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

Question

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?

Answer 1

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 $\mathbb{R}^3$.