Eigenvalues of product of two complex matrices Let $A$ be a $n\times n$ complex matrix. Assume that $A$ is self-adjoint and $B$ denote the inverse of $A+iI$, where $I$ is identity matrix of order $n\times n$.  Then all eigenvalues of $(A-iI)B$ are


*

*purely imaginary

*real

*of modulus one

*of modulus less than one
My attempt: I discarded options $1,4$ by taking $A=O$ $($null matrix$)$. Let '$a$' be an eigenvalue of $A$ then the eigenvalues of $(A-iI)B$ are $(a-i)(a+i)^{-1}$. But I'm not sure about it. Please help me.
 A: With $A$ self-adjoint and
$Av = \lambda v, \tag 1$
we have
$\lambda \in \Bbb R \tag 2$
and
$(A + iI)v = (\lambda + i)v; \tag 3$
since every eigenvalue of $A$ is real, no eigenvalue of $A + iI$ vanishes, hence it is invertible; we use this fact to write (3) as
$(A + iI)^{-1}v = \dfrac{1}{\lambda + i} v = \dfrac{\lambda - i}{\lambda^2 + 1}v; \tag 4$
then
$(A - iI)(A + iI)^{-1}v = (A - iI) \dfrac{\lambda - i}{\lambda^2 + 1}v =  \dfrac{\lambda - i}{\lambda^2 + 1}(A - iI)v$
$= \dfrac{\lambda - i}{\lambda^2 + 1}(\lambda - i)v  = \dfrac{(\lambda - i)^2}{\lambda^2 + 1}v; \tag 5$
we have
$\left \vert \dfrac{(\lambda - i)^2}{\lambda^2 + 1} \right \vert = \dfrac{1}{\lambda^2 + 1}  \vert(\lambda - i)^2 \vert$
$= \dfrac{1}{\lambda^2 + 1}  \vert(\lambda - i) \vert^2 = \dfrac{1}{\lambda^2 + 1}  (\sqrt{\lambda^2 + 1})^2 = \dfrac{\lambda^2 + 1}{\lambda^2 + 1} = 1; \tag 6$
thus option (3) is correct.
The transformation
$A \to (A - iI)(A + iI)^{-1} \tag 7$
is a useful one, since in maps self-adjoint $A$ to the unitary $(A - iI)(A + iI)^{-1}$, yielding additional insights into such operators.  This is especially true since it may be extended to unbounded $A$ on a Hilbert space, and allows $A$ to be addressed in terms of a bounded (unitary) operator.
A: Since $A$ is self-adjoint, the matrix $B=(A-iI)(A+iI)^{-1}$ is unitary. Now let $ \mu \in \mathbb C$, $x \ne 0$ and $Bx= \mu x$. Then
$||x||^2=(x,x)=(B^*Bx,x)=(Bx,Bx)=(\mu x,\mu x)=|\mu|^2 ||x||^2.$
This gives $|\mu|=1.$
( $ (\cdot, \cdot)$ denotes the usual inner product on $\mathbb C^n$.)
