Is it possible to find a $3 \times 3$ matrix satisfying $p(t)=t^2+t+1$

Question

Is it possible to find a $3 \times 3$ real matrix satisfying

1) $p(t)=t^2+t+1$

2) $p(t)=t^2+1$

My intuition says no for both the polynomial but how do I prove that?

Answer 1

No, it is not possible. $p(t)$ in both cases has complex conjugate roots, thus, due to the fundamental theorem of algebra, the third eigenvalue $\lambda$ of the original matrix $A$ exists and it must be real. It means that $$ \exists x\ne 0:\quad Ax=\lambda x, $$ but $$ (A^2+A+I)x= A^2x+Ax+Ix=\lambda^2 x+\lambda x+x=(\lambda^2+\lambda+1)x, $$ i.e. $\lambda^2+\lambda+1$ (which is $\ne 0$) is an eigenvalue of $A^2+A+I$. This is impossible for $A^2+A+I=0$.