If a Power of a Matrix is the Identity, then the Matrix is Diagonalizable

Diagonalization Problems and Solutions in Linear Algebra

Problem 84

Let $A$ be an $n \times n$ complex matrix such that $A^k=I$, where $I$ is the $n \times n$ identity matrix.

Show that the matrix $A$ is diagonalizable.
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Contents

Hint.

Use the fact that if the minimal polynomial for the matrix $A$ has distinct roots, then $A$ is diagonalizable.

Proof.

Since the matrix $A$ satisfies the equation $x^k-1$, the minimal polynomial of $A$ divides $x^k-1$.
Since
\[x^k-1=\prod_{j=0}^{k-1}(x-e^{2\pi i j/k}),\] the roots of $x^k-1$ are all distinct.

Hence the roots of the minimal polynomial are also distinct.
Therefore the matrix $A$ is diagonalizable.


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1 Response

  1. 08/27/2016

    […] Remark that if $A$ is a square matrix over $C$ with $A^k=I$, then $A$ is diagonalizable. For a proof of this fact, see If a power of a matrix is the identity, then the matrix is diagonalizable […]

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