Let $A,B\in\mathbb{C}^{n\times n}$ be such that $A^*B=B^*A$. Show that $\text{rank} (A,B)=n$ iff $\det(B+i A)\neq0$

Question

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 $A^*B=B^*A$.

Answer 1

The statement is not true - the matrices $A=\left(\begin{smallmatrix} 1 & 0\\ 2 & 0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix} 1 & 0\\ 0 & 0\end{smallmatrix}\right)$ provide a counterexample.

Answer 2

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 :

$$rank(C)=n \ \iff \ \det(CC^*) \neq 0 \ \iff \ |\det(A+iB)| \neq 0.$$

Answer 3

Let $C=\begin{bmatrix} A \\ B \end{bmatrix}$; it is known that $\operatorname{rank}C=\operatorname{rank}C^*C$ (proof below) and that $\operatorname{rank}C=\operatorname{rank}C^*$.

Since $$ C^*C=\begin{bmatrix} A^* & B^* \end{bmatrix}\begin{bmatrix} A \\ B \end{bmatrix}=A^*A+B^*B =(A+iB)^*(A+iB) $$ (which follows from $A^*B=B^*A$), we are done.

Suppose $X$ is an $m\times n$ matrix; we want to prove that $\operatorname{rank}X=\operatorname{rank}X^*X$; this follows from the fact that $Xv=0$ if and only if $X^*Xv=0$ and the rank-nullity theorem. One direction is obvious; for the other one, suppose $X^*Xv=0$; then also $0=v^*X^*XV=(Xv)^*(Xv)$, so $Xv=0$. Now apply this to $X=C$.