# How to find eigenvalues from a set of equations of a linear transformation

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. :)

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).

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.