Can Hermitian Matrices be Traceless?

[SgrA*]

Hello,
Here's a textbook question and my solution, please check if it is correct, I'm slightly doubtful about the second part.

Consider Hermitian matrices $M_1, M_2, M_3,\ and\ M_4$ that obey:
$M_i M_j+M_j M_i = 2 \delta_{ij} I \hspace{10mm} for\ i,\ j\ = 1,\ ... ,4$

(1) Show that the eigenvalues of $M_i$ are $±1$.
Solution: When $i = j, \hspace{20mm} M_i M_i = I$.
Since $M_i$ are Hermitian, this equation implies that they are unitary. The eigenvalues of a unitary matrix are complex numbers of unit modulus, since it's also Hermitian, they have to be real. So, the eigenvalues have to be $±1$. $\hspace{20mm} Q.E.D.$

(2) Show that $M_i$ are traceless.
Solution: When $i ≠ j, \\ M_i M_j = -M_j M_i\\ \Rightarrow M_i M_j M_i = -M_j M_i M_i\\ \Rightarrow M_i M_j M_i = -M_j I\\ \Rightarrow Tr(M_i M_j M_i) = -Tr(M_j) \\ \Rightarrow Tr(M_j M_i M_i) = -Tr(M_j)\\ \Rightarrow Tr(M_j) = -Tr(M_j)\\ \Rightarrow Tr(M_j) = 0. \hspace{20mm} Q.E.D.$

(3) Show that $M_i$ cannot be odd-dimensional matrices.
Solution: For some eigenbasis $U, D = U^†M_iU, D$ is a diagonal matrix with $Tr(D) = Tr(M_i)$ and the diagonal elements of $D$ are the eigenvalues, $±1$. Also, $Tr(D) = Tr(M_i) = 0.$ An odd dimension cannot result in a traceless matrix, hence by argument of parity, $M_i$ are even dimensional matrices. $\hspace{20mm}Q.E.D.$

Thanks!

 

[wabbit]

Yes, I think the three proofs are fine. Nothing suspicious about (2) in particular.

 

[SgrA*]

Thanks for verifying, I'm still not very confident. :)

 

[wabbit]

I can't see why, you've spelled out every step in detail, which part gives you doubt ?

 

[SgrA*]

I mean, I got it right, but I'm not confident that I can get math right, yet. Hopefully more problems and I'll feel confident about my solutions. Thanks for asking! :)

 

[wabbit]

Ah yes I feel that way about physics all the time : )

 

[goun]

I think there is a mistake on the 2nd question. Between line 4 and 5 you comutate Mi with Mj which you can't do it. You should put a (-) in front so the final result would be 0=0 . There is another way ;) .

 

[ca11289]

No, it's okay - it's a cyclic permutation, just cycled the "wrong" way (or cycled twice)