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This question already has an answer here:

I have a true/false question:

Every n × n-matrix A with real entries has at least one real eigenvalue.

I am thinking that this is true but I would like to hear other opinions.

Thanks

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marked as duplicate by Martin Sleziak, Andrés E. Caicedo, Claude Leibovici, Avitus, Marc van Leeuwen Apr 14 '14 at 7:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ I wonder if the next T/F question asks the same for odd $n$. $\endgroup$ – Git Gud Apr 13 '14 at 23:42
  • $\begingroup$ On the practice test? I am asking the ones that I am unsure of. $\endgroup$ – A A Apr 13 '14 at 23:44
  • $\begingroup$ Yes. ${{{{}}}}$ $\endgroup$ – Git Gud Apr 13 '14 at 23:45
  • $\begingroup$ Why do you think this is being marked? $\endgroup$ – A A Apr 14 '14 at 0:03
  • $\begingroup$ I don't think that. I'm just wondering what I mentioned in the first comment. $\endgroup$ – Git Gud Apr 14 '14 at 0:08
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Nope. Try a rotation in $\Bbb R^2$.

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    $\begingroup$ I tried the identity rotation and I found the answer is yes:-)+1 $\endgroup$ – user63181 Apr 13 '14 at 23:49
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Since the eigenvalues of a matrix $A\in\mathcal M_n(\Bbb R)$ are roots of its characteristic polynomial then the question is equivalent to say: "every polynomial with real coefficients has a real root" which is wrong.

Remark: The answer is yes if we more assume that $n$ is odd (why?).

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    $\begingroup$ Is there an easy way to see that every polynomial is the characteristic polynomial of some linear transformation? $\endgroup$ – Seth Apr 13 '14 at 23:52
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    $\begingroup$ Do you know the companion matrix? $\endgroup$ – user63181 Apr 13 '14 at 23:55
  • $\begingroup$ Oh yes, thank you, I forgot. $\endgroup$ – Seth Apr 13 '14 at 23:55
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Apr 13 '14 at 23:56
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Here's a counterexample:

\begin{bmatrix} 1 & 1 \\[0.3em] -4 & 1 \\[0.3em] \end{bmatrix}

Being clever, we can construct any number of $2 \times 2$ matrices such that its characteristic polynomial has negative discriminant, and thus has no real roots (eigenvalues).

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Eigenvalues of a $2\times 2$ matrix $M$ are the roots of the characteristic polynomial $\lambda^2 - \text{tr}(M)\lambda + \det(M)$. All you need to do is find a $2\times 2$ matrix where $\text{tr}(M)^2-4\det(M)<0$ to find a counterexample.

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