Show that the eigenvalues of a unitary matrix have modulus $1$.
I know that a unitary matrix can be defined as a square complex matrix $A$, such that
$$AA^*=A^*A=I$$
where $A^*$ is the conjugate transpose of $A$, and $I$ is the identity matrix. Furthermore, for a square matrix $A$, the eigenvalue equation is expressed by $$Av=\lambda v$$
If I use the relationship $u v=v^*u$ and take the conjugate transpose of this equation then
$$v^*A^*=\lambda^*v^*$$
But now I got stuck. Can someone help?
Best Answer
You multiply your two relations to obtain
\begin{align} v^*A^*Av &=\lambda^* v^*\lambda v \\ v^*Iv &=\left(\lambda^*\lambda\right) v^*v \\ v^*v &=\left(\lambda^*\lambda\right) v^*v \\ ||v||^2 &= |\lambda|^2 ||v||^2 \\ \sqrt{1} &=|\lambda| \\ 1 &=|\lambda| \end{align}
Recall that the modulus of a complex number $\lambda = a + bi$, also called the "complex norm", is denoted $|\lambda|$ and defined by $|\lambda| = |a + bi| = \sqrt{a^2 + b^2}$ and $\lambda^*\lambda = (a -bi)(a + bi) = a^2 + b^2$. Hence $\lambda^*\lambda = |\lambda|^2.$