Knowing $M^2 + M = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, find the eigenvalues of $M$ 
Let $M \in M_2(\mathbb R)$ and $$A = \begin{pmatrix} 1 & 1 \\  1 & 1 \end{pmatrix}$$ such that $$M^2 + M = A$$ Find the eigenvalues of $M$.

Here is what I did:
Let $\lambda$ be an eigenvalue of $M$, and $x \in \operatorname{Ker}(\lambda I_2 - M)$.
Thus $$Ax = (\lambda ^2 + \lambda)x.$$
We found that $\lambda ^2 + \lambda$ is an eigenvalue of $A$. 
We can calculate $\operatorname{Sp}(A)$, the eigenvalues of $A$: those are the roots of $\det(XI_2 - A)$, i.e: $$ \operatorname{Sp}(A) = \left\{0, 2\right\}.$$
By solving $\lambda ^2 + \lambda =0$ and $\lambda ^2 + \lambda =2$, we find that $\lambda \in S = \left\{-2, -1,0, 1\right\}$.
I can't figure out what are the eigenvalues of $M$, all I know if that they are element of $S$, any help would be appreciated.
 A: We know more than just the fact that $M$ has eigenvalues in $\{-2,-1,0,1\}$. If $M$'s eigenvalues are $\lambda_1, \lambda_2$, we know that $M^2+M$ has eigenvalues $\lambda_1^2+\lambda_1$ and $\lambda_2^2+\lambda_2$. Since $M^2+M$ has eigenvalues $0$ and $2$, we know that one of $M$'s eigenvalues (say, $\lambda_1$) has $\lambda_1^2+\lambda_1 = 0$ and the other eigenvalue has $\lambda_2^2+\lambda_2 = 2$.
So one eigenvalue is $0$ or $-1$, and the other eigenvalue is $-2$ or $1$. 
All four combinations are possible. Write the matrices in the basis $\mathcal B$ of the eigenvectors $(1,1)$ and $(1,-1)$: in this basis, we have
$$
   (M_{\mathcal B})^2 + M_{\mathcal B} = \begin{bmatrix}2 & 0 \\ 0 & 0\end{bmatrix}
$$
and so any of the possibilities
$$
   M_{\mathcal B} \in \left\{\begin{bmatrix}-2 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}-2 & 0 \\ 0 & -1\end{bmatrix}, \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\right\}
$$
will work. Convert these to the standard basis and you find four possibilities for $M$, each corresponding to one of the four cases for $M$'s two eigenvalues.
(It's worth spelling out in detail why this argument still makes sense without assuming that $M$ is diagonalizable. We don't need to assume $\lambda_1 \ne \lambda_2$ for the statement about the eigenvalues of $M^2+M$ to hold: if $\lambda_1 = \lambda_2$ and this eigenvalue has algebraic multiplicity $2$, then  $M^2+M$ has $\lambda_1^2+\lambda_1 = \lambda_2^2+\lambda_2$ as an eigenvalue with algebraic multiplicity $2$. This is then contradicted by our knowledge of the eigenvalues of $M^2+M$.)
