# Every n × n-matrix A with real entries has at least one real eigenvalue. [duplicate]

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 Caicedo, Claude Leibovici, Avitus, Marc van LeeuwenApr 14 '14 at 7:18

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

Nope. Try a rotation in $\Bbb R^2$.

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I tried the identity rotation and I found the answer is yes:-)+1 – user63181 Apr 13 '14 at 23:49

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

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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