Eigenvalues of matrix $A$ $n \times n$, knowing that $A^2+A-6I_n=O_n$

Question

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?

Answer 1

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.

Answer 2

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 $2$.

Answer 3

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 $\lambda \in \{-3,2\}.$

Answer 4

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.