Matrix such that $A^n=aA$ Let $A\in M_n(\mathbb{C})$ be a matrix such that $A^n=aA$,where $a\in \mathbb{R}-\{0,1\}$.
I wanted to find $A$'s eigenvalues and I thought that they are the roots of the polynomial equation $x^n=ax$. Is this correct?

It is correct: if $ \mu $ is an eigenvalue of $A$ with corresponding eigenvector $x$, then $A^nx= \mu^n x$, hence $\mu^nx=a Ax=a \mu x$. Since $x \ne 0$, we get $ \mu^n=a \mu.$
