Finding the eigenvalues of a matrix from self adjoint matrix Let $A$ be an $n\times n$ matrix complex matrix. Assume that $A$ is self-adjoint and let $B$ denote the inverse of $A+iI$. Then all eigenvalues of $(A-iI)B$ are :
$(1)$ purely imaginary $(2)$ of modulus one $(3)$ real $(4)$ of modulus less than one
Here's my approach: $A$ is Hermitian i.e $A^\theta = A$ $\implies$ all eigenvalues of $A$ are real. Let $\lambda \in \mathbb{R}$ be an eigenvalue of $A$ then $\dfrac{1}{\lambda + i}$ is an eigenvalue of $B$
[Let $X$ be the eigenvector corresponding to eigenvalue $\lambda$ and $(A+iI)X = AX + iX = \lambda X + iX = (\lambda + i)X$  so $\lambda + i$ is an eigenvalue of $A+iI$ but the inverse of $A+iI$ is $B$ so the it has $(\lambda + i)^{-1}$ as an eigenvalue of $B$]
we can simplify it further as $\dfrac{\lambda - i}{\lambda^2 + 1}$. So an arbitrary eigenvalue of $(A-iI)B$ is $(\lambda - i)\dfrac{\lambda - i}{\lambda^2 + 1} = \dfrac{(\lambda - i)^2}{\lambda^2 + 1} = \dfrac{\lambda^2 - 1 -2\lambda i}{\lambda^2 + 1}= \dfrac{\lambda^2 -1}{\lambda^2 + 1}- \dfrac{2\lambda i}{\lambda^2 + 1}$. (Call it $w$)
So $|w| = \dfrac{(\lambda^2 - 1)^2}{(\lambda^2 + 1)^2} + 4\dfrac{\lambda^2 }{(\lambda^2 + 1)^2}= \dfrac{\lambda^4 + 1 -2\lambda^2 + 4 \lambda^2}{(\lambda^2+1)^2}= \dfrac{(\lambda^2 + 1)^2}{(\lambda^2 + 1)^2} = 1$
So Its modulus $1$.
Am I correct?
 A: Suppose $x\ne 0$ and suppose that there exists $\lambda$ such that
$$
              (A-iI)(A+iI)^{-1}x=\lambda x.
$$
Then
$$
             (A+iI-2iI)(A+iI)^{-1}x=\lambda x \\
                  x-2i(A+iI)^{-1}x=\lambda x \\
                    -2i(A+iI)^{-1}x=(\lambda-1)x \\
                     -2ix = (\lambda-1)(A+iI)x \\
                      -2ix-i(\lambda-1)x=(\lambda-1)Ax \\
                      -i(1+\lambda)x=(\lambda-1)Ax.
$$
$\lambda\ne 1$, because that would contradict the assumption that $x\ne 0$. Therefore,
$$
                       Ax = i\frac{1+\lambda}{1-\lambda}x
$$
Because $A$ is self-adjoint, that forces $i\frac{1+\lambda}{1-\lambda}=\mu$ to be real, which gives $\lambda$ as
$$
                  1+\lambda=i\mu(\lambda-1) \\
                   \lambda(1-i\mu)=-(i\mu+1) \\
                       \lambda = \frac{i\mu+1}{i\mu-1}
$$
Therefore $|\lambda|=1$ because $|i\mu+1|=|i\mu-1|$ for all real $\mu$.
A: Your answer is correct, but your instructor would probably like you to explain why each eigenvector of $(A-iI)B$ originates from an eigenvector of $A$.
Here is a simpler solution. Let $U=(A-iI)(A+iI)^{-1}$. Since $A$ is self-adjoint,
$$
U^\ast
=\left[(A+iI)^\ast\right]^{-1}(A-iI)^\ast
=(A-iI)^{-1}(A+iI)
=U^{-1}.
$$
Hence $U$ is unitary and its eigenvalues have moduli $1$.
