# Normal matrix as sum of self adjoint and commuting matrices

1. Sep 18, 2014

### diegzumillo

1. The problem statement, all variables and given/known data
I need to show that any normal matrix can be expressed as the sum of two commuting self adjoint matrices

2. Relevant equations
Normal matrix A: $[A,A^\dagger]=0$
Self Adjoint matrix: $B=B^\dagger$

3. 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.

2. Sep 19, 2014

### 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..

3. Sep 19, 2014

### 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)

4. Sep 19, 2014

### DEvens

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

5. Sep 19, 2014

### DEvens

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

6. Sep 19, 2014

### Fredrik

Staff Emeritus
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?