The problem of interest from Gilbert Strang's Introduction to Linear Algebra 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.

  • $\begingroup$ 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 $\mathbb{C}$). $\endgroup$ – darij grinberg Mar 31 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 $\det\big(Q\big) = -1$

| cite | improve this answer | |
  • $\begingroup$ 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 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

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


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

from which it immediately follows that

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

| cite | improve this answer | |
  • 1
    $\begingroup$ 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 at 5:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.