Finding trace of a matrix by minimal polynolmial

Question

Let $M\in M_{10}\mathbb(R)$ be real matrix which satisfy the below equation: $$A^{2}=A+2I.$$ If the rank of $A+I$ is 3, find $tr(A)$.

Answer 1

$A$ satisfies a squarefree polynomial $\lambda^2 - \lambda - 2 = (\lambda - 2)(\lambda + 1), \; $ so this must be the minimal polynomial, and the eigenvalues are $2,-1.$ Note that $A$ is diagonalizable, we might as well assume it is diagonal.

It says the rank of $A+I$ is three. This means $2$ occurs 3 times, while $-1$ occurs 7 times.

$$ 3 \cdot 2 + 7 \cdot (-1) = 6 - 7 = -1 $$

Answer 2

$\quad$ Minimal polynomial of $A$ divides the polynomial $x^2-x-2$, which splits completely over $\mathbb{R}$ with two roots -1 and 2, both with algebraic multiplicity one.
So, $A$ is diagonalizable. Let $D=P^{-1}AP$, where $D$ is diagonal.
Then $tr(A)=tr(D)$ and diagonal entries of $D$ are either -1 or 2.
But, since $rank(D+I)=rank(A+I)=3$, so the number of -1 has to be 7.
$$\text{Hence,$\quad$} tr(A)=tr(D)=7\times(-1)+3\times 2 = -1.$$