The problem of interest from Gilbert Strang's Introduction to Linear Algebra Section 5.1 is as follows.
- Prove that every orthogonal matrix ($Q^TQ = I$) has determinant $1$ or $-1$.
(b) Use only the product rule. If $|\det(Q)| > 1$, then $\det(Q^n) = (\det(Q))^n$ blows up. How do you know this can't happen to $Q^n$.
Anyone who has even sniffed a Strang textbook knows that the words inside are filled with ambiguity; this problem is no exception. From what I can interpret, Strang is giving the first steps to prove the desired result. Assume $|\det(Q)| \neq 1$. If $|\det(Q)| > 1$, then $\det(Q^n) = (\det(Q))^n$ and thus $|\det(Q^n)| \to \infty$ as $n \to \infty$. And this contradicts... what exactly? In the solution manual (by Strang), this is the solution.
$Q^n$ stays orthogonal so its determinant can't blow up as $n \to \infty$.
This sounds like circular logic to me. Since $Q$ is orthogonal, so is $Q^n$. Since $Q^n$ is orthogonal, $|\det(Q^n)| \not\to \infty$ as $n \to \infty$. The only way this is true is if $|\det(Q^n)| \le 1$, which is what we are trying to prove in the first place.
Is my interpretation correct? Is Strang using circular logic here or am I missing something? Also, what is Strang getting at with his hint? Thanks for the help.