Isometry/Decomposition/Section
Let be a real finite-dimensional vector space, and let
denote an endomorphism. Then there exists a -invariant linear subspace of dimension
or .We may assume , and that is described by the matrix with respect to the standard basis. If has an eigenvalue, then we are done. Otherwise, we consider the corresponding complex mapping, that is,
which is given by the same matrix . This matrix has a complex eigenvalue , and a complex eigenvector . In particular, we have
Writing
with , this means
Comparing the real part and the imaginary part we can deduce that . Therefore, the real linear subspace is invariant.
Let
be an isometry on the euclidean vector space . Then is a orthogonal direct sum
of -invariant linear subspaces,
where the are one-dimensional, and the are two-dimensional. The restriction of to the is the identity, the restriction to is the negative identity, and the restriction to is a rotation without eigenvalue.We do induction over the dimension of . The one-dimensional case is clear, due to fact. Let . The determinant has, because of fact either the value or . In case it is , the characteristic polynomial has two distinct zeroes, and these zeroes are, due to fact, and . Then we have a reflection at an axis, and
If the determinant is , then we are in the situation of fact, and we have a rotation. If the angle of rotation is , then we have the identity, and we can decompose . If the angle of rotation is , then we have a point reflection ; therefore, we have a decomposition . For all other angles, there is no eigenvector.
Let now be arbitrary, and suppose that the statement is proven for smaller dimensions. Because of fact, there exists a -invariant linear subspace of dimension or , and, because of fact, there exists an invariant orthogonal complement, that is,
The induction hypothesis, applied to , yields the result.
In this decomposition, is the eigenspace for the eigenvalue , and is the eigenspace for the eigenvalue ; the decompositions are not unique. The Isometry is proper if and only if is even.