Proof that both eigenvalues lie inside the unit circle iff $\det < 1$ and $\operatorname{tr }< 2$

Question

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.

Answer 1

Your proof attempt lacks a few things. I suggest write your assumptions and what do you want to show with this assumptions to avoid using $y \Rightarrow x$ statements instead of $x \Rightarrow y$, which are very different things.

For the proof it is a good idea break it to cases if you think it is useful (as you already did).

1. Assume $\lambda_{1,2} = \alpha \mp i\beta$ with $\beta \neq 0$.
   1. ($\Rightarrow$) Assume $|\lambda_i|<1$. Then,
      1. $\det(A)= \lambda_1 \lambda_2 = \lambda_1 \bar{\lambda}_1 = |\lambda_1|^2 < 1$.
      2. $|\operatorname{tr}(A)| = |\lambda_1 + \lambda_2| = 2|\alpha| < 1 + \alpha^2 < 1 + \alpha^2 + \beta^2 = 1 + \det(A) $ where we used the fact that $0 < (1 - |\alpha|)^2$. (Note: showing $|\operatorname{tr}(A)|<2$ is not enough in this step.)
   2. ($\Leftarrow$) Assume $\det(A) < 1$ and $|\operatorname{tr}(A)| < 1 + \det(A)$. Then, $\det(A)= \lambda_1 \lambda_2 = \lambda_1 \bar{\lambda}_1 = |\lambda_1|^2 < 1$, which implies $|\lambda_i|<1$.
2. Assume $\lambda_{1,2} \in \mathbb{R}$.
   1. ($\Rightarrow$) Assume $-1 < \lambda_i < 1$. Then,
      1. $\det(A)= \lambda_1 \lambda_2 < 1$.
      2. $|\operatorname{tr}(A)| = |\lambda_1 + \lambda_2| < 1 + \lambda_1 \lambda_2=1+\det(A)$, where we used $(1-\lambda_1)(1-\lambda_2) > 0$ or $(1+\lambda_1)(1+\lambda_2) > 0$ depending on the sign of $\lambda_1 + \lambda_2$.
   2. ($\Leftarrow$) Assume $\det(A) < 1$ and $|\operatorname{tr}(A)| < 1 + \det(A) \Rightarrow |\lambda_1 + \lambda_2| = 1 + \lambda_1 \lambda_2$.
      1. Assume $\operatorname{tr}(A)\geq 0$, which implies $(1-\lambda_1)(1-\lambda_2) > 0$. So either $\lambda_{1,2}<1$ or $\lambda_{1,2}>1$. But since $\det(A)<1$ it must be $\lambda_i < 1$.
      2. Assume $\operatorname{tr}(A)<0$, which implies $(1+\lambda_1)(1+\lambda_2) > 0$. So either $\lambda_{1,2}<-1$ or $\lambda_{1,2}>-1$. But since $\det(A)<1$ it must be $\lambda_i > -1$.

Answer 2

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 $x^2y^2 \le 1$.