Finding spectrum with Schur decomposition

Question

let $\alpha$ be a real number and $A_n(\alpha)$ the set of $n\times n$ matrices such that $$\alpha A+A^HA+A^H=I$$

I'd like to find the set of all eigenvalues for the matrices in $A_n(\alpha)$ using Schur decomposition.

Answer 1

Let's denote $A^H$ by $A^*$.

First let's consider the case $\alpha\neq-1$ (condition whose importance is pointed out by @user1551 below).

$$\alpha A+A^*A+A^*=I$$

then $A^*(I+A)+\alpha(I+A)=(1+\alpha)I$ or $(A^*+\alpha I)(I+A)=(1+\alpha)I$ or $(I+A)(A^*+\alpha I)=(1+\alpha)I$ or $AA^*+\alpha A+A^*=I$ which means $A^*A=AA^*$, which means the Schur Decomposition's result of $A$ is a diagonal matrix, $D$. Denote $D_{ii}=d_i$.

The condition is equivalent to $d_i^2+(1+\alpha)d_i=1$, which determines the possible eigenvalues.

For the case $\alpha=-1$, the equality reads $A^*(I+A)=I+A$. For any eigenvalue $\lambda$ with eigenvector $v$, multiply by $v^*$ on the left and $v$ on the right leaves us with: $$\bar{\lambda}(1+\lambda)=1+\lambda\iff(\bar{\lambda}-1)(1+\lambda)=0$$ and thus there are only two possible eigenvalues: $-1$ and $1$, which correspond duly to, for example, the matrices $-I$ and $I$.