Showing $| \lambda_i | < 1 \implies \operatorname{tr}(A) - 1 < \det(A) <1$

Question

Prove that every eigenvalue of $A \in \mathbb{R}^{2 \times 2}$ has modulus less than 1 if and only if $$ \operatorname{tr}(A)-1 < \det(A) <1. $$

I can prove the forward direction, but not through steps that can be reversed so I'm stuck on proving the other directing, i.e. assuming the inequality holds and proving that the eigenvalues have modulus less than 1.

I know $\operatorname{tr}(A)=\lambda_1 + \lambda_2$ and $\det(A) = \lambda_1 \lambda_2$, assuming the matrix has two eigenvalues.

Answer 1

One way:

Let $a,b$ be eigenvalues, then we have to prove $$a+b-1\leq ab$$

but this is the same as $$0\leq (a-1)(b-1)$$ which is true if $a,b$ are real.

If $a,b$ are nonreal then $b=\overline{a}$ then we have to prove $$0\leq (a-1)\overline{(a-1)} = |a-1|^2$$ which is also true.

And $$\det A\leq |\det A| = |ab| < 1$$

Vice versa: Since $$0\leq (a-1)(b-1)$$ is true, then if $a,b $ are bot real then both are $>1$ or $<1$. Say both are $>1$. Then $ab>1$ which is not true since $ab = \det A <1$. So both are $<1$.

If $a,b$ are nonreal, then $b=\overline{a}$, so $$|a|^2 =|b|^2 = ab = \det A <1$$ and we are done.

Answer 2

Hint: $$\operatorname{tr} A - 1 - \det A = (1-\lambda_1)(\lambda_2 - 1).$$ Also, the eigenvalues can be complex and $A$ must have have precisely two eigenvalues. Note that these inequalities can hold when both of them are complex.