# Properties of hermitian matrices

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1. Nov 8, 2015

### shinobi20

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.

2. Nov 8, 2015

### blue_leaf77

b) Multiply $M^i M^j = - M^j M^i$ with $(M^j)^{-1}$ from the left. This will give $(M^j)^{-1} M^i M^j = - M^i$. Then calculate the trace of both sides of the equation and apply the cyclic permutation property of a trace. (Note that you might have made a typo in writing the hint. That relation should follow cyclic permutation, which is not in the way you wrote it).
c) It follows from b) and use the invariant property of a trace under change of basis. The trace is zero, and the eigenvalues of any of those matrices can only be either 1 or -1, so?

3. Nov 8, 2015

### shinobi20

For part b) Tr((Mj)-1 Mj Mi) = -Tr(Mi) imples 2Tr(Mi) = 0 which means Tr(Mi) = 0. For part c) they cannot be odd dimensional because it would not produce a trace that is 0, suppose it is a 3x3 matrix then the eigenvalues would be degenerate and produce a trace either 1 or -1. Is this correct?

4. Nov 8, 2015

### blue_leaf77

Yes, that's correct.

5. Nov 8, 2015

### shinobi20

How about part a)? I think I did something wrong because I forgot the nonlinearity of determinants.

6. Nov 8, 2015

### blue_leaf77

That relation is valid in all basis. You can then assume that they are already diagonalized. Since for $i=j$, $(M^i)^2= I$, this means any eigenvalue $\lambda_k$ will satisfy $\lambda_k^2 = 1$.

7. Nov 8, 2015

### shinobi20

So you mean I can just assume that they are diagonal and set i=j then add the matrices, that is Mi Mi + Mi Mi = 2I ⇒ 2Mi Mi = 2I ⇒ Mi Mi = I ⇒ det(Mi Mi) = det(I) ⇒ (det(Mi))2 = 1 ⇒ det(Mi)=±1. Since Mi is diagonal ±1 are exactly the eigenvalues of Mi. Is this correct?

8. Nov 8, 2015

### blue_leaf77

You shouldn't use determinant because determinant is not more unique than the element-by-element comparison. For example I can have the eigenvalues to be 4 and 1/4 (2 by 2 matrix) and their determinant is still one.

9. Nov 8, 2015

### shinobi20

So maybe I can just manipulate each arbitrary diagonal element mi and show that mi2=±1. Then conclude that EACH eigenvalue can only have values ±1?

10. Nov 8, 2015

### blue_leaf77

Arbitrary diagonal element in the diagonalized version of the matrix, remember the question asks you about the eigenvalues.

11. Nov 8, 2015

### shinobi20

I just thought of now that how can (Mj)-1(Mj) be the identity matrix, M is just stated to be hermitian, not unitary?

12. Nov 8, 2015

### blue_leaf77

$(M^j)^{-1}$ is the inverse matrix of $M^j$.

13. Nov 8, 2015

### shinobi20

AM very sorry for that lousssssy question. Thanks!

14. Nov 8, 2015

### shinobi20

I'm a little confused in this part, part a just says that the eigenvalues can only have values 1 or -1 right? So if the M's are not odd dimensional, what if the eigenvalues are all 1's, then it would not add up to 0.

15. Nov 8, 2015

### blue_leaf77

Actually you don't need part a) to work out part b). In b) you found that the trace must vanish, thus the eigenvalues cannot be all 1 nor all -1.

16. Nov 8, 2015

### shinobi20

So how will I argue with part c)?

17. Nov 8, 2015

### blue_leaf77

From part a) and b), you can write $\textrm{Tr} (M) = \sum_{k=1}^N \lambda_k = 0$ with $\lambda_k$ being either -1 or 1. Think of how $N$ will be if you were to satisfy that equation.