Eigenvalues from the relation $2M + I = M^2$ $$M=
        \begin{bmatrix}
        0 & 1  \\
        1 & 2  \\
        \end{bmatrix}
$$
How can we use $2M + I = M^2$ to find the eigenvalues of M.
 A: $$det(M - \lambda I) = 0$$
$$ \iff(-\lambda)(2-\lambda) - 1 = 0$$
$$\iff -2\lambda + {\lambda}^2 -1 = 0$$
$$\iff {\lambda}^2 = 2\lambda + 1$$
Notice this is the same as replacing $M$ by $\lambda$ in the relation you wrote.
I'll leave the rest up to you.

Edit
I've just learned that this is a consequence of the Cayley-Hamilton Theorem.
It states that any square $\left (n \times n \right)$ matrix $A$ satisfies its own characteristic polynomial $p(\lambda)$.
A: If we are using row vectors, then $\lambda$ is an eigenvalue if and only if there exists $\vec x\ne\vec 0$ such that
$$\vec x M = \lambda \vec x$$
From this you get:
$\vec x M^2 = \lambda \vec x M = \lambda^2 \vec x$
$\vec x (2M+I) = 2\vec x M + \vec x = 2\lambda\vec x+\vec x = (2\lambda+1)\vec x$
Since we have $(2\lambda+1)\vec x=\lambda^2 \vec x$ for a non-zero vector $\vec x$, we get that
$$2\lambda+1=\lambda^2.$$ 

There is nothing special about the choice of the polynomial in this particular example. To learn more about this, you can read about relation of eigenvalues to minimal polynomial and characteristic polynomial of a matrix.
