# How to prove that every orthogonal matrix has determinant $\pm1$ using limits (Strang 5.1.8)?

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.

• 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 $\mathbb{C}$). Mar 31, 2020 at 6:29

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 $$\det\big(Q\big) = -1$$

• Great argument. I wonder if this is what Strang had in mind for students barely exposed to Linear Algebra :) Thanks for the answer! Mar 31, 2020 at 16:38

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

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

• 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! Mar 31, 2020 at 5:39

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 $$1$$.