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Let $A$ a $n \times n$ matrix with real entries such that $A^3 = I$ but $A \ne I$.

a) Give an example that satisfies this conditions.

b) what are the eigenvalues ​​of $A$?

Well for $a)$ i construct this matrix , let $A=\begin{bmatrix} 1 & -\frac{3}{2} & \frac{\sqrt{3}}{2} \\ 0 & -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}$. Then $A^3=I$, and $A \ne I$ also $A$ is not orthogonally.

For $b)$ I can't found the eigenvalues for $A$ I suppose I need to find the characteristic polynomial and then find it but in general for a matrix $n \times n$ I can't find this eigenvalues, some help please.

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  • $\begingroup$ What exactly is the issue with finding the eigenvalues of a general $n \times n$ matrix? What have you tried with the characteristic polynomial? $\endgroup$ – Joshua Mundinger Jun 16 '14 at 6:14
  • $\begingroup$ i have a matrix $n \times n$ with unknown entries this is my problem... if a have a matrix n x n i have a polynomial with degree less than n and then n eigenvalues but how can i find it ? $\endgroup$ – Knight Jun 16 '14 at 6:16
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if $\lambda$ is an eigenvalue with eigenvector $v$, $A v = \lambda v$. Then, $A^2 v = A (Av) = A (\lambda v) = \lambda A v = \lambda \lambda v = \lambda^2 v$. Finally, $v = I v = A^3 v = A (A^2 v) = A (\lambda^2 v) = \lambda^2 A v = \lambda^2 \lambda v = \lambda^3 v$.

So, $\lambda^3 = 1$.

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consider the $n \times n$ permutation matrix $\sigma.I$ corresponding to a three cycle $\sigma$ in $S_n$ for $n \ge 3$. This matrix satisfies the required property. These matrices are orthogonal and you can easily find the eigen values. Let $\sigma$ be a k-cycle in $S_n$ where $k \le n$. Then the permutation matrix $\sigma.I$ satisfies $A^k = I$.

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Given

$$ \mathbf{A}_n^3 = \mathbf{I}_n \hspace{1em} \textrm{and} \hspace{1em} \mathbf{A}_n \ne \mathbf{I}_n. \tag {OP} $$

Eigenvalues

Let $\lambda$ be an eigenvalue of $\mathbf{A}_n$. Therefore $$ \mathbf{A}_n \vec{x}_\lambda = \lambda \vec{x}_\lambda. $$ Whence $$ \mathbf{A}_n^k \vec{x}_\lambda = \lambda^k \vec{x}_\lambda. $$ From (OP) follows $$ \lambda^3 = 1. $$


It is also clear that $$ \lambda^3 = 1 \Longrightarrow \big( \lambda^k \big)^3 = 1. $$ Therefore we obtain

$$ \lambda \ne 1, p + q > 0, p+q+r = n : \mathbf{A}^\flat_n = \left[ \begin{array}{ccc} \lambda \mathbf{I}_p & 0 & 0\\ 0 & \lambda^2 \mathbf{I}_q & 0\\ 0 & 0 & \mathbf{I}_r \end{array} \right]. \tag {A1} $$

Property of eigenvalues

Let us define $$ z = 1 + \lambda + \lambda^2. $$ We have $$ \lambda z = z \Rightarrow \big( \lambda - 1 \big) z = 0 \Rightarrow \lambda = 1 \vee z = 0. $$ As $\lambda = 1$ excludes (OP), we have $z=0$, thus

$$ \lambda + \lambda^2 = -1. \tag{p1} $$

Conjugation

Let $\mathbf{B}_n$ be any invertible $n \times n$ matrix. We obtain $$ \mathbf{I}_n = \mathbf{B}_n \mathbf{I}_n \mathbf{B}^{-1}_n = \mathbf{B}_n \mathbf{A}^3_n \mathbf{B}^{-1}_n = \left( \mathbf{B}_n \mathbf{A}_n \mathbf{B}^{-1}_n \right)^3. $$

From (A) follows that $\mathbf{A}^\flat_n$ is (not yet) a real matrix. Let us define $$ \mathbf{A}^\sharp_n = \mathbf{B}^\flat_n \mathbf{A}^\flat_n \big( \mathbf{B}^\flat_n \big)^{-1}. $$

If $\mathbf{A}^\sharp_n$ is a real $n \times n$ matrix, then at least we have that BOTH the trace $(\chi)$ and the determinant $(\Delta)$ are real. Thus $$ \chi\big(\mathbf{A}^\sharp_n\big) \in \mathbb{R},\\ \Delta\big(\mathbf{A}^\sharp_n\big) \in \mathbb{R}. $$ As $$ \chi\big(\mathbf{P}\mathbf{Q}\mathbf{P}^{-1}\big) = \chi(\mathbf{Q}),\\ \Delta\big(\mathbf{P}\mathbf{Q}\mathbf{P}^{-1}\big) = \Delta(\mathbf{Q}), $$ we obtain $$ p \lambda + q \lambda^2 + r \in \mathbb{R}, \tag 1 $$ $$ \lambda^{p+2q} \in \mathbb{R}. \tag 2 $$ Using (p1) we can write (1) as $$ (p-q) \lambda + (r-q) \in \mathbb{R}. $$ As $\lambda \ne \mathbb{R}$, we obtain $$ p=q. \tag {c1} $$ And (2) becomes $$ \lambda^{p+2q} = \lambda^{3p} = 1^p = 1 \in \mathbb{R}. $$ Therefore we obtain

$$ \lambda \ne 1, p > 0, 2p+r = n : \mathbf{A}^\flat_n = \left[ \begin{array}{cc} \left[ \begin{array}{cc} \lambda & 0 \\ 0 & \lambda^2 \end{array} \right] \mathbf{I}_p & 0\\ 0 & \mathbf{I}_r \end{array} \right]. \tag {A2} $$

Real solution

As $$ \left[ \begin{array}{cc} \lambda & 0 \\ 0 & \lambda^2 \end{array} \right]^3 = \mathbf{I}_2, $$ we can consider the conjugation $$ \left[ \begin{array}{cc} -\frac{1}{2} & \pm \frac{\sqrt{3}}{2} \\ \mp \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{array} \right]^3 = \mathbf{I}_2. $$ The general solution can be written as

$$ p > 0, 2p+r = n, \mathbf{B}_n \in \mathbb{R}^n, \exists \mathbf{B}^{-1}_n \in \mathbb{R}^n: $$ $$ \mathbf{A}_n = \mathbf{B}_n \left[ \begin{array}{cc} \left[ \begin{array}{cc} -\frac{1}{2} & \pm \frac{\sqrt{3}}{2} \\ \mp \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{array} \right] \mathbf{I}_p & 0\\ 0 & \mathbf{I}_r \end{array} \right] \mathbf{B}^{-1}_n. $$

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Hint: If we know that $A\vec v = \lambda \vec v$, what else can we deduce about $\vec v$? We don't know anything about $\vec v$ other than that it's an eigenvector, so what else can we manipulate?

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Geometrically, consider a matrix describing a rotation by $2\pi/3$, so that $A^3$ is a rotation by $2\pi=Id$. Then the eigenvalues of a rotation are known --and, since $3$ is odd, you have at least one Real eigenvalue; there are just three possible choices for them: http://mathworld.wolfram.com/RotationMatrix.html

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Note that it's minimal polynomial is $x^3-1=0$. Now use the fact that characteristic and minimal polynomial have the same roots, except for multiplicities.

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