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I´m taking linear algebra, and the professor asked us to get back next class with many methods to solve this exercise. I can´t find even one after 2 hours of thinking, really sad. Could you please help?

I´ve the following matrix "A", which represent a linear transformation:

 a   0  -2a 0 
 2a -3a  0 -2a 
-a   0   3a 0
 0   a  -a  2a

and I need to find what the title says.

Thanks for your help!!!

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  • $\begingroup$ Yes, but how would that help? $\endgroup$ – Luis Oct 6 '14 at 2:47
  • $\begingroup$ Module less than 1 (or 1). $\endgroup$ – Luis Oct 6 '14 at 2:48
  • $\begingroup$ Absolute value. $\endgroup$ – Luis Oct 6 '14 at 2:49
  • $\begingroup$ I think that could work! But requires a lot of computation. I guess there must be some other (more elegant way) to solve this, don´t you? $\endgroup$ – Luis Oct 6 '14 at 2:54
  • $\begingroup$ I´ll try to do something with theorem 2.1, seems to be near... thanks! $\endgroup$ – Luis Oct 6 '14 at 3:01
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Notice that we can write the matrix as $aM$ for some other matrix $M$. Then use the defining equation of an eigenvalue or your favorite computational methods to determine how to compute the eigenvalues of $M$, and apply your restriction.

A priori, there's no reason to believe this will not require quite a bit of computation; though there is at least one invariant subspace that you might be able to see by inspection (or not; it's fairly well-hidden).

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