How to find the eigenvalue of matrix A? We have: $$A\left(\begin{array}{l}\xi \\ \eta\end{array}\right) = \left(\begin{array}{l}a\xi+b\eta \\ a\xi-b\eta \end{array}\right)$$
How to find the eigenvalue of matrix $A$?
 A: Hint: if $\vec{x}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A \vec{x} = \lambda \vec{x}$. 
Thus, you need to solve $\begin{bmatrix} \xi \\ \eta \end{bmatrix} = \lambda \begin{bmatrix} a \xi + b \eta \\ a \xi - b \eta \end{bmatrix}$ for $\xi, \eta$. Without loss of generality, you can assume $\xi^2+\eta^2 = 1$ (i.e. $\vec{x}$ is a unit vector) as well so you have 3 equations with 3 unknowns: 
$ \xi = \lambda ( a \xi + b \eta)$
$ \eta = \lambda (a \xi - b \eta)$
$ \xi^2 + \eta^2 = 1$
Solve for $\lambda, \xi, \eta$ in terms of $a$ and $b$. $\lambda$ is the eigenvalue with corresponding eigenvector $\begin{bmatrix} \xi \\ \eta \end{bmatrix}$.
Alternatively, you can write find the matrix representation of $A$ from the equation, and then use the usual formula for eigenvalues. This is probably faster.  
A: Well, now that we see that
$A \begin{pmatrix} \xi \\ \eta \end{pmatrix} = \begin{pmatrix} a \xi + b \eta \\ a \xi - b\eta \end{pmatrix}͵ \tag{1}$
it is clear that the matrix $A$ is given by
$A = \begin{bmatrix} a & b \\ a & -b \end{bmatrix}, \tag{2}$
so to find the eigenvalues we first obtain the characteristic polynomial $p_A(x)$ of $A$:
$p_A(x) = \det(A - xI) =  x^2 - (\text{Tr} A) x + (\det A)I, \tag{3}$
where $\text{Tr} A$ is the trace of $A$.  (3) holds for any $2 \times 2$ matrix $A$; in the specific case at hand we have
$\text{Tr} A = a - b \tag{4}$
and
$\det A = -2ab, \tag{5}$
so the eigenvalues $\lambda$ of $A$ satisfy
$\lambda^2 + (b - a) \lambda - 2ab = 0. \tag{6}$
From (6), by the quadratic formula, 
$\lambda_{\pm} = \dfrac{1}{2}((a - b) \pm \sqrt{(b - a)^2 + 8ab}) = \dfrac{1}{2}((a - b) \pm \sqrt{a^2 + b^2 + 6ab}). \tag{7}$
(7) gives the eigenvalues of $A$͵ hence the question is answered at this point; we note that (7) implies, after some easy algebra, that as long as $a \ne 0 \ne b$ we have $\lambda_{\pm} \ne 0$.   In addition,  the eigenvectors are also easily found, since we have from (1) that
$a \xi + b \eta = \lambda \xi, \tag{8}$
$a \xi - b\eta = \lambda \eta \tag{9}$
for $\lambda = \lambda_{\pm}$; if $b \ne 0$, we may then pick $\xi$ arbitrarily and then by (8) set
$\eta = b^{-1} (\lambda - a) \xi, \tag{10}$
whereas if $a \ne 0$, we may use (9), pick $\eta$ freely, and find $\xi$ via
$\xi = a^{-1} (\lambda + b) \eta. \tag{11}$
In the event that $a = b = 0$, then $A = 0$ and we may choose any two linearly independent vectors as eigenvectors with eigenvalue $0$.
Hope this helps; Cheerio,
and as always,
Fiat Lux!!!
