For a matrix ($2 \times 2$) $A$ satisfing (1.), (2.), (3.), show that $-2\leq a+d\leq 2$ I would appreciate if somebody could help me with the following problem:
Question: For a matrix ($2 \times 2$) $A$  satisfing (1), (2), (3):  
(1.) $A=\begin{pmatrix} a& b\\ 
c&d\end{pmatrix},(a,b,c,d \in\mathbb{R});$
(2.) $ad-bc=1;$ 
(3.) $A^n=\begin{pmatrix} 1& 0\\
0&1\end{pmatrix},$ (for some integer $n$), 
Show that 
$$-2\leq a+d\leq 2.$$
I tried by using C-H formula ($A^2-(a+d)A+(ad-bc)E=\mathbb{O}$) but couldn’t get it that way.
 A: Let $\lambda_1$, $\lambda_2$ be the eigenvalues of $A$.  The characteristic polynomial $p_A(\lambda)$ of $A$ is
$p_A(\lambda) = \det(A - \lambda I) = \lambda^2 -(a + d)\lambda + (ad - bc) = \lambda^2 -(a + d)\lambda + 1, \tag{1}$
since we are given that $ad - bc = 1$.  Furthermore, we know that $p_A(\lambda)$ factors as
$p_A(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) = \lambda^2 - (\lambda_1 + \lambda_2) + \lambda_1 \lambda_2 = 0; \tag{2}$
comparing (1) and (2) we see that
$\lambda_1 + \lambda_2 = a + d \tag{3}$
and
$\lambda_1 \lambda_2 = 1; \tag{4}$
suppose next that $\lambda_1 = \lambda_2$; then each must be real, since otherwise we would have $\lambda_2 = \bar \lambda_1 \ne \lambda_1$ (this since the coefficients of $p_A(\lambda)$ are real).  Then (4) becomes
$\lambda_1^2 = \lambda_2^2 = 1, \tag{5}$
from which we infer, since $\lambda_1 = \lambda_2$, 
$\lambda_1 = \lambda_2 = \pm 1, \tag{6}$
and thus, by (3),
$a + d = \pm 2; \tag{7}$
the desired conclusion holds in this case.  Proceeding to the case $\lambda_1 \ne \lambda_2$, we note from the hypothesis
$A^n = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I, \tag{8}$
that for any eigenvector $v \ne 0$
$v = Iv = A^n v = A^{n - 1} Av = A^{n - 1} (\lambda v) = \lambda A^{n - 1} v = \ldots = \lambda^{n - 1} Av = \lambda^n v; \tag{9}$
thus
$(\lambda^n - 1) v = 0, \tag{10}$
forcing
$\lambda_i^n = 1, \; \; i = 1, 2. \tag{11}$
For real $\lambda_i$, (11) together with the assumption $\lambda_1 \ne \lambda_2$ implies
$\lambda_2 = - \lambda_1 = \pm 1 \tag{12}$
for $n$ even; for $n$ odd there are no real solutions of (11) with  $\lambda_1 \ne \lambda_2$.  In this event, we have from (3) that
$a + d = 0, \tag{13}$
and the desired conclusion holds once again.  Finally, for $\lambda_1 \ne \lambda_2$ complex, we have $\lambda_2 = \bar \lambda_1$ whence by (4)
$\vert \lambda_1 \vert^2 = \lambda_1 \bar \lambda_1 = 1; \tag{14}$
$\lambda_1$, thus being unimodular, may be written
$\lambda_1 = e^{i\theta} = \cos \theta + i \sin \theta \tag{15}$
for some real $\theta$; then
$\lambda_2 = e^{-i\theta} = \cos \theta - i \sin \theta, \tag{16}$
whence from (3)
$a + d = 2\cos \theta, \tag{17}$
and hence, since $-1 < \cos \theta < 1$,
$-2 < a + d < 2. \tag{18}$
This final result shows the desired conclusion binds in all possible cases.  QED.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
