Eigenvalues of matrix $A$ $n \times n$, knowing that $A^2+A-6I_n=O_n$ How can I find ALL eigenvalues of matrix $A$, $n \times n$, $n \ge 3$, knowing that 
$$A^2+A-6I_n=O_n.$$
I applied the Cayley-Hamilton theorem, and obviously two of the eigenvalues are $-3$ and $2$.
However, how do I find the rest of them?
 A: I will use a small result related to minimal polynomial. 
Result: The characteristic polynomial and the minimal polynomial have the same roots, possibly with different multiplicities. 
Now, $x^2+x-6=(x+3)(x-2)$ annihilates $A$. So, the minimal polynomial of $A$ divides $(x+3)(x-2)$. So, we have three options for the minimal polynomial : $(x+3)(x-2), (x+3)$ and $(x-2)$. In all the cases we get $-3$ and $2$ are the only possible eigenvalues.
A: Let's try to replicate the minimal polynomial theorem, though in this problem, nothing says that the given polynomial is minimal.
Let $\lambda$ be an eigenvalue of $A$ and $v$ an eigenvector of $\lambda$. Then 
$$ A^i v = \lambda ^i v, \text{ for } i=0,1,2,\dots$$
Thus
$$\lambda^2 + \lambda -6 =0.$$
Thus the only two possible eigenvalues are $-3$ and $2$. On the other hand, since
$$(A+3I)(A-2I) = 0,$$
So if $\mathop{\mathrm{Im}}(A-2I) \neq 0$, that is $A\neq 2I$, then $\mathrm{Ker}(A+3I)\neq 0$ or $-3$ is an eigenvalue. Similarly argument for eigenvalue $2$.
A: If $ \lambda$ is an eigenvalue of $A$, then $\lambda^2+\lambda-6$ is an eigenvalue of $A^2+A-6I_n$, hence $\lambda^2+\lambda-6=0$, which gives $\lambda \in \{-3,2\}.$
A: It helps here to know the basic result that if a matrix $A \in M_n(\Bbb F)$ for some field $\Bbb F$ satisfies the polynomial
$p(x) = \displaystyle \sum_0^n p_i x^i \in \Bbb F[x], \tag 1$
so that
$p(A) = 0, \tag 2$
then the eigenvalues $\lambda$ of $A$ also satisfy $p(x)$:
$p(\lambda) = 0; \tag 3$
this is easy to see for if
$A \vec x = \lambda \vec x, \tag 4$
then it is easy to further see that
$A^n \vec x = \lambda^n \vec x, \; n \in \Bbb N, \tag 5$
which easily leads to
$\displaystyle \sum_0^n p_i \lambda^i \vec x = \sum_0^n p_i A^i \vec x = p(A) \vec x = 0; \tag 6$
in the present case this indicates that the eigenvalues of $A$ satisfy
$\lambda^2 -3\lambda + I = 0; \tag 7$
since the entire set of roots of this quadratic is $\Lambda = \{2, -3 \}$, the only possible eigenvalues of $A$ are in this set; but membership in $\Lambda$ doesn't force a number to be an eigenvalue; for example, if 
$A = -3I, \tag 8$
$A$ obeys (7) but the only eigenvalue is $-3$; if, on the other hand,
$A = \begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix}, \tag 9$
then every element of $\Lambda$ is an eigenvalue.
In short, we can find every possible eigenvalue by solving (2), which in this case is (7); but some of those so generated may be extraneous.  To discover which numbers $\lambda$ actually are eigenvalues, we need to solve the characteristic polynomial equation
$\det(A - \lambda I) = 0, \tag{10}$
which contains complete information about which numbers are, and which are not, eigenvalues.
