Show that there is eigenvector that all composition is real number

Question

If $A$ is a real matrix and $\lambda$ is a real eigenvalue of $A$, then there is a real eigenvector corresponding to $\lambda$.

How can I prove this?

Answer 1

Use that $$\det(A-\lambda\operatorname{Id})=0.$$

Answer 2

Use the Row-echelon form on $A-\lambda I$.

Answer 3

If $z\in\Bbb C^n$ is an eigenvector then $z=x+iy$ where $x,y\in\Bbb R$. If the eigenvalue $\lambda$ is real this shows that $$Ax+iAy=\lambda x+i\lambda y,$$hence $Ax=\lambda x$ and $Ay=\lambda y$.