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Linear Algebra Proof, Hermitian Matrices

  • #1

Homework Statement



Let A, B, C, D be nxn complex matrices such that AB and CD are Hermitian, i.e., (AB)*=AB and (CD)*=CD.

Show that AD-B*C*=I implies that DA-BC=I

The symbol * indicates the conjugate transpose of a matrix, i.e., M* is the conjugate transpose of M.

I refers to the identity.

Homework Equations



Wikipedia says that if A and B are Hermitian, AB is only Hermitian if AB=BA. I'm not sure if this is supposed to be a biconditional, and AB=BA indicates that AB is Hermitian.

I don't know what else may be relevant to proving this.

The Attempt at a Solution



I'm not sure where to start. Just because AB and CD are Hermitian, do we know that A, B, C, and D individually are hermitian? I suppose that if we knew that, and we knew (somehow) that AD=DA, the proof is trivial. But alas, we don't know those things, do we?

Thank you for your help. I appreciate it.

Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
jambaugh
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It would help to group matrices when conjugating them together:
i.e. (AB)* = AB.

You should make use of the rule: (XY)* = Y*X*.

Also w.r.t. your wikipedia quote, you don't need to assume A and B are individually Hermitian here, just that their product is. (For example if A = B* their product is Hermitian even if B* isn't equal to B).
 
  • #3
Sorry about forgetting the brackets. I fixed that.

Okay, (XY)*=Y*X*. So we can rewrite as AD-(CB)*=I. But how does that help?

My main problem is that we don't know that any of the individual matrices A, B, C, D are Hermitian. If we could prove that, we're done. B*C* is the same as BC, and then we just need to know that AD=DA. But that's easy, I suppose. If we know that B and C are Hermitian, then we can say that since (CD)(AB) is Hermitian, and [tex]C^{-1}[/tex] and [tex]B^{-1}[/tex] are Hermitian, showing that DA is Hermitian is easy. Then DA=AD.

Thank you for your help, I appreciate it.

Also, wikipedia seems to indicate that A and B have to be Hermitian and AB must commute before we know that AB is Hermitian:

The sum of any two Hermitian matrices is Hermitian, and the inverse of an invertible Hermitian matrix is Hermitian as well. However, the product of two Hermitian matrices A and B will only be Hermitian if they commute, i.e., if AB = BA. Thus An is Hermitian if A is Hermitian and n is an integer.
http://en.wikipedia.org/wiki/Hermitian_matrix
 
  • #4
I just realized something: If AB=(AB)*=B*A*=BA, then AB commutes. But that is only true if A, B, and AB are all Hermitian.

Alternatively, we can say that if A, B are both Hermitian and AB commutes, then AB=BA=B*A*=(AB)*, so we know that AB is Hermitian. That's why the property I cited from wikipedia is true, but that only holds true if the individual matrices are Hermitian on their own.
 
  • #5
jambaugh
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Hmmm... you shouldn't assume any of A, B, C, and D are Hermitian by themselves.

Some tools you have are the identity properties of I plus that it too is Hermitian.

But since you brought up commutativity, often when you have something that's close to commutative or would be simpler if things were commutative, it is useful to utilize the commutator bracket to swap things around. For example:

XY = [X,Y] + YX

Try playing with that. (And I'll do likewise. I haven't yet figured it out either.)
 
  • #6
jambaugh
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I've found a counter example to what you have written.

Let:
[tex] B^\dagger = A = \left( \begin{array}{c c} 1 & 1\\ 0 & 1\end{array}\right)[/tex]
and
[tex] D^\dagger = C = \left( \begin{array}{c c} 0 & 1\\ -1 & 0\end{array}\right)[/tex]

I think you get AD - B*C*= I but DA-BC = -2C not I. Check my work and if correct check your copy of the problem.

Could it have read AD and BC are Hermitian?

EDIT[I think it must be the case. This makes the problem simple using my first suggestion.]
 
  • #7
I didn't check all your work, but my initial reaction: all four matrices have to be complex matrices, that's the initial condition. If they're not complex, the conjugate transpose is the same as the transpose, and they're just simple symmetric matrices.

Am I wrong?
 

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