Any matrix in c can written as sum of two self adjoint matrices
Answers
The reference to the MathOverflow question is a good one. If AA is a complex matrix, you can normalize so that ∥A∥≤1‖A‖≤1. Then
A=B+iC
A=B+iC
where BB, CC are selfadjoint and given by
B=12(A+A⋆),C=12i(A−A⋆).
B=12(A+A⋆),C=12i(A−A⋆).
These selfadjoint operators also satisfy ∥B∥≤1‖B‖≤1 and ∥C∥≤1‖C‖≤1, which means that their eigenvalues--which must be real--are in [−1,1][−1,1]. Then you can decompose BB and CC as
B=12(UB+VB),C=12(UC+VC)
B=12(UB+VB),C=12(UC+VC)
where UB,VB,UC,VCUB,VB,UC,VC are unitary and given by
UB=B+iI−B2−−−−−−√,VB=B−iI−B2−−−−−−√UC=C+iI−C2−−−−−−√,VC=C−iI−C2−−−−−−√
UB=B+iI−B2,VB=B−iI−B2UC=C+iI−C2,VC=C−iI−C2
This makes sense becaue I−B2I−B2 and I−C2I−C2 are selfadjoint with their eigenvalues in [0,1][0,1]; so the square roots are defined that also have eigenvalues in [0,1][0,1]. You can check that
UBU⋆B=U⋆BUB=(B−iI−B2−−−−−−√)(B+iI−B2−−−−−−√)=B2+(I−B2)=I.
UBUB⋆=UB⋆UB=(B−iI−B2)(B+iI−B2)=B2+(I−B2)=I.
Then 12(UB+VB)=B12(UB+VB)=B, 12(UC+VC)=C12(UC+VC)=C and A=B+iCA=B+iC is a linear combination of unitary matrices.