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Given two invertible matrices $A,B\in M_{2}(\mathbb R)$ such that $B^{-1}AB=A^{2}$, and $1$ is NOT an eigenvalue of $A$.

(1) Find the eigenvalues of A, and

(2) Find $A$ and $B$ satisfying the given conditions.

What I tried is as follows: since $B^{-1}AB=A^{2}$, then $A$ is similar to $A^{2}$, so the set of eigenvalues of $A$ and $A^{2}$ coincide. So, $\{\lambda_{1},\lambda_{2}\}=\{\lambda^{2}_{1},\lambda^{2}_{2}\}$, and we have two cases:

  1. $\lambda_{1}=\lambda^{2}_{1}, \lambda_{2}=\lambda^{2}_{2}$, so $\lambda_{1}=0,1$, but since $A$ is invertible then $\lambda_{1}\neq 0$, so $\lambda_{1}=1$ !! I'm getting confused here!!

  2. $\lambda_{1}=\lambda^{2}_{2}, \lambda_{2}=\lambda^{2}_{1}$, so $\lambda_{1}=\lambda^{4}_{1}$ is a 3rd primitive root of unity.

I don't Know how to find such $A$ and $B$, any help?!

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So case 2. occurs. The eigenvalues of A are j and j^2, an example of such a matrix A is A=[0,-1|1,-1], then A^2=[-1,1|-1,0], hence B=... –  Did Aug 15 '11 at 21:46

1 Answer 1

For part (1), you've shown that case 1 cannot happen.

For part (2), let $\lambda_1$ be some primitive third root of unity, then $\lambda_2 = \lambda_1^2$ is the other primitive third root of unity. The sum of all third roots of unity is $0$, so $\lambda_1 + \lambda_2 = -1$. Also, $\lambda_1 \lambda_2 = \lambda_1^3 = 1$. This gives us the trace and determinant of $A$. It is very easy to construct a matrix (having real entries!) with trace $-1$ and determinant $1$. You can then solve for $B$ in various ways.

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