determine the nature of eigenvalues of given matrix whose entries are complex number under a given condition Let $Q$ be a matrix ($n \times n$), $P$ be its  complex conjugate transpose.    $PQ$ equals identity.    


*

*What can we say about eigenvalues of Q?  

*For example - are they    all    real or all complex?  

*Are some real and some complex?
My view: 
I tried for $2$ by $2$ case. 
I concluded that determinant of the matrix must be $1$ or $-1$.
I solved it for $1$ and $-1$ separately. 
But it became very tough to solve equations. 
I took entries of the form $a+ib$.
 A: Let $Q$ be an $n\times n$ matrix, and let $Q^\dagger$ denote its conjugate transpose. Matrices satisfying your condition $Q^\dagger Q = I$ are called unitary matrices. Let $\mathbf{v}\in\mathbb{C}^n$ be a unit eigenvector for $Q$ under eigenvalue $\lambda$. Then we must have
$$Q\mathbf{v} = \lambda \mathbf{v},$$
and taking the conjugate transpose, also
$$\mathbf{v}^\dagger Q^\dagger = \lambda^*\mathbf{v}^\dagger.$$
Multiplying the two equations together, we have
$$\mathbf{v}^\dagger Q^\dagger Q \mathbf{v} = \lambda \lambda^* \mathbf{v}^\dagger \mathbf{v}=|\lambda|^2\|\mathbf{v}\|^2=|\lambda|^2,$$
where the last equality comes from the fact that $\mathbf{v}$ is a unit eigenvector so that $\|\mathbf{v}\|=1$. Since $Q^\dagger Q = I$, the left hand simplifies to 
$$\mathbf{v}^\dagger Q^\dagger Q \mathbf{v} = \mathbf{v}^\dagger I \mathbf{v} = \|\mathbf{v}\|^2 = 1.$$
Therefore we conclude that
$$1=\|\lambda\|^2,$$
which means that the eigenvalues of $Q$ must lie on the unit circle, i.e. $\lambda = e^{i\theta}$ for some $\theta \in \mathbb{R}$. This is the most general possible constraint, since for any set of such eigenvalues $S=\{e^{i\theta_1}, \cdots ,e^{i\theta_n}\}$, there exists an $n\times n$ unitary matrix with eigenvalues in $S$, namely the diagonal matrix 
$$Q=\mathrm{diag}(e^{i\theta_1},\cdots, e^{i\theta_n}).$$
