1. The problem statement, all variables and given/known data Consider hermitian matrices M1, M2, M3, M4 that obey the property Mi Mj + Mj Mi = 2δij I where I is the identity matrix and i,j=1,2,3,4 a) Show that the eigenvalues of Mi=+/- 1 (Hint: Go to the eigenbasis of Mi and use the equation for i=j) b) By considering the relation Mi Mi= -Mj Mi for i≠j. Show that Mi are traceless (Hint: Tr(ABC)=Tr(CBA) c) Show that they cannot be odd dimensional matrices 2. Relevant equations 3. The attempt at a solution For part a), det(Mi) det(Mj) + det(Mj) det(Mi) = 2δij det(I) implies 2det(Mi)2 = 2det(I). This means det(Mi)=+/-1. Since Mi is hermitian and diagonalizable. The diagonalized matrix will have the same determinant with eigenvalues +/-1 therefore the eigenvalues of Mi are +/-1. This is how I think it can be proved, I'm not sure what the hint is telling. For part b) and c) I really don't know how to start.