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We are given vectors $a_1, ... , a_n$, which are linearly independent and a set of equations about a linear map $A: \mathbb{R}^n \rightarrow \mathbb{R}^n$;

$Aa_1 = a_2$, $A^2a_1 = a_3$, ... , $A^na_1 = a_1$

and are supposed to find its eigenvalues, eigenvectors and the characteristic polynomial from the given information.

We can see that the set of equations can also be written as:

$Aa_1 = a_2$, $Aa_2 = a_3$, ... , $Aa_n = a_1$

and by summing all of them we find the first eigenvalue to be 1 (and the corresponding eigenvector $a_1 + a_2 + ... + a_n$).

I'm now having a problem finding the other eigenvalues. I wrote down $A$ for a general $x$, which can be written as $x = b_1a_1 + b_2a_2 + ... + b_na_n$, where $b_1, ..., b_n \in \mathbb{R}$, since $a_1, ... , a_n$ are linearly independent in a $n$-dimensional space.

From that I got:

$Ax = A(b_1a_1 + b_2a_2 + ... + b_na_n) = b_1Aa_1 + ... + b_nAa_n = b_1a_2 + ... + b_{n-1}a_n + b_na_1$

so for x to be an eigenvector it has to be

$b_na_1 + b_1a_2 + ... + b_{n-1}a_n = \lambda (b_1a_1 + b_2a_2 + ... + b_na_n)$

which gives another system of equations to find $\lambda$, the eigenvalues. It seems all eigenvalues will be $1$ or $-1$ (I'm just assuming, based on the equations I got), I just have trouble proving it.

Thank you for any help with my problem. :)

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Actually,you can denote $A$ as a special circulant matrix as $J$,e.g. \begin{pmatrix} 0 & 1 & 0 &\cdots & 0\\ 0 & 0 & 1 &\cdots & 0\\ \vdots & \vdots & \vdots &\ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 0 & \cdots & 0 \end{pmatrix} The eigenvalues of $J$ is obvious. The characteristic polynomial would be $f(\lambda)=\lambda^n+(-1)^n$

Hope that my short answer may help.

Correction: $A^T$ $=J$,but it doesn't matter because $A \sim A^T$.($\sim$ means similar).

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First let me answer your question over $\mathbb{C}$.

You have that $A^n=I$ and $A^{n-1}\neq I$ so the minimal and the characteristic polynomials are equal to $X^n-1$.

The eigenvalues are therefore $\zeta^k$ for $k=0,\dots, n-1$, where $\zeta:=\exp\frac{2\pi i}{n}$.

The eigenvector corresponding to $\zeta^k$ is $\sum_{l=1}^{n} \zeta^{k(l-1)}a_l$; this is easily checked.

Now over $\mathbb{R}$ there are only one or two eigenvalues: always $+1$, and $-1$ when $n$ is even. The corresponding eigenvectors are unchanged.

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