# Characteristic polynomial roots of $3\times 3$ orthogonal matrix.

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.

• 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. – Bernard Aug 30 '18 at 0:06
• @Bernard: then the claim holds for all 3x3 matrices? – b_choi Aug 30 '18 at 0:09
• Over the field $\mathbf R$, yes. – Bernard Aug 30 '18 at 0:10
• @Bernard: I see, thanks! – b_choi Aug 30 '18 at 0:12
• @TheoreticalEconomist: That's a nicer, but more sophisticated argument: it supposes you've proved $\mathbf C$ is algebraically closed – which isn't exactly trivial. – Bernard Aug 30 '18 at 8:23