- #1
dykuma
- 55
- 6
- Homework Statement
- Consider two hermitian matrices who do not commute. Show that the result of the commutation equals a hermitian matrix times the complex number I.
- Relevant Equations
- [A,B] = AB- BA = iC
I think I roughly see what's happening here.
> First, I will assume that AB - BA = C, without the complex number.
>Matrix AB equals the transpose of BA. (AB = (BA)t)
>Because AB = (BA)t, or because of the cyclic property of matrix multiplication, the diagonals of AB equals the diagonals of BA. Thus, when you subtract the two matrices, the diagonals are zero.
>After subtracting AB- BA, because AB = (BA)t, the off diagonal elements (AB)i,j will equal the off diagonal elements (AB-BA)j,i in magnitude, but will have a sign difference. (so element (AB-BA)1,2 = -(AB-BA)2,1 )
>Further, (AB-BA) equals the negative of it's transpose. So (AB-BA)= -(AB-BA)t
>Thus, for matrix C (AB - BA) to be hermitian, all that's required is it would need to be multiplied by the complex number -I (so that way the complex conjugate transpose equals the original matrix). As such, for C to be Hermitian, C = -I(AB-BA)
>However, there is a rule stating that "the product of two hermitian matrices A and B is also hermitian if and only if AB = BA". But I originally assumed that AB- BA = C, meaning that C can not be Hermitian. As such, AB- BA must equal C times a complex number if matrix C is to be hermitian. Doing so ensures the final result of AB - BA is not hermitian, and that C alone is hermitian.What I'm having trouble with is showing this. I'm not great at doing proofs in linear algebra, and I don't how to begin to show this. I feel like all I've shown up to this point is that AB-BA = I(-I(AB-BA)).
> First, I will assume that AB - BA = C, without the complex number.
>Matrix AB equals the transpose of BA. (AB = (BA)t)
>Because AB = (BA)t, or because of the cyclic property of matrix multiplication, the diagonals of AB equals the diagonals of BA. Thus, when you subtract the two matrices, the diagonals are zero.
>After subtracting AB- BA, because AB = (BA)t, the off diagonal elements (AB)i,j will equal the off diagonal elements (AB-BA)j,i in magnitude, but will have a sign difference. (so element (AB-BA)1,2 = -(AB-BA)2,1 )
>Further, (AB-BA) equals the negative of it's transpose. So (AB-BA)= -(AB-BA)t
>Thus, for matrix C (AB - BA) to be hermitian, all that's required is it would need to be multiplied by the complex number -I (so that way the complex conjugate transpose equals the original matrix). As such, for C to be Hermitian, C = -I(AB-BA)
>However, there is a rule stating that "the product of two hermitian matrices A and B is also hermitian if and only if AB = BA". But I originally assumed that AB- BA = C, meaning that C can not be Hermitian. As such, AB- BA must equal C times a complex number if matrix C is to be hermitian. Doing so ensures the final result of AB - BA is not hermitian, and that C alone is hermitian.What I'm having trouble with is showing this. I'm not great at doing proofs in linear algebra, and I don't how to begin to show this. I feel like all I've shown up to this point is that AB-BA = I(-I(AB-BA)).
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