Matrix such that $A^n=aA$

Question

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?

Answer 1

If $A$ has eigenvalue $b$, and $v$ is a corresponding (non-zero) eigenvector, then $$0 = (A^n-aA)v = A^nv - aAv = b^nv-abv = (b^n-ab)v$$This means $b^n-ab = 0$, which does make $b$ a root to the polynomial equation $x^n = ax$.

Apart from that, there is not much we can say. $A$ could have one, some, or all of those roots as eigenvalues, in any combination and multiplicity (as long as the total multiplicity is $n$). For instance, by being a diagonal matrix with the desired combination of eigenvalues along the diagonal.

Answer 2

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.$

Answer 3

You have

• $A^n-aA=O_{n\times n}$

So, based on Cayley-Hamilton theorem you can conclude that any matrix $A$ satisfies the above equation, iff its minimal polynomial $m_A(\lambda)$ satisfies

• $m_A(\lambda)|\lambda(\lambda^{n-1}-a)$