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Let A be a $5 \times 5$ matrix with 4 distinct eigenvalues. Two of its eigenspaces are unidimensional and one of its eigenspaces is bidimensional.

Does the characteristic polynomial of A has a root of multiplicity two?


I think it has. My reasoning is that because an eigenspace is bidimensional, one eigenvalue has geometric multiplicity of two. This implies that the algebraic multiplicity of this eigenvector is (at least) two. From this we may conclude that a root of the characteristic polynomial has multiplicity of at least two. As there are 5 roots and 4 distinct roots, the root which we know has multiplicity of at least two has multiplicity of exactly 2.

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    $\begingroup$ Your reasoning is perfectly fine. Actually the information is about the eigenspaces is redundant. If a polynomial of degree $n$ has $n-1$ distinct roots, then by trivial combinatorics there is precisely one root with multiplicity two. $\endgroup$ – MooS Nov 29 '16 at 8:50
  • $\begingroup$ I do think that the teacher who gave this as an exercise left to the reader in the board meant more "at least 4 distinct eigenvalues" than the text suggests. But even if he meant "at least 4", it's good to see that what I wrote is good enough. Thanks. $\endgroup$ – Bernardo Sulzbach Nov 29 '16 at 8:57
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    $\begingroup$ Ok, fair enough. If there at least 4 distinct eigenvalues, then the informtion about the two-dimensional eigenspace is of course important, $\endgroup$ – MooS Nov 29 '16 at 9:00

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