6

If $A$ has $k$ distinct eigenvalues $\{\lambda_1,\lambda_2,...,\lambda_k\}$ then does $A^n$ have only $\{\lambda^n_1,\lambda^n_2,...,\lambda^n_k\}$ as eigenvalues?

It is well know that if $\lambda$ is an eigenvalues of $A$ then $\lambda^n$ is an eigenvalues of $A^n$ for $n\in \mathbb{N}$. But I am wondering whether $A^n$ has only $\lambda^n$ as eigenvalues, or are there more?

Math Lover
  • 15,153
Student
  • 9,196
  • 8
  • 35
  • 81

1 Answers1

4

Yes as long as it is diagonlizable. $$A = P \Lambda P^{-1}$$ Then, $$A^n = P \Lambda^n P^{-1}$$ I am not sure if that is possible in case it is not diagonlizable.

edit: it is possible even if not diagonlizable $$Ax=\lambda x$$ $$AAx=\lambda Ax$$ $$A^2x=\lambda^2 x$$ and so on.

Kareem Metwaly
  • 162
  • 2
    $A$ just need to be triangularizable – Baconaro Jan 09 '18 at 21:17
  • @Aaron: no. $\pmatrix{0&1\0&0}$ is not diagonalizable over any field. Presumably you mean "triangularizable". – Ben Grossmann Jan 09 '18 at 21:32
  • @Omnomnomnom yes. That was a typo. Proof shows triangularizabolity (which is why the last word is triangular). Will delete and repost with correction (since i'ts been too long to edit) – Aaron Jan 09 '18 at 21:41
  • @Baconaro But over an algebraically closed field, every matrix is triangularizable. Indeed, Let $A:V\to V$. The characteristic equation of $A$ has at least one root, and hence we have at least one eigenvector $v$. Then $A$ descends to an endomorphism of $V/\mathbb kv$, and an induction on dimension of $V$ shows that $v$ is part of a basis in which $A$ is triangular. – Aaron Jan 09 '18 at 21:45
  • @Baconaro according to the posted question $A$ has distinct eigen-values thus it is diagonlizable. – darkmoor Jan 10 '18 at 15:54