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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, Andres Caicedo, Claude Leibovici, Avitus, Marc van Leeuwen Apr 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.

4  
I wonder if the next T/F question asks the same for odd $n$. –  Git Gud Apr 13 at 23:42
    
On the practice test? I am asking the ones that I am unsure of. –  A A Apr 13 at 23:44
    
Yes. ${{{{}}}}$ –  Git Gud Apr 13 at 23:45
    
Why do you think this is being marked? –  A A Apr 14 at 0:03
    
I don't think that. I'm just wondering what I mentioned in the first comment. –  Git Gud Apr 14 at 0:08

4 Answers 4

up vote 12 down vote accepted

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

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3  
I tried the identity rotation and I found the answer is yes:-)+1 –  Sami Ben Romdhane Apr 13 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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2  
Is there an easy way to see that every polynomial is the characteristic polynomial of some linear transformation? –  Seth Apr 13 at 23:52
3  
Do you know the companion matrix? –  Sami Ben Romdhane Apr 13 at 23:55
    
Oh yes, thank you, I forgot. –  Seth Apr 13 at 23:55
    
You're welcome. –  Sami Ben Romdhane Apr 13 at 23:56
    
Well done, Sami! –  amWhy Apr 14 at 11:38

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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