Determinant of an orthogonal matrix

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Discussion Overview

The discussion centers on the properties of the determinant of an orthogonal matrix, specifically exploring why the determinant is either +1 or -1. Participants examine the mathematical reasoning behind this property, including relationships involving the identity matrix and the transpose of the matrix.

Discussion Character

  • Technical explanation

Main Points Raised

  • One participant proposes that for an orthogonal matrix Q, the relationship det(I) = det(Q^T Q) leads to the conclusion that (det(Q))^2 = 1, suggesting that det(Q) must be ±1.
  • Another participant later reinforces this by stating that if Q is orthogonal, then the equation A^T A = I_n implies that det(A) det(A^T) = (det(A))^2 = 1, leading to the same conclusion about the determinant being ±1.

Areas of Agreement / Disagreement

There is no explicit disagreement noted, but the discussion does not reach a consensus on sharing the derived knowledge with the broader forum community.

Contextual Notes

Some assumptions about the properties of determinants and orthogonal matrices are present, but not all steps in the reasoning are fully detailed or resolved.

Who May Find This Useful

Readers interested in linear algebra, particularly those studying the properties of matrices and determinants, may find this discussion relevant.

squenshl
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How is it the determinant of an orthogonal matrix is [tex]\pm1[/tex].
Is it:
Suppose Q is an orthogonal matrix [tex]\Rightarrow[/tex] 1 = det(I) = det(QTQ)
= det(QT)det(Q) = ((det(Q))2
and if so, what is it for -1.
Thanks.
 
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Never mind, got it.
 
Maybe you can share your new knowledge with the forum so that when other users see the question they can also see the answer!
 
Sweet.
Suppose Q is orthogonal, then ATA = In
[tex]\Rightarrow[/tex] det(A)det(ATA) = det(A)det(A) = (det(A))2 = 1
Which implies that det(A) = [tex]\pm\sqrt{1}[/tex] = [tex]\pm1[/tex]
 

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