Determinant of Orthogonal Matrix is Plus or Minus One
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Theorem
Let $\mathbf Q$ be an orthogonal matrix.
Then:
- $\det \mathbf Q = \pm 1$
where $\det \mathbf Q$ is the determinant of $\mathbf Q$.
Proof
- $\det \mathbf Q^\intercal = \det \mathbf Q$
Then:
\(\ds \mathbf Q \mathbf Q^\intercal\) | \(=\) | \(\ds \mathbf I\) | Product of Orthogonal Matrix with Transpose is Identity | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \det {\mathbf Q \mathbf Q^\intercal}\) | \(=\) | \(\ds \det \mathbf I\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \det {\mathbf Q \mathbf Q^\intercal}\) | \(=\) | \(\ds 1\) | Determinant of Unit Matrix | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \det \mathbf Q \det \mathbf Q^\intercal\) | \(=\) | \(\ds 1\) | Determinant of Matrix Product |
Hence the result.
$\blacksquare$
Sources
- 1980: A.J.M. Spencer: Continuum Mechanics ... (previous) ... (next): $2.1$: Matrices: $(2.13)$