If $A^2 + A +I = 0$, then A is diagonalizable. True or false?

Question

Let $A$ be a real square matrix. We assume $A^2 + A +I = 0$. Then A is diagonalizable.

Is this statement true or false? Prove it.

I don't know which could be the first step to prove that. Could you give me any hint?

Thanks in advance.

Answer 1

$A$ cannot be diagonalizable. If $P^{-1}AP=D$, where $D$ is diagonal matrix, one easily derives that $D^2+D+I=0$, so for every element $\lambda$ on the diagonal we would have $\lambda^2+\lambda+1=0$. However, this equation has no real solutions.

Answer 2

By definition, the minimal polynomial of $A$ divides $x^2+x+1$. But this last polynomial is irreducible over $\Bbb R$. So $m_A(x) = x^2+x+1$ and $A$ is not diagonalizable, as $m_A(x)$ is not the product of distinct linear factors.

Answer 3

Hint:

Consider $A=\pmatrix{-\dfrac12&-\dfrac{\sqrt3}2\\\dfrac{\sqrt3}{2}&-\dfrac12},$ which rotates $120^\circ$.

Answer 4

Yes, A is diagonalizable with non-real eigenvalues.

For example, consider $$A=\begin{bmatrix} -2 & -3 \\ 1 & 1 \end{bmatrix}$$that satisfies $A^2+A+I=0$, it is diagonalizable. It being non-symmetric (non-Hermitian) has non-real eigenvalues as $w, w^2$ where $w$ is one cube root of unity.