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Given $A: n\times n$ matrix with eigenvector $w$ for eigenvalue $c$, does $B$, where $B^2 = A$ have $w$ as an eigenvector?

I.e, $A*w = B*B*w = c*w$. Is $w$ an eigenvector of $b$ with eigenvalue $\sqrt{c}$? I know that $A^2*w = A*A*w = A*c*w = c^2*w$ implies $A^2$ has eigenvector $w$.

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  • $\begingroup$ As Crostul said in their answer, this isn't in general true. But it is true for positive definite matrices $A$ and $B$. $\endgroup$ – Josephine Moeller Nov 4 '14 at 21:09
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No. An example is the real square matrix $$B= \left[ \begin{matrix}0 &1 \\ -1 & 0 \end{matrix} \right]$$

which has no real eigenvalues (its characteristic polynomial is $x^2+1$ which has no real roots). However, $B^2 = -1$, so it has $-1$ as an eigenvalue, and all vectors are eigenvectors.

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  • $\begingroup$ I think you made a computation error. B has characteristic equation $x^2-1$ and eigenvalues $1$ and $-1$. $\endgroup$ – Zach Effman Nov 4 '14 at 21:05
  • $\begingroup$ You are right! I will edit. $\endgroup$ – Crostul Nov 4 '14 at 21:06
  • $\begingroup$ You have another mistake: now $B^2=-I$ with eigenvalue $-1$. $\endgroup$ – Reinstate Monica Nov 4 '14 at 21:11
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Crostul's answer is correct but I would like to give some intuition about how this situation is possible. If a matrix $B$ rotates vectors by a right angle then it has no eigenvectors. However when this matrix is applied twice, it inverts all vectors so $-1$ is an eigenvalue.

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