Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $P$ and $Q$ be two Hermitian complex idempotent matrices ($i.e.$ $P^* = P$ and $P^2 = P$ and likewise for $Q$) such that $PQ - QP \neq 0$. Define the matrices \begin{equation} A = P + Q \hspace{0.5cm} \text{ and } \hspace{0.5cm} B = P - Q. \end{equation} Because of the idempotency of $P$ and $Q$, we can write \begin{equation} B^2 = 2A - A^2 \end{equation} Now let $u$ be an eigenvector of $A$ with eigenvalue $\lambda \neq \{ 0, 1, 2\}$ so that $Au = \lambda u$. This implies then that $u$ is also an eigenvector of $B^2$ because $B^2 u = (2\lambda - \lambda^2) u$. Here is my problem: I learned in linear algebra that squaring a matrix does not changes the eigenvectors; however, $u$ is not an eigenvector of $B$. I verified this numerically; in octave, I computed the quotients $u^*B^2u/(u^*u)$ and $u^*Bu/(u^*u)$ and by doing this for all eigenvectors of $A$ I get the correct spectrum only for $B^2$ but not for $B$ (only the eigenvectors associated with eigenvalues equal to one or zero are correct).

Can someone explain me why is $u$ an eigenvector of $B^2$ but not of $B$? What is wrong in my analysis? Thanks very much.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

"Squaring a matrix does not changes the eigenvectors" means that if $u$ is an eigenvector of $B$, then it is also an eigenvector of $B^2$. The converse is not true, however, as shown in your findings. Here is another counterexample. Let $B=\begin{pmatrix}1&1\\0&-1\end{pmatrix}$. Then $B^2=I$. Clearly every eigenvector of $B$ is also an eigenvector of $B^2$ (in fact, every nonzero vector is an eigenvector of $B^2=I$), but the converse is patently false.

share|improve this answer
    
+1. I think now I understand exactly where my problem was: since all my matrices are Hermitian, I was looking at $B^2$ as $B^2 = U^*\Lambda U$ and $B = U^* \Lambda^{1/2} U$. However, I failed to recognize that (in contrast with $\Lambda$) $U$ and $\Lambda^{1/2}$ would not be uniquely defined. I guess the lesson from here is that is never safe to assume $B^2 u = \lambda u \implies Bu = \lambda^{1/2} u$. –  Goku Feb 6 '13 at 16:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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