I was wondering, suppose you have a matrix of the form $A=B+iCC^\dagger$ where $^\dagger$ denotes the hermitian conjugate. $B$ is hermitian and $CC^\dagger$ is obviously hermitian positive semi-definite.
Is it true that if $A \psi = 0$, then $B \psi = 0$ and $C^\dagger \psi = 0$? If so, how can I prove it? My idea was to look at $\psi^\dagger A \psi$, but to no avail.
If it is not true, what would be a counter-example?
On
Here, I use $*$ to denote the conjugate transpose.
Thoughts so-far:
Lemma: for a matrix $M$ and vector $x$, $Mx = 0 \iff M^*Mx = 0$
Note that $$ \psi^*(B + iCC^*)\psi = (\psi^*B\psi) + i(\psi^*CC^*\psi) $$ Noting that $(\psi^*B\psi),(\psi^*CC^*\psi) \in \Bbb R$, we conclude that $$ \psi^* A \psi = 0 \iff (\psi^*B\psi) = (\psi^*CC^*\psi) = 0\\ \Longleftarrow B\psi = C^*\psi = 0 $$ We then note that $$ A\psi = 0 \implies \psi^*A\psi = 0 $$ (though the converse does not necessarily hold).
Notice that if $D$ is a Hermitian matrix and $v$ is a vector then $v^\dagger D v\in\mathbb{R}$.
Thus, if $Av=0$ then $0=v^\dagger Av=v^\dagger Bv+i(v^\dagger CC^\dagger v)$.
Since $v^\dagger Bv \in \mathbb{R}$ and $v^\dagger CC^\dagger v\in \mathbb{R}$ then $v^\dagger Bv=v^\dagger CC^\dagger v=0$. But this implies that $C^\dagger v=0$, since $CC^\dagger$ is positive semidefinite.
Therefore, $Av=0=Bv+iCC^\dagger v=Bv$.