Let $A,B\in \mathcal{M}_n(\mathbb{C})$.

It is known that if $AB=BA$ and $\lambda_1, \lambda_2, \dots, \lambda_n $ are the eigenvalues of $A$ and $\beta_1, \beta_2, \dots, \beta_n$ are the eigenvalues of $B$ then for any polynomial $P(x, y)$, the eigenvalues of $P(A, B)$ are $P(\lambda_i, \beta_i)$.

My question is, does the theorem also hold true if $AB=-BA$ instead of $AB=BA$?

  • $\begingroup$ If you know the proof of previous result then you can check it by yourself if you can replace the condition by $AB=-BA$ $\endgroup$ – user87543 Feb 10 '16 at 15:47
  • $\begingroup$ I don't know it. It's a theorem and I think it's pretty difficult. $\endgroup$ – sagregravsky Feb 10 '16 at 15:53
  • $\begingroup$ It is reasonable to think about some variation only when you know the proof of present theorem... It is just an advice... $\endgroup$ – user87543 Feb 10 '16 at 15:55

There are several things to be said here, but first: a similar statement cannot hold for anti-commuting matrices. For example, take $$ A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. $$ Then $A$ and $B$ anti-commute and they both have $1$ and $-1$ for eigenvalues. Take $P(x,y)=xy$. Then $P(A,B)=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ has $i$ and $-i$ for eigenvalues, which cannot be obtained by evaluating $xy$ at $\pm 1$.

Now, even if we wanted a statement for anti-commuting matrices, we would need more information. In the commuting case, the proof of the result relies on the fact that $A$ and $B$ can be simultaneously triangularized, i.e. there is an invertible matrix $P$ such that $PAP^{-1}$ and $PBP^{-1}$ are both triangular, with their eigenvalues on the diagonal. This gives us an ordering on the eigenvalues $\lambda_1,\ldots, \lambda_n$ and $\beta_1,\ldots, \beta_n$ which is implicit in your question. In the anti-commuting case, we do not have this, so the (implicit) ordering used in the statement does not exist anymore. Thus, without more information on the ordering of the eigenvalues, the statement is not completely well-defined.

That said, the example above shows that the statement cannot hold, whatever the ordering of the eigenvalues is.


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.