# Prove that a 3x3 matrix always has an eigenvector in $\mathbb R^3$ [duplicate]

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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?

## marked as duplicate by José Carlos Santos linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 13 at 22:20

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• This is not the same question as the linked one. Please read more carefully before markiing a question as duplicate. – amsmath Apr 13 at 22:25

## 1 Answer

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 $$\mathbb{R}^3$$.

• @amsmath -- Read the title again – uniquesolution Apr 13 at 22:24
• @amsmath -- Sorry, I don't understand your comments. Are you a machine? Please prove that you are not. – uniquesolution Apr 13 at 22:28
• 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? – uniquesolution Apr 13 at 22:35
• 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 $\mathbb R^3$. – amsmath Apr 13 at 22:39
• @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. – Mark Apr 13 at 22:46