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How would you approach this problem?

Let $A$ be an orthogonal $3$ by $3$ matrix. That is, $A^TA = AA^T=I_3$. Prove that the characteristic polynomial $\textit{p}_A$ has a real root.

I am not familiar with how to prove a third degree polynomial has a real root. I started the problem by noticing that $\det(A-tI) = \det(A-tAA^T)=\det(A)\det(I-tA^T)$, but this isn't getting me anywhere.

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  • $\begingroup$ Welcome to Maths SX! All cubic polynomials have (at least) one real root, and this has nothing to do with orthogonal matrices: it relies on the Intermediate value theorem. $\endgroup$ – Bernard Aug 30 '18 at 0:06
  • $\begingroup$ @Bernard: then the claim holds for all 3x3 matrices? $\endgroup$ – b_choi Aug 30 '18 at 0:09
  • $\begingroup$ Over the field $\mathbf R$, yes. $\endgroup$ – Bernard Aug 30 '18 at 0:10
  • $\begingroup$ @Bernard: I see, thanks! $\endgroup$ – b_choi Aug 30 '18 at 0:12
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    $\begingroup$ @TheoreticalEconomist: That's a nicer, but more sophisticated argument: it supposes you've proved $\mathbf C$ is algebraically closed – which isn't exactly trivial. $\endgroup$ – Bernard Aug 30 '18 at 8:23
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As you mentioned the characteristic polynomial of your matrix is a third degree polynomial and every third degree polynomial has a real root.

You do not have to prove that every third degree polynomial has a real root to solve your problem because it is a well-known fact by just looking at the end behavior of the polynomial.

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