How to prove the following?
Lemma. Let $C=[A,A^{\star}]$. $A$ is normal iff $[A,C]=0$.
One direction is trivial. The other direction reduces to showing that $A^2 A^\star+A^\star A^2=0$ implies that $A$ is normal, but I don't see why that holds.
Asked
Active
Viewed 130 times
5
-
1Shouldn't it be $[A,C]=AC-CA=A(AA^-A^A)-(AA^-A^A)A=A^2A^-2AA^A+A^*A^2=0$ ? – Christiaan Hattingh Apr 28 '14 at 08:08
1 Answers
2
Suppose $[A,C]=0$. As $C$ is Hermitian, it is normal and possesses a complete eigensystem. If $(\lambda,x)$ is an eigenpair of $C$, then $C(Ax) = ACx = \lambda Ax$. In other words, every eigenspace of $C$ is an invariant subspace of $A$.
Pick any eigenvalue $\lambda$ of $C$ and denote the associated eigenspace by $V$. Since eigenspaces of $C$ (being a normal matrix) are orthogonal to each other and they are invariant subspaces of $A$, the restriction of $C$ on $V$ is still the commutator of $A$ and $A^\ast$ restricted on $V$. But every commutator is traceless. Therefore $\lambda$ must be zero. Hence all eigenvalues of $C$ are zero and $C=0$.
EDIT. Ahhh, I've finally found it! Initially, I wanted to offer a one-line proof using a more general fact, but I couldn't locate the sources and ended up with the above. Now the proof can be greatly simplified.
A one-line proof. As $A$ commutes with $C=[A,A^\ast]$, $C$ is nilpotent (see also q227984 and q299640); since $C$ is also Hermitian, it must be zero and hence $A$ is normal.
-
$C = AA^* - A^A$ is actually Hermitian (not skew-Hermitian) since both $AA^$ and $A^*A$ are Hermitian (in fact, positive). – Mike F Apr 28 '14 at 20:31
-