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The problem of interest from Gilbert Strang's Introduction to Linear Algebra Section 5.1 is as follows.

  1. 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.

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  • $\begingroup$ Even if this argument can be fixed, it is a pointlessly roundabout way to prove a simple fact (and in lesser generality than it deserves to be proven; the result holds over any field, but Strang's argument can only work over $\mathbb{R}$ or $\mathbb{C}$). $\endgroup$ Mar 31, 2020 at 6:29

3 Answers 3

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Here's one way of completing the 'blowup' proof:

we are dealing with $\text{m x m}$ matrices.
$Q^TQ = I$ tells us each column has length 1 which implies each component of $Q$ has modulus $\leq 1$,

Thus we can very crudely bound $\big\vert \det\big(Q\big)\big\vert$ as being a sum with $m!$ permutations and apply triangle inequality to get, for any orthogonal $Q \in \mathbb R^\text{m x m}$

$\big\vert \det\big(Q\big)\big\vert \leq m!$

now qualitatively, since the product of finitely many orthogonal matrices is an orthogonal matrix we have, by applying the product rule
$LHS= \big\vert \det\big(Q\big)\big\vert^n = \big\vert \det\big(Q^n\big)\big\vert \leq m!$
for all natural numbers $n$. But if there is some $\delta \gt 0$ such that
$1 + \delta = \big\vert \det\big(Q\big)\big\vert$
then the LHS may be made arbitrarily large by selecting large enough $n$

(e.g. to make this explicit use the Bernouli Inequality or if that isn't known, just look at the first 2 terms of the binomial expansion of $(1+\delta)^n$ and use that as a lower bound, then set: $N= \frac{m!-1}{\delta}$ which tells us $m!\lt \big\vert \det\big(Q\big)\big\vert^n \leq m!$ for all $n\geq N$ which is a contradiction ). This qualitative issue tells us that
$\big \vert\det\big(Q\big)\big\vert \leq 1$

now use the fact that $Q^T$ is also orthogonal so the above tells us $\big \vert\det\big(Q^T\big)\big\vert \leq 1$

but, again using the product rule:
$1 =\det\big(I\big) = \det\big(Q^TQ\big) = \det\big(Q^T\big) \det\big(Q\big) = \big \vert \det\big(Q^T\big) \big \vert\big \vert\det\big(Q\big)\big\vert \leq 1$
i.e. the inequality is met with equality so
$\big \vert\det\big(Q\big)\big\vert=1$

finally since $Q \in \mathbb R^\text{m x m}$ and the determinant is real with modulus 1, thus $\det\big(Q\big) = +1$ or $\det\big(Q\big) = -1$

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    $\begingroup$ Great argument. I wonder if this is what Strang had in mind for students barely exposed to Linear Algebra :) Thanks for the answer! $\endgroup$
    – HiMatt
    Mar 31, 2020 at 16:38
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    $\begingroup$ Yes, this is nice. $\endgroup$
    – user1551
    Mar 27, 2023 at 19:55
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What is the "product rule" in the context of determinants?

I assume it refers to the well-known fact that

$\det(AB) = \det(A) \det(B). \tag 1$

We also have

$\det(A) = \det(A^T); \tag 2$

thus

$(\det(A))^2 = \det(A) \det(A^T)$ $= \det(AA^T) = \det(I) = 1, \tag 3$

from which it immediately follows that

$\det(A) = \pm 1. \tag 4$

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    $\begingroup$ Yes, the product rule is $\det(AB) = \det(A)\det(B)$ (apparently). Your solution is the proof required for part (a) of this problem, not part (b). Part (a) allows the "transpose rule" ($\det(A) = \det(A^T)$), but part (b) doesn't. Thanks anyway! $\endgroup$
    – HiMatt
    Mar 31, 2020 at 5:39
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I am not sure what he was getting at, but here is a topological argument.

Observe that the set of all orthogonal matrices is compact: it is bounded because each element of an orthogonal matrix has absolute value $\le1$, and closed because it is the inverse image of $\{I\}$ under the continuous function $f:Q\mapsto Q^TQ$. Now, given any $Q$, since $\{Q^k\}_{k\in\mathbb N}$ is an infinite sequence in a compact set, there is a subsequence $\{Q^{k_n}\}$ that converges to some orthogonal matrix $Q_0$. Then $\lim_{n\to\infty}|\det Q|^{k_n}$ exists and is equal to $|\det Q_0|$. Hence $|\det Q|$ cannot be greater than $1$.

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