I did the following question:
Show that if $A$ is a square matrix and $A^n = I$, the identity matrix, for some $n>0$, then $A$ has a basis of eigenvectors.
The solution actually shows that A has no generalized eigenvectors that are not genuine/ordinary eigenvectors. That is, $A$ only has "ordinary" eigenvectors. But I don't understand why showing this can conclude that $A$ has a basis of eigenvectors?
Thanks in advance for any help.
Let $p(x)$ be the minimal polynomial of $A$, then since $A^n-I=0$, we see $p(x)|(x^n-1)$. Since $x^n-1$ has no repeated root, we see $p(x)$ also has no repeated root. Then there is a theorem saying that a matrix is diagonalizable if and only if its minimal polynomial has no repeated root, which is our case.
For your information, this is the theorem in Linear Algebra, Hoffman and Kunze, Page 204.