# What is $[M_1,M_2]$ equal to? ($M_1$ and $M_2$ are matrices)

This is an old exercise that I had a year ago:

$$M_1 = \dfrac{1}{\sqrt{2}} \begin{bmatrix}0 & 1 &0\\1 & 0 & 1\\0 & 1 & 0\end{bmatrix}$$

$$M_2 = \dfrac{1}{\sqrt{2}} \begin{bmatrix}0 & -i &0\\i & 0 & -i\\0 & i & 0\end{bmatrix}$$

$$M_3 = \begin{bmatrix}1 & 0 &0\\0 & 0 & 0\\0 & 0 & -1\end{bmatrix}$$

The exercise asks to prove that $[M_l,M_m] = i \sum \epsilon_{lmk} M_k$

($\epsilon_{lmk}$ is the permutation symbol/tensor of Levi-Civita.)

I already have the correction of the exercise. The problem is that I don't understand what $[M_l,M_m]$ means.

Here is the beginning of the correction:

$[M_1,M_2] = \dfrac{1}{2} ( \begin{bmatrix}i & 0 &-i\\0 & 0 & 0\\i & 0 & -i\end{bmatrix} - \begin{bmatrix}-i & 0 &-i\\0 & 0 & 0\\i & 0 & i\end{bmatrix} ) = i M_3$

I see that $\dfrac{1}{2} \begin{bmatrix}i & 0 &-i\\0 & 0 & 0\\i & 0 & -i\end{bmatrix} = M_1 M_2$

But I don't know where $\dfrac{1}{2} \begin{bmatrix}-i & 0 &-i\\0 & 0 & 0\\i & 0 & i\end{bmatrix}$ comes from (it's not equal to $M_1 M_2^*$).

Therefore it would seem that $[M_1,M_2] = M_1 M_2 - something$

So my question is: what is $[M_1,M_2]$ equal to? And does this thing have a name so that I can Google it and find more information about it?

The Lie bracket notation $[M_1, M_2]$ is standard notation for the commutator $M_1M_2 - M_2M_1$.
HINT: The notation $[A,B]=AB-BA$ refers to the commutator of $A$ and $B$; this object is especially prevalent in Quantum Mechanics, where observables can be modeled as the eigenvalues of matrix operators in a Hilbert Space.
The matrices that you have there are in fact the spin matrices for a spin-$1$ particle multiplied by a constant factor.
$[M_1, M_2]$ is actually the commutator in the algebra of matrices: $$[M_1, M_2] = M_1 M_2 - M_2 M_1$$