[M_{01},P_0\right] = -i
\begin{pmatrix}
 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 \\
 0 & 0 & 0 & 0 & 0  \\
 0 & 0 & 0 & 0 & 0  \\
 0 & 0 & 0 & 0 & 0 
\end{pmatrix} = P_1.
$$
And so on!
