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This question already has an answer here:

I am trying to prove this statement. I am assuming that if a 3x3 matrix always has an eigenvector, then it also always has an eigenvalue. I tried to prove this looking at a general 3x3 case and trying to calculate det(A-$\lambda$I)=0, but it does not get me anywhere. Is there something intuitive that I am missing?

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marked as duplicate by José Carlos Santos linear-algebra Apr 13 at 22:20

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    $\begingroup$ This is not the same question as the linked one. Please read more carefully before markiing a question as duplicate. $\endgroup$ – amsmath Apr 13 at 22:25
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The characteristic polynomial is a cubic polynomial. Every cubic polynomial with real coefficients has at least one real root. Hence every real $3\times 3$ matrix has at least one real eigenvalue, and obviously, a corresponding eigenvector in $\mathbb{R}^3$.

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    $\begingroup$ @amsmath -- Read the title again $\endgroup$ – uniquesolution Apr 13 at 22:24
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    $\begingroup$ @amsmath -- Sorry, I don't understand your comments. Are you a machine? Please prove that you are not. $\endgroup$ – uniquesolution Apr 13 at 22:28
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    $\begingroup$ omg. We are in $\mathbb{R}^3$. Right? (read the question). The definition of the concept "eigenvalue" involves the concept of an eigenvector, right? So what is it you want from me, dude? $\endgroup$ – uniquesolution Apr 13 at 22:35
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    $\begingroup$ Ok, I give it up. @briiibriiiii If $Ax = \lambda x$ with $x\in\mathbb C^3$ and $\lambda\in\mathbb R$, then consider the vector $\overline x$, where $\overline x$ has the complex conjugates of $x$ as entries. Since $A$ has only real entries, $A\overline x = \overline{Ax} = \overline{\lambda x} = \lambda\overline x$. So, $A(x+\overline x) = Ax + A\overline x = \lambda x + \lambda\overline x = \lambda(x+\overline x)$. Hence $x+\overline x$ is an eigenvector of $A$ in $\mathbb R^3$. $\endgroup$ – amsmath Apr 13 at 22:39
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    $\begingroup$ @amsmath You don't need to prove that. There is a general theorem: if $A$ is a matrix in $M_n(F)$ (when $F$ is a field) then a scalar $\lambda\in F$ is an eigenvalue of $A$ if and only if it is a root of the characteristic polynomial. By $\lambda$ is an eigenvalue I mean it is an eigenvalue in in the field $F$, which by definition means that it has an eigenvector in $F^3$. You don't need the field to be algebraically closed or something. $\endgroup$ – Mark Apr 13 at 22:46

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