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For orthogonal matrix, $Q$, $QQ^T=I\tag{1}$ Wikipedia states that $1=\det(I)=\det\left(Q^\mathrm{T}Q\right)=\det\left(Q^\mathrm{T}\right)\det(Q)=\bigl(\det(Q)\bigr)^2$

But I can't understand how the last equality (on the right) follows: $\det\left(Q^\mathrm{T}\right)\det(Q)=\bigl(\det(Q)\bigr)^2$

Is it the case that $\det\left(Q^T\right)=\det\left(Q\right)$? This would imply $Q^T=Q$, but taking the inverse of $(1)$ implies that $Q^T=Q^{-1}\ne Q$ which is a contradiction.

Before asking this question I searched this site and read this strongly related question. In one of the comments and the answer it is stated right from the start that $\det M^T=\det M$, but I still don't understand how this can be true.

Electra
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    Does this help? https://math.stackexchange.com/questions/1468064/determinant-of-transpose-intuitive-proof – razivo Jul 10 '21 at 19:46
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    Why should $\det(Q^{T})=\det(Q)$ imply $Q^{T}=Q$ ? – Tobsn Jul 10 '21 at 19:53
  • @razivo Yes, that does help thanks, the only problem is there are typo's in the answer and I don't understand what the answer means in writing that $D=CAC^{-1}$ – Electra Jul 10 '21 at 19:56
  • @Tobsn Hi, I was thinking about it in reverse, $Q^T=Q$ then take the determinant of both sides implies $\det(Q^{T})=\det(Q)$. But you're saying the converse is not true right? – Electra Jul 10 '21 at 19:58
  • the converse is obviously correct, but irrelevant since $Q^T\ne Q$. It's a simple fact that $\det(Q^T)=\det(Q)$, and no there's simply no reason that should imply $Q^T=Q$. – David C. Ullrich Jul 10 '21 at 20:04
  • @DavidC.Ullrich 'Converse is obviously incorrect', that's what you meant to write isn't it? If so many thanks. – Electra Jul 10 '21 at 20:07
  • By converse I meant what I see you called the "reverse": If $Q^T=Q$ then $\det(Q^T)=\det(Q)$. that's obviously correct. – David C. Ullrich Jul 10 '21 at 20:09
  • @DavidC.Ullrich Ah yes, now I understand you completely, thanks for the clarification :-) – Electra Jul 10 '21 at 20:11

2 Answers2

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It's true that $\det\left(Q^T\right)=\det\left(Q\right)$ but this doesn't imply that $Q^T=Q$ (take any non-symmetric matrix) (see here for more details).

Michelle
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  • Thanks for the answer, is there a simple proof that $\det\left(Q^T\right)=\det\left(Q\right)$? – Electra Jul 10 '21 at 20:00
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    The simplest proof I know is through the uniqueness of the determinant. Show that the map $f : M_n (\mathbb{R}) \to \mathbb{R}$ given by $f(Q) = \det (Q^T)$ has all the properties of a determinant and hence is the determinant. – Charles Hudgins Jul 10 '21 at 20:06
  • @CharlesHudgins I never thought of this elegant proof, I would have instead suggest to look at the messy sum and a reordering etc. Thank you for that comment! – Didier Oct 12 '22 at 10:57
  • @Didier Somewhere along the way you have to do that, say in proving uniqueness. Just depends what you're willing to take for granted. – Charles Hudgins Oct 14 '22 at 20:41
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Yes, for any square matrix $A\in M_n(K)$, it is true that $\operatorname{det} A = \operatorname{det} A^T$, and this doesn’t imply that $A=A^T$. In general, two matrices can be distinct and yet have the same determinant, for example: \begin{gather*} A = \begin{pmatrix} 1 & 3 \\ 1 & 2 \end{pmatrix} \\ B = \begin{pmatrix} 2 & -3 \\ -3 & 4 \end{pmatrix} \end{gather*} $A$ and $B$ are distinct matrices but $\operatorname{det} A = \operatorname{det} B = -1$.

Now, if you are looking for a proof of the equality $\operatorname{det} A = \operatorname{det} A^T$ (which is widely known) take a look here

Samuel M. A. Luque
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