Proving Invertibility and Eigenvalues

Solution 1:

Write $A^{2}-A=2I$. Factor out $A$ to get $A(A-I)=2I$.

For the eigenvalues, factor the original polynomial $A^{2}-A-2I$. What do the factors tell you?

Edit: Some details.

Divide both sides of $A(A-I)=2I$ by $2$. We have $$A\cdot \frac{1}{2}(A-I)=I$$ Hence $A$ is invertible with inverse $\frac{1}{2}(A-I)$

For the eigenvalues

$$0=A-A-2I=(A+I)(A-2I)$$

Since the matrix $0$ is singular, either $A+I$ or $A-2I$ must be singular. This means one of them has a vector in the nullspace. Hence,

$$(A+I)v=0$$ or $$(A-2I)v=0$$ for some $v\neq 0$.

Therefore either $1$ or $2$ is, by definition, an eigenvalue