The table below summirizes the four different geometries, the transformations allowed in each, and the invariants under those transformations.
Geometry | ||||
---|---|---|---|---|
Transformation | Euclidean | Similarity | Affine | Projective |
rotation | x | x | x | x |
translation | x | x | x | x |
uniform scaling | x | x | x | |
non-uniform scaling | x | x | ||
shear | x | x | ||
perspective proj. | x | |||
icomposition of proj. | x | |||
Invariants | ||||
lengths | x | |||
angles | x | x | ||
ratio of lengths | x | x | ||
parallelism | x | x | x | |
incidence | x | x | x | x |
cross-ratio | x | x | x | x |
Examples in 1-dim (projective space P1)
Translations | y = x + t | abelian, unit is t=0 |
Dilations | y = k x | abelian, unit k=1 |
Similarity | y = k x + t | non-abelian, semidirect product of transl. and dilat. |
Symmetry | y = x0 + (x - x0) | |
Projectivity | y = (ax+b)/(cx+d) | with ad-bc!=0; non-abelian |
The Kronecker delta is
The antisymmetric symbol (also called Ricci or Levi-Civita tensor), in 3-dim, is
More generally in n-dimension the symbol has n indices and its value is 0 if the indices are not all different, +1 if they are an even permutation of 123...n, -1 if they are an odd permutation of 123...n.
The basic identity is the square of the symbol, (here for n=4)
| d.im d.in d.ir d.is | e.ijkl e.mnrs = | d.jm d.jn d.jr d.js | | d.km d.kn d.kr d.ks | | d.lm d.ln d.lr d.ls |
From this one obtains the products with contractions on one or more indices. For example in n=4
eijkl eijrs = 2 ( dkr dls - dks dlr )
eijkl eijks = 3! dls
eijkl eijkl = 4! = 24
In particular in 3-dim we have the important identity
Homogeneous Coordinates
Point | P = ( X, Y, W ) = (px, py, pw) | |
Line | U = ( a, b, c ) = (ux, uy, yw) | |
incidence | pi ui = 0 | scalar product |
join of 2 pts | ui = eijk p1j p2k | cross product |
intersect of 2 lines | pi = eijk u1j u2k | |
collinearity | eijk p1i p2j p3k = 0 | null determinant |
concurrence | eijk u1i u2j u3k = 0 | null determinant |
ideal pts | (px, py, 0) | |
ideal line | (0, 0, uw) |
Duality points-lines: points are contravariant: P = pi Vi where Vi unit vector of i-th axis; lines are covariant: U = ui. Under change of coords xi=xi(y), with derivatives gij = dxi/dyj the coordinates of a point change as
Points transform like the differentials dxi; lines transform like the derivative df/dxi.
Example: coordinate shear:
The unit vectors of the axes becomes
Therefore a point P is written
and the points coordinate transformation is
The inverse of gij is gij = dyj/dxi,
Lines tranform with the inverse of the matrix g (covariantly)
The antisymmetric symbol is invariant under change of coords.
Indices are raised and lowered with the metric tensor gij
The metric tensor is symmetric, and we will consider only spaces in which it is also positive definite (all eigenvalues are strictly positive). Under change of coordinates the metric tensor changes covariantly,
infact g(y)ij dyi dyj = g(x)kj dxk/dyi dxl/dyj dyi dyj. The inverse of gij is the contravariant form of the metric tensor (it is also symmetric positive definite)
If the coordinates yi are cartesian
Isomorphisms of P2
Equivalent representations of the projective plane:
Simple-ratio
The simple ratio of three collienar points is defined as
The values of the six simple ratios of three points are related:
C(1,2,3) = r C(2,1,3) = 1/r C(1,3,2) = 1-r C(3,1,2) = 1/(1-r) C(2,3,1) = (r-1)/r C(3,2,1) = r/(r-1)
Given two points p, q, and their ratios with respect to the fixed points a, b, m = C(a,b,p) n = C(a,b,q) these determine the ratio of the distances p-q and a-b,
Using homogeneous coords (a, b are non-homogeneous coordinates), the ratio of p=(p1, p2) with respect to a, b is
Remark. The point a is mapped into 0, b into infinity (k2=0), and infinity (p2=0) into 1.
Cross-ratio
The cross-ratio of four collienar points is
(p1 - p3) (p2 - p4) C( p1,p2,p3,p4 ) = ----------- ----------- (p2 - p3) (p1 - p4)
where the distances are "with sign": consider one coordinate index such that the points have different coordinate values for that index, say x, then (p - q) = px - qx. It does not matter which coordinate is chosen. Of course it must be the same for all four distances.
Four points can be arranged in 24 ( = 4! ) ways, but there are only six different values of the cross-ratios:
C(1,2,3,4) = C(2,1,4,3) = C(3,4,1,2) = C(4,3,2,1) = r C(1,2,4,3) = C(2,1,3,4) = C(3,4,2,1) = C(4,3,1,2) = 1/r C(1,3,2,4) = C(2,4,1,3) = C(3,1,4,2) = C(4,2,3,1) = 1-r C(1,3,4,2) = C(2,4,3,4) = C(3,1,2,4) = C(4,2,1,3) = 1/(1-r) C(1,4,2,3) = C(2,3,1,4) = C(3,2,4,1) = C(4,1,3,2) = (1-r)/r C(1,4,3,2) = C(2,3,4,1) = C(3,2,1,4) = C(4,1,2,3) = r/(1-r)
The cross-ratio of lines is defined
sin(u1-u3) sin(u2-u4) C( u1,u2,u3,u4 ) = ------------ ------------ sin(u2-u3) sin(u1-u4)
This turn so out to be equal to the cross-ratio of the angular coefficients
of the lines, C( u1,u2,u3,u4 ) = (m1-m3)/(m2-m3) (m2-m4)/(m1-m4).
