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It seems, although I cannot prove, that if we have a matrix $A$ with the eigenvalue $a$ that the eigenvalues $\lambda$ of $$(A+A^n)x = \lambda x \implies \lambda = a+a^n$$

The last term follows by $Ax=\lambda x \implies A^nx=\lambda^nx$.

This however, seems not to be true for $(A+B)x=\lambda x$ iff $B\not=A$.

  • Is there any theorem to prove this? If not, what's an example where this is not true?

Added own thoughts:

Is the above perhaps the results of the fact that $A^n$ has the same eigenvectors as $A$? Whereas $A$ and $B$ does not need to have the same eigenvectors.

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  • $\begingroup$ If $A$ is a matrix with eigenvalue $\lambda$ then for any polynomial $f(T)$ the matrix $f(A)$ will have eigenvalue $f(\lambda)$. Your counterexample is useless because you're adding a random different matrix $B$ to $A$ and trying to keep the same eigenvalue. $\endgroup$ – KCd Mar 7 '15 at 20:15
  • $\begingroup$ Yes, sorry. I wasn't trying to provide a counter example, just making observations that might show to become clearer for me once I learned why the first case behaves as it does. However, I'm still not quite convinced. $\endgroup$ – B. Lee Mar 7 '15 at 20:17
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    $\begingroup$ Your statement would be true for any pair of simultaneously diagonalizable matrices. $\endgroup$ – Michael Burr Mar 7 '15 at 20:20
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    $\begingroup$ Ah, you are asking a harder question than I thought. Theorem 2.1 of math.uconn.edu/~kconrad/blurbs/galoistheory/tracenorm2.pdf is related to your question. $\endgroup$ – KCd Mar 7 '15 at 20:20
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Your last thought is indeed accurate. We would need to use something like "$A$ and $A^n$ have the same eigenvectors" before concluding that $(A + A^n)x = \lambda x$ implies that $x$ is an eigenvector of both $A$ and $A^n$.

In fact, $A$ and $A^n$ might not have all the same eigenvectors ($A^n$ might have some extras), so we're in even more trouble than you might have thought.

The usual way of bypassing this mess is to use the fact that any matrix $A$ can be upper-triangularized, either using Schur factorization or Jordan canonical form. Barring that, you could trace through an argument like this one.

Another typical (though perhaps sophisticated) argument that may work here is to use the fact that the set of the diagonalizable matrices is dense in $\Bbb C^{n \times n}$ and the continuous dependence of eigenvalues on matrix entries.

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