I do not know.
Orthonormal Matrices, Euclidean Geometry, Studies on Congruence and Linear maps.
Euclid decided that if a set of point, with specific set of distances and angles between them, are the same for 2 sets of points, but they don't share the same position on a grid, then they are still the same as One of the sets just has to be rotated and translated to the other figure to superimpose the other one. These Figures are called Plane figures, this property between 2 plane figures is called Congruence. If there is a Map, does not have to be linear, That exists which can do this, its an isometry, an isometric map. And isometric map is distance preserving, this means that if 2 point in R^n exists, Isometric map is a linear transformation (a homomorphism to itself) where distance(p-q)=distance(f(p)-f(q)). These maps allow congruence. The other properties which occur after the definition of it being distance preserving are, if F(x) represents an isometric map:
In ABC, is the point are collinear then F(A),F(B),F(C) are also collinear. This is because if they are not collinear then the distance from each point will be different or even if they are same, the size of the distances will differ and hence needs to preserve collinearity.
If plane figure ABC is a triangle, then plane figure F(A),F(B),F(C) is also a triangle with same sides and angles but potentially different orientation. they are 3 point and hence must form triangle, if there is a difference within lengths of sides, its obvious it is not distance preserving. If there is a difference in angle, then one of the side lengths must change so one of the distances between 2 point must change and hence not isometric. And so angles and sides have to be preserved.
If C is the mid point of A,B then F(C) is the mid point of F(A),F(B). This is true because the distances must be preserved. This is not only true for all Isometric functions but also true for All linear functions (isometric or not) due to the definition of linear functions: aF(b)+cF(d)=F(ab+cd).
If a set of points form a circle, under the Isometric Map. The set of points will also have to be a circle. This is because if it is not a circle then at least one of the points will not have the same distance from each other as the unmapped version did and hence not isometric, so cirlces have to be preserved.
Congruence is an equivalence relation. This is because Reflextivity is true as A is automatically congruent it to itself, Isometric map is the identity map, what it is represented as a matrix though depends on domain and codomain matrix. Symmetric is true, If A is congruent to B, B is congruent to A as if A is rotate, reflected and translated. Reverse that process and you get A,this also shows Isometric functions have an inverse. They are transistive, If A is congruent to B, B is congruent to C then A is congruent to C by composite Isometric maps.
An example of A isometric function is the translation, Such as translating vectors by units up, down, across a diagonal, However these are not Linear maps as if F is a translation. then F(x)=x+c, so F(x+a)=a+x+c, but F(x)+F(a)=a+x+2c, if c is non zero then linear combinations are not preserved.
Finding points and distances to points in an isometric map will lead to 1 point. take for example, e1,e2 and origin. only 1 point is a distance: d1,d2,d0 in this order, and so only 1 point in an isometric map can be d1,d2,d0, if not then either the 2 points are equal or the map isn't isometric.
If an Isometric Functions, F(x), is a linear map then the Origin must map to itself. Consider the Standard basis in, e1,e2,...,en. The F(ei) dont have to map to themselves However their Magnitudes (all magnitude of ALl vectors) must be same and as in standard basis, all ei are orthogonal, the F(ei) must be orthogonal. F(x) also must be an isomorphism too. And Matrix representing this is an Orthonormal Matrix.
Linear Isometric Map, F(x). If Origin is its self. The distance from a point V to the origin is A. Then distance of F(V) to f(0) is A but f(0)=origin so F(V)'s size/magnitude/distance-from-origin is A.
If Triangles and their angles must be preserved, all the ei's have 90 degrees angles so if finding functions with respect to standard basis, the f(ei) must have 90 degrees too.
If F(x) is not an isomorphism, then 2 points will map to 1 and not all points will be in the codomain, this is not isometric as 2 originally non-0 points will have 0 distance and if points in domain is omitted, its obvious their distances are not preserved.
If You are going to want a matrix to represent this map. And if the Matrix's codomain AND domain are with respect to Natural Basis THEN it will be orthonormal as the way Matrices are defined, coloumns represent the mapped version of the domain's basis and is put under the representation function with respect to codomain. As this one is both Standard basis, The coloumns will equal the Mapped version, no extra multiplying of coefficients and Actual Vector in basis needed. So, The mapped set will be size 1 because of magnitudes of Standard basis, and the set has to be mutually orthogonal because that is how the original basis is. Hence orthonormal matrix.
In R^2. If F(e1)=<a,b> and F(e2)=<c,d>. the a2+b2=1 and same with c2 and d2. orthogonal vectors have dot product to 0 so ac+bd=0.
And so: b=(+or-)sqrt(1-a2), -a/b=d/c, (-or+)a/(sqrt(1-a2))=d/c, a2/1-a2=d2/c2. c2=1-d2. so d=(+or-)a, c=(-or+)sqrt(1-a2).
And hence F(e1)=<a,(+or- sqrt(1-a2))>, F(e2)=<(+or- sqrt(1-a^2)), -or+ a>. And then you can see the matrix is orthonormal is you made it respect to natural. For larger natural basis, you would use same method. magnitudes are one. All dot products are 0, use this to find solution.
Examples of Linear Isometric maps are reflections and rotations. A vector distance A from origin will map onto a point of circle centred at origin with distance A.
You can Combine reflections/rotations with Translations. Making a combination of Glided Reflections and rotations.
Similar Shapes differ from each other with each side being scaled, so in vector matrix computations:
If a,b congruent. And C is similar to b my a scalar to K. c=bk, and a=kM+v0. v0 represent tranlsation. M represents Map, most likely with respect to Natural basis, K is scalar. |