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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!

Amy
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3 Answers3

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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 $p(\lambda) = 0$.

Robert Israel
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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.

Davide Giraudo
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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.

user2566092
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  • Why do the characteristic roots satisfy the same equation? – Amy Aug 30 '13 at 14:55
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    You can see this e.g. for diagonalizable matrices by diagonalizing as $MDM^{-1}$ and seeing what happens when you plug the diagonalization into a polynomial $p(x)$. You get $M\cdot p(D) \cdot M^{-1}$. So if $p(A) = 0$ then $p(D) = 0$ as well, so every characteristic root $\lambda$ on the diagonal $D$ must satisfy $p(\lambda) = 0$. The same general argument can be used by using Jordan normal form if the matrix is not diagonalizable. – user2566092 Aug 30 '13 at 15:00