How to calculate the eigenvalues of $A^\ast A$ where $A$ is a companion matrix? Let $$A=\left[\begin{array}{cc}0&-a_0\\I_{n-1}&\xi\end{array}\right]$$ be a companion matrix where $\xi=[-a_1\ \cdots\ -a_{n-1}]^T\in \mathbb{C}_{n-1}$. Hence, $$A^\ast A=\left[\begin{array}{cc}I_{n-1}& \xi \\ \xi^\ast & s\end{array}\right]$$ where $s=\sum\limits_{i=0}^{n-1} |a_i|^2$. How to calculate the eigenvalues of $A^\ast A$?

I see some relations between this and Find The Eigenvalues and Eigenvectors of the Hermitian Matrix. However, I'm not sure can the conclusion apply to my problem.
I tried calculating the eigenvalues in software and found that all the eigenvalues of $A^\ast A$ (then they're singular values of $A$) are 1 despite the first and the last one, and these two eigenvalues happen to be $$ \frac{1}{2} \left( (s+1) \pm \sqrt{(s+1)^2-4|a_0|^2} \right) $$ So my second problem is: why do eigenvalues form in this way here? Does it have any intuitive/visual explanation?
 A: Not sure what kind of intuition you are looking for, but it might be clearer to find the eigenvectors directly. The eigenvalue equation looks like
$$
\begin{bmatrix}I_{n-1}&\xi\\ \xi^*&s\end{bmatrix}\begin{bmatrix}x\\ t\end{bmatrix}=\lambda \begin{bmatrix}x\\ t\end{bmatrix}.
$$
This gives the two equations
\begin{align}
x+t\xi&=\lambda x\\
\xi^*x+ts &=\lambda t
\end{align}
When $t=0$, we have $x\ne0$. This forces $\lambda=1$ and $\xi^*x=0$. As the orthogonal $\{\xi\}^\perp$ has dimension $n-2$, the eigenspace for $\lambda=1$ is $(n-2)$-dimensional.
When $t\ne0$, we can substitute the first equation into the second one to get (note that $t\ne0$ implies $\lambda\ne1$)
$$
t(\lambda-s)=\frac t{1-\lambda}\,\xi^*\xi.
$$
Using that $t\ne0$ and that $s=|a_0|+\xi^*\xi$, this simplies to
$$
\lambda^2-(1+s)\lambda+|a_0|^2=0.
$$
This gives the other two eigenvalues as
$$
\lambda_\pm=\frac{1+s\pm \sqrt{(1+s)^2-4|a_0|^2}}{2}.
$$
The corresponding eigenvectors, normalized to $t=1-\lambda_\pm$, are
$$
\begin{bmatrix}-\xi\\ 1-\lambda_\pm\end{bmatrix}.
$$
