<s>[INST] Proof that both eigenvalues lie inside the unit circle iff $\det < 1$ and $\operatorname{tr }< 2$ Let $A = \begin{pmatrix}a&b \\c &d \end{pmatrix}$, prove that both eigenvalues lie within the unit circle if and only if $\det(A) < 1$ and $|\operatorname{tr}(A)| < 1 +\det(A)$. 
This is what I have done so far:
$|\operatorname{tr}(A)| < 1 +\det(A) \Rightarrow |tr(A)| < 2$.
The characteristic polynomial of A is
$$\lambda^2 - (a+b)\lambda +ab+bc = 0$$
and the solutions are
$$\lambda_{1,2} = \frac{a+d}{2} \pm \sqrt{\frac{(a+d)^2}{4}-(ad-bc)} \qquad (\star)$$
If the eigenvalues $x$ and $y$ are complex then they are also conjugated. So $\det(A) = xy < 1$ is true only if the real part is less then 1 and $|\operatorname{tr}(A)| = |x+y| < 2$ is obviously true then as well (the two complex parts take out each other since conjugated). 
This is the part that I'm a bit uncertain about. If $x$ and $y$ are real then we could express $(\star)$ as $k \pm t$ where both $k$ and $t$ are constants. 
$\det(A) =xy < 1$  gives us two possibilities, either $x$ and $y$ are both less then 1 or if say $x  > 1$ then $y < \frac{1}{x}$.
So if $x = k+t > 1$ and $y = k-t < \frac{1}{k+t} \Leftrightarrow y = \frac{1}{k-t} > k+t \Rightarrow y > 1$ which is a contradiction. Thus both $x$ and $y$ must be less then 1.
Thanks in advance.
 [/INST] Hint:
Assume $x,y \in \mathbb{R}$.
We have $0 \le |x+y| < 1+xy$ which implies that actually $ -1 \le xy < 1$, or $x^2y^2 \le 1$.
Squaring the relation $|x+y| < 1+xy$ gives
$$x^2+y^2 < 1+x^2y^2 \implies (1-x^2)(1-y^2) > 0$$
so $x^2, y^2 > 1$ or $x^2, y^2 < 1$. However, the first possibility contradicts $x^2y^2 \le 1$.
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