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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=...2011-08-15

1 Answers 1