Isometries of linear euclidean spaces
A linear mapping , where , are linear euclidean spaces is an isometry if the following equivalent conditions hold
- is a linear isomorphism and preserves the inner product
- is a linear isomorphism and preserves the lengths of vectors
- takes an orthonormal basis of onto a orthonormal basis of ,
- takes any orthonormal basis of onto a orthonormal basis of ,
A matrix is orthogonal if .
In other words, a matrix is orthogonal if its rows (equivalently columns) form an orthonormal basis with respect to the standard inner product.
Orthogonal matrices are invertible and for such matrices.
Notice that is an isometry if and only if is orthogonal for some (equivalently, any) orthogonal bases , of respectively spaces and .
Rotations and perpendicular reflection
If is a perpendicular reflection across , and if consists of an orthonormal basis of after which we have an orthonormal basis of we get
Obviously a perpendicular reflection is an isometry.
If is a two-dimensional subspace of , then the rotation around by has the following matrix
where consists of an orthonormal basis of after which we have an orthonormal basis of . A rotation is an isometry.
It is easy to show that every isometry of a linear euclidean two-dimensional space is either a perpendicular reflection or a rotation.
Isometries of affine euclidean spaces
A mapping on affine euclidean spaces is called an isometry if one of the following equivalent conditions hold
- is an affine isomorphism and is a isometry of linear spaces,
- is an affine mapping and is orthogonal in orthonormal bases of and ,
One can prove that is an isometry iff it preserves the distance between points , i.e. for any .
Mapping preserving measure of parallelepipeds
Notice that if is a -dimensional parallelepiped and is an affine isomorphism, then is also a -dimensional parallelepiped.
Moreover, if is a -dimensional parallelepiped in a -dimensional space, then . Indeed, if , then is a basis , and is a basis . Then for an orthonormal basis , we get
Thus, an affine isomorphism on -dimensional affine euclidean spaces preserves measure of -dimensional parallelepipeds if and only if .
An endomorphism of a linear Euclidean space is self-adjoint, if for any , .
Notice that then if is an orthonormal basis of , the matrix is symmetrical! Indeed, it is so because is the -th coordinate of with respect to . Reversly, if for an orthonormal basis , matrix is symmetrical, then is self-adjoint.
It is easy to notice, that if is an eigenvector and , then . Moreover, if are eigenvectors for different eigenvalues, then . Indeed if those values are , then
Also it can be proved every eigenvalue of a symmetrical matrix is real! Moreover, by an easy inductive argument using the fact of preserving perpendicularity to an eigenvector, we can prove that every symmetrical matrix (thus every self-adjoint mapping) has a basis consisting of eigenvectors. This basis is orthogonal, and can be made orthonormal. In other words, for every symmetrical matrix there exists an orthogonal matrix , such that is diagonal!