Are Hermitian Matrices with Specific Properties Traceless and Even-Dimensional?

[shinobi20]

Homework Statement

Consider hermitian matrices M¹, M², M³, M⁴ that obey the property Mⁱ M^j + M^j Mⁱ = 2δ^ij 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= -M^j Mⁱ for i≠j. Show that Mⁱ are traceless (Hint: Tr(ABC)=Tr(CBA)
c) Show that they cannot be odd dimensional matrices

Homework Equations

The Attempt at a Solution

For part a), det(Mⁱ) det(M^j) + det(M^j) det(Mⁱ) = 2δ^ij 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.

 

[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?

 

[shinobi20]

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?

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?

For part b) Tr((M^j)^-1 M^j 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?

 

[blue_leaf77]

Yes, that's correct.

 

[shinobi20]

Yes, that's correct.

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

 

[blue_leaf77]

Mi Mj + Mj Mi = 2δij

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 $\lambda_k^2 = 1$.

 

[shinobi20]

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 $\lambda_k^2 = 1$.

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?

 

[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.

 

[shinobi20]

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.

So maybe I can just manipulate each arbitrary diagonal element m_i and show that m_i²=±1. Then conclude that EACH eigenvalue can only have values ±1?

 

[blue_leaf77]

each arbitrary diagonal element

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

 

[shinobi20]

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?

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

 

[blue_leaf77]

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

 

[shinobi20]

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

AM very sorry for that lousssssy question. Thanks!

 

[shinobi20]

For part b) Tr((M^j)^-1 M^j 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?

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.

 

[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.

 

[shinobi20]

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.

So how will I argue with part c)?

 

[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.