Does anybody know how to prove the following statement: Let $A,B\in\mathbb{C}^{n\times n}$ be such that $A^*B=B^*A$. Show that $\det(B+iA)\neq0$ if and only if $\operatorname{rank} (A,B)=n$. It seems to be quite simple but I am not able to finish the proof. I think that I am missing something trivial.
The direction "$\Rightarrow$" is easy: If $\det(B+iA)\neq0$, then also $\det(B^*-iA^*)\neq0$, which yields that $\ker B^*\cap\ker A^*=\{0\}$, which is equivalent to the equality $\operatorname{rank} (A,B)=n$.
But the opposite implication? Let $\operatorname{rank} (A,B)=n$ or equivalently $\ker B^*\cap\ker A^*=\{0\}$ and assume that $\det(B+iA)=0$. Then we have also $\det(B^*-iA^*)=0$, which means that there is a vector $h\in\mathbb{C}^n\setminus\{0\}$ such that $(B^*-iA^*)h=0$, i.e., $B^*h=iA^*h$. If $h\in\ker B^*$ then also $h\in\ker A^*$, and so $h=0$ (a contradiction). Similarly, if $h\in\ker A^*$ then also $h\in\ker B^*$, and so $h=0$ (a contradiction). It remains to get a contradiction in the case when $h$ is such that $0\neq B^*h=iA^*h$. But how to do it?? By the way, so far I did not use the fact that $A^*B=B^*A$.
Best Answer
I assume hereafter that notation $M^*$ is for the conjugate transpose of $M$.
The main remark is that, if $A^*B=B^*A$, then :
$$(A+iB)(A+iB)^*=(A+iB)(A^*-iB^*)=AA^*+i\underbrace{(BA^*-AB^*)}_{= \ 0}+BB^*$$
Thus :
$$(A+iB)(A+iB)^*=AA^*+BB^*=\underbrace{(A \ \ B)}_C \underbrace{\binom{A^*}{B^*}}_{C^*}\tag{1}$$
(matrix $C$ is $n \times 2n$).
As a consequence of (1), $$|\det(A+iB)|^2=\det(CC^*).\tag{2}$$
Let $C=U\Sigma V^*$ be the Singular Values Decomposition (SVD) of $C$ (https://en.wikipedia.org/wiki/Singular_value_decomposition) where $U$ ($n \times n$) and $V$ ($2n \times 2n$) are unitary. We have therefore
$$CC^*=U\Sigma\underbrace{(V^*V)}_I \Sigma^*U^*=USU^*$$
where $S:=\Sigma \Sigma^*$ ; matrices $U$ and $S$ are $n \times n$ with $$S=diag(\sigma_1^2,\sigma_2^2,\cdots,\sigma_r^2\underbrace{,0,0, \cdots}_{\text{optional}})$$
where $r=rank(S)=rank(C)=rank(CC^*)$
As a consequence, using (2), we get :
$$rank(C)=n \ \iff \ \det(CC^*) \neq 0 \ \iff \ |\det(A+iB)| \neq 0.$$