Normal matrix as sum of self adjoint and commuting matrices

[diegzumillo]

Homework Statement

I need to show that any normal matrix can be expressed as the sum of two commuting self adjoint matrices

Homework Equations

Normal matrix A: [A,A^\dagger]=0
Self Adjoint matrix: B=B^\dagger

The Attempt at a Solution

A is a normal matrix. I assume I can write any matrix as a sum of two other matrices (no conditions imposed yet). So: A=B+C
But A is normal so
AA^\dagger=A^\dagger A
Expanding the AA^\dagger term we have
AA^\dagger=(B+C)(B^\dagger +C^\dagger )
AA^\dagger=BB^\dagger +BC^\dagger+CB^\dagger+CC^\dagger
This will only be equal to A^\dagger A if B and C are self adjoints and commute with each other. (I could write the other steps but I think you got the point).

The problem is that I'm not sure if this proves that ANY normal matrix can be written like the sum of two self adjoint and commuting matrices. I'm not sure what this proves at all. This would still be valid if B=I and C=0, right? Also, I've read some proofs using complex numbers and I just don't see where that's coming from.

 

[BvU]

In fact, I find the question rather questionable. If you succeed, haven't you proven that any normal matrix is self-adjoint as well ? Namely via $A^\dagger=(B^\dagger +C^\dagger )=B+C = A$ ? Or am I ranting ?

(Was that what you meant when you said "if B=I and C=0" ?)

I am puzzled, so I just try something: $A = \begin{pmatrix} 1 & i\\ i & 1\end{pmatrix}$ so that $AA^\dagger = \begin{pmatrix} 1 & 1\\ 1 & 1\end{pmatrix} = A^\dagger A \ \$
Hence A is normal but clearly not self-adjoint.
Pick out a part that IS self-adjoint:
$A+A^\dagger =2{\mathbb I}$ which is clearly self-adjoint, leaving
$A-A^\dagger$ which unfortunately is anti-self-adjoint (a term I invent right here and now)
The nice thing is that these two do commute (since the first one is the identity matrix).

Don't see a proof of the original claim shining through..

 

[diegzumillo]

Honestly, I believe the statement is supposed to say 'linear combination' instead of sum. If I'm allowed to have scalars multiplying the matrices then they can be commuting and self adjoint without making A be self adjoint as well, and the proof can be valid for a larger class of matrices (normal, maybe? I have to check it)

 

[DEvens]

$A = \begin{pmatrix} 1 & i\\ i & 1\end{pmatrix}$ so that $AA^\dagger = \begin{pmatrix} 1 & 1\\ 1 & 1\end{pmatrix} = A^\dagger A \ \$

Isn't $AA^\dagger = \begin{pmatrix} 2 & 0\\ 0 & 2\end{pmatrix}$ ?

 

[DEvens]

Honestly, I believe the statement is supposed to say 'linear combination' instead of sum. If I'm allowed to have scalars multiplying the matrices then they can be commuting and self adjoint without making A be self adjoint as well, and the proof can be valid for a larger class of matrices (normal, maybe? I have to check it)

For example A = B + i C ? Then B is real(A) and C is imaginary(A).

 

[Fredrik]

I agree that the best result you can hope for is to be able to rewrite A as B+iC, where B and C are self-adjoint commuting matrices. You can think of them as the real and imaginary part of A.

One way to find B and C is to see what A=B+iC tells you about $A^\dagger$, and then solve for B and C. Another is to just fiddle around with A and $A^\dagger$ for a while. Is there a way to combine $A$ and $A^\dagger$ into a self-adjoint matrix?