Editing Euler's rotation theorem (section)

==Matrix proof==
A spatial rotation is a linear map in one-to-one correspondence with a {{nowrap|3 × 3}} [[rotation matrix]] {{math|'''R'''}} that transforms a coordinate [[Vector (mathematics and physics)|vector]] {{math|'''x'''}} into {{math|'''X'''}}, that is {{math|'''Rx''' {{=}} '''X'''}}. Therefore, another version of Euler's theorem is that for every rotation {{math|'''R'''}}, there is a nonzero vector {{math|'''n'''}} for which {{math|'''Rn''' {{=}} '''n'''}}; this is exactly the claim that {{math|'''n'''}} is an [[eigenvector]] of {{math|'''R'''}} associated with the [[eigenvalue]] 1. Hence it suffices to prove that 1 is an eigenvalue of {{math|'''R'''}}; the rotation axis of {{math|'''R'''}} will be the line {{math|''μ'''''n'''}}, where {{math|'''n'''}} is the eigenvector with eigenvalue 1.

A rotation matrix has the fundamental property that its inverse is its transpose, that is
:<math>
\mathbf{R}^\mathsf{T}\mathbf{R} = \mathbf{R}\mathbf{R}^\mathsf{T} = \mathbf{I},
</math>
where {{math|'''I'''}} is the {{nowrap|3 × 3}} identity matrix and superscript T indicates the transposed matrix. 

Compute the determinant of this relation to find that a rotation matrix has [[determinant]] ±1. In particular,
:<math>\begin{align}
  1 = \det(\mathbf{I}) &= \det\left(\mathbf{R}^\mathsf{T}\mathbf{R}\right) = \det\left(\mathbf{R}^\mathsf{T}\right)\det(\mathbf{R}) = \det(\mathbf{R})^2 \\
  \Longrightarrow\qquad \det(\mathbf{R}) &= \pm 1.
\end{align}</math>

A rotation matrix with determinant +1 is a proper rotation, and one with a negative determinant −1 is an ''improper rotation'', that is a reflection combined with a proper rotation.

It will now be shown that a proper rotation matrix {{math|'''R'''}} has at least one invariant vector {{math|'''n'''}}, i.e., {{math|'''Rn''' {{=}} '''n'''}}. Because this requires that {{math|('''R''' − '''I''')'''n''' {{=}} 0}}, we see that the vector {{math|'''n'''}} must be an [[eigenvector]] of the matrix {{math|'''R'''}} with eigenvalue {{math|''λ'' {{=}} 1}}. Thus, this is equivalent to showing that {{math|det('''R''' − '''I''') {{=}} 0}}.

Use the two relations
:<math> \det(-\mathbf{A}) = (-1)^{3} \det(\mathbf{A}) = - \det(\mathbf{A}) \quad</math>
for any {{nowrap|3 × 3}} matrix '''A''' and
:<math> \det\left(\mathbf{R}^{-1} \right) = 1 \quad</math>
(since {{math|det('''R''') {{=}} 1}}) to compute
:<math>\begin{align}
        &\det(\mathbf{R} - \mathbf{I}) = \det\left((\mathbf{R} - \mathbf{I})^\mathsf{T}\right) \\
  {}={} &\det\left(\mathbf{R}^\mathsf{T} - \mathbf{I}\right) = \det\left(\mathbf{R}^{-1} - \mathbf{R}^{-1}\mathbf{R}\right) \\
  {}={} &\det\left(\mathbf{R}^{-1}(\mathbf{I} - \mathbf{R})\right) = \det\left(\mathbf{R}^{-1}\right) \, \det(-(\mathbf{R} - \mathbf{I})) \\
  {}={} &-\det(\mathbf{R} - \mathbf{I}) \\[3pt]
 \Longrightarrow\ 0 ={} &\det(\mathbf{R} - \mathbf{I}).
\end{align}</math>

This shows that {{math|''λ'' {{=}} 1}} is a root (solution) of the [[Characteristic polynomial|characteristic equation]], that is,
:<math>
\det(\mathbf{R} - \lambda \mathbf{I}) = 0\quad \hbox{for}\quad \lambda=1.
</math>

In other words, the matrix {{math|'''R''' − '''I'''}} is singular and has a non-zero [[Kernel (matrix)|kernel]], that is, there is at least one non-zero vector, say {{math|'''n'''}}, for which
:<math>
(\mathbf{R} - \mathbf{I}) \mathbf{n} = \mathbf{0} \quad \Longleftrightarrow \quad \mathbf{R}\mathbf{n} = \mathbf{n}.
</math>

The line {{math|''μ'''''n'''}} for real {{mvar|μ}} is invariant under {{math|'''R'''}}, i.e., {{math|''μ'''''n'''}} is a rotation axis. This proves Euler's theorem.
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This result is often used as a showcase for the elementary theory of zeroes of polynomials, proving by a process of elimination that there are no other possibilities than that of the characteristic polynomial having 1 as a zero, but that ultimately results in a longer argument than the one above. First, since the matrix is real and has side 3, its characteristic polynomial must be real and of odd degree, hence it has a real zero. Second, multiplication by the matrix preserves the length of all vectors, and thus all eigenvalues must have absolute value 1. This leaves only 1 and −1 as candidates for being the real eigenvalue. Because complex eigenvalues occur in conjugate pairs, they give no net contribution to the determinant. Hence, the fact that the determinant is 1 means the multiplicity of −1 as a zero of the characteristic equation must be even. If the multiplicity is 2 then the third zero must be 1. If the multiplicity of −1 is 0 then the real zero must be 1. Either way, 1 is a zero of the characteristic equation.
-->