Proving that $\det (A^2 - I) < 0 \Rightarrow \lambda \in (-1,1)$

Question

Let $A$ be real square matrix. If $\det (A^2 - I) < 0$, then $A$ has an eigenvalue $\lambda \in (-1,1)$.

How to prove this?

Answer 1

Hint :

$\det(A^2-I)<0\Rightarrow \det(A+I)\cdot \det(A-I)<0$

Consider $f(x)=\det(A+xI)$ and consider case of $\det(A+I)>0$ and $\det(A-I)<0$

• What is $f(1)$
• What is $f(-1)$

Use Intermediate value theorem for this $f(x)$ and complete the rest...

Answer 2

Hint: We have, if $\lambda_i$, $1 \le i \le n$ denote the eigenvalues of $A$, $$ \det(A^2 - 1) = \prod_{i=1}^n (\lambda_i^2 - 1) $$