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Let $A$ and $B$ be real $n\times n$ matrices with $ABA=A$. How can I prove $\operatorname{tr} AB\le n$?

  • Welcome to Math.SE, hami, if this is your first visit. Are you able to provide any more context to this Question? Were you able to solve any special cases, or could an application of this be of use to you? – hardmath Feb 10 '15 at 00:21
  • Thanks for the reply, but I'd prefer you added some information to the question itself. Unless you simply copied a problem you found on the Internet here, without giving any thought to how you could solve it, you will have some of your own efforts to add to the problem statement. – hardmath Feb 10 '15 at 01:05
  • @hardmath If you check my question is a more general question deduced and my thoughts can be found in comments below. So one can regard it as a request for hints. –  Feb 10 '15 at 01:15
  • Your Question is but a single line. I think you are asking a lot of your Readers to regard it as a request for hints. More about how to ask good questions is in the Help Center FAQ. – hardmath Feb 10 '15 at 01:17

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$$ABAB = AB{{{{{{{{}}}}}}}}$$hence $$ tr AB = rank( AB )\le n $$

mookid
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Hint: show that the eigenvalues of $AB$ are all $0$ or $1$.

Dustan Levenstein
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  • OK, I proved all eigenvalues satisfy $(1-x)^n=0$. Is trace the sum of them? –  Feb 09 '15 at 22:45
  • @hami yes, since the trace is invariant under base change. – hjhjhj57 Feb 09 '15 at 22:55
  • @hjhjhj57: Do you mean $AB$ similar to a diagonal matrix with eigenvalues on diagonal? –  Feb 09 '15 at 23:01
  • @hami if the matrix is diagonalizable, yes. – hjhjhj57 Feb 09 '15 at 23:02
  • @hjhjhj57 How if $AB$ is not diagonalizble. Is the trace still the sum of eigenvalues? –  Feb 09 '15 at 23:04
  • @hami Yes; you can always put a matrix into upper-triangular form if you allow for complex entries, and then the eigenvalues can be read right off the diagonal, and the trace is their sum. – Dustan Levenstein Feb 09 '15 at 23:14
  • Btw, it seems any idempotent matrix is diagonalizable. –  Feb 09 '15 at 23:32
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    Yes, that's correct, because the minimal polynomial has no multiple roots. The particular case of an idempotent has a cute proof that I think you linked to below (I'm on my phone). – Dustan Levenstein Feb 10 '15 at 00:31