characteristic roots of a matrix I have a question from linear algebra which I need to understand over the vacation. 

If $A$ is a square matrix over $\mathbb C$ and $A^n=I$, then why is every characteristic root of $A$ an $n$-th root of $1$ in $\mathbb C$? 

I've learnt what a characteristic root of a matrix is (a root of the determinant of $XI-A)$ but I really don't know what to do next. Can anyone help me please??
Thank you! 
 A: More generally, if $p$ is a polynomial and $\lambda$ is a characteristic root of $A$, then $p(\lambda)$ is a characteristic root of $p(A)$.  One way to see this
is that we can write the polynomial $p(x)$ as $p(\lambda) + g(x) (x - \lambda)$ for some polynomial $g(x)$ (i.e. $p(x) - p(\lambda)$ is divisible by $x - \lambda$), and then $p(A) = p(\lambda) I + g(A) (A - \lambda I)$.  So
$p(A) - p(\lambda) I = g(A) (A - \lambda I)$ must be singular (if you want to understand this on the level of determinants, $\det(p(A) - p(\lambda) I) = \det(g(A)) \det(A - \lambda I) = 0$), which says
$p(\lambda)$ is a characteristic root of $p(A)$.  
Now in your case $p(x) = x^n - 1$, and $p(A) = 0$.  The only characteristic root of the $0$ matrix is $0$, so for every characteristic root $\lambda$ of $A$ we must have $p(\lambda) = 0$.
A: If $\lambda$ is a characteristic root of $A$,  there is $x\neq 0$ such that $Ax=\lambda x$. We thus have $A^2x=A(\lambda x)=\lambda Ax=\lambda^2x$, and by induction over $p$, $A^px=\lambda^px$ for each positive integer $p$. In particular, for $p=n$, we get $A^nx=\lambda^nx$, hence $x=\lambda^nx$. Since $x\neq 0$, $\lambda$ is necessarily a $n$th root of unity.
A: By definition, the matrix $A$ satisfies the polynomial equation $X^n = 1$ (where $I$ is $1$ for matrices).  Any time a matrix satisfies a polynomial equation where $1$ is considered to be $I$, the characteristic roots of $A$ must satisfy the same polynomial equation.  Thus $\lambda^n = 1$ for every characteristic root.
