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 $w$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.