In order to show the formula of the cross-ratio consider four lines
of a pencil (centered at the point O),
a, b, c, and d, and their points of intersection with a line
not belonging to the pencil, A, B, C, and D, respectively.
Draw the the perpendicular
lines from A, B to the lines OC, OD. Let K,M and H,N be the
basepoints. Then AC/BC = AK/BH = [OA sin(ac)]/[OB sin(bc)].
Do the same for BD/AD, and put them together. The length OA
and OB cancel out, and we have the formula above.
Laguerre formula
Given two lines u1, u2 (intersecting the ideal line in p1, p2 respectively) the angle between the lines is
Proof. The angle between the two lines is the same as the angle between
the parallel lines passing through the origin, a1 x - y = 0 and
a2 x - y = 0 respectively.
The normal vectors to these lines are v1=(1,a1) and v2=(1,a2), therefore
tan( theta ) = (v2 X v1) / (v2 . v1) = (a1 - a2) / (1 + a1 a2).
Now p1 = ( a1, -1, 0) and p2 = (a2, -1, 0)
then is just a metter of algebra ... .
This formula can also be derived writing p=(a,b,0) as a complex number
z=a+ib. The angle between the two comples numbers z1 and z2 is
2i theta = log( z1/z1* z2*/z2 ) where z* denotes the complex conjugate
of z.
Four elements are harmonic if their cross ratio is -1.
Given three elements, a, b, and c, the croos-ratio defines a coordinate for a fourth element x
k = (a,b,c,x) = (c-a)/(c-b) (x-b)/(x-a) x = [k a (c-b) - b (c-a)] / [ k (c-b) - (c-a) ]
Remark. In this coordinate, the point a corresponts to infinity, b to 0, and c to 1.
Remark. Cross-ratios are invariant under projectivity, hence a projectivity in P1 is completely determined by three points in correspondence, and has equation C(a', b', c', y) = C(a, b, c, x). The general form of the equation of a projectivity is
a yx + b x + c y + d = 0 (ad-bc!=0)
The limiting points of a pojectivity in P1 are the points mapped into improper points, ie, x = -c/a and y = -b/a. The power of the projectivity is the product of the distances of two corresponding points from the limiting points, (bc - ad) a-2.
Remark
Given four points on a line such that the first two and the second two
are equispaced, it is possible to compute the cross-ratio
of three of them with the ideal point (at infinity).
Let a be the distance between the first two points and b the distance of the third from the second. We can assume that the points have coordinates 0, a, a+b, 2a+b. Let r denote the cross-ratio of the four points,
r = C(p0,p1,p2,p3) = (a+b)/b (a+b)/(2a+b)
From this we can compute the ratio b/a (it is a second order equation, but we keep only the positive solution since we assume a>0 and b>0),
from which one can compute C(p0,p1,p2,pinf) = (a+b)/b.
Conics (ellipses, hyperbolas, parabolas)
A conic has equation
C is a 3x3 symmetric matrix (2-nd order covariant tensor) and defines a "reciprocity relation" in the plane P2, that maps points into lines (polar line) and viceversa, lines into points (pole). The polar line of p with respect to C is ui = Cij pj. The pole of the line u with respect to C is pi = (C-1)ij uj. Notice that C-1 = C* = transpose of C-1 since C is symmetric; C-1 is the dual representation of the conic.
A conic is a self-dual figure. It can be considered as the set of points that belong to their polar line, or as the set of lines that contan their pole (= envelope of tangent lines).
Absolute points: any circle intersect the ideal line in the same two points:
I = (i,+i,0) J = (1,-i,0)
The abosolute points are invariant under similarity transf.
Collineation
Collineations are mapping of P2 in itself. They are represented by 3x3 real matrix (actually only 8 params) Tij,
Here T* is the transpose of T-1. A collineation preserves the incidence relations: p u = 0 implies q v = 0.
The affine plane is projective plane minus the ideal line. Therefore affine transf. must preserve the ideal line: T_aff (X,Y,0) = a (X,Y,0), and have the form
| t11 t12 t13 | T_aff = | t21 t22 t23 | | 0 0 t33 |
Similarity transf. must preserve angles and ratio of lengths
| cos(t) sin(t) t13 | T_simil = | - sin(t) cos(t) t23 | | 0 0 t33 |
Proof. T is a similarity transf. iff it preserves the absolute points. Thus T (1,+/-i,0) = exp(+/-i t)(1,+/-i,0). The converse is a straightforward computation.
Euclidean transf. must preserve also distances: they are rotation and translations ( must divide by third coord.)
x' = cos(t) x + sin(t) y + tx y' = - sin(t) x + cos(t) y + ty
Homology
Theorem of Desargues (Theorem of homologous triangles).
Given two triangles (sides a,b,c, vertices A,B,C) and (sides a',b',c', vertices A',B',C'), let
D = intersection of a, a' E = intersection of b, b' F = intersection of c, c' d = line joining A, A' e = line joining B, B' f = line joining C, C'
then the three lines d,e,f are concurrent iff the three points D,E,F are collinear. Two triangles that satisfy these conditions are called homologous. Remark. This is a configuration of 10 points and 10 lines; each line contains three points and each point belongs to three lines.
Proof. Suppose the D,E,F, are collienar and belong to the line w. Then a,a',w belong to one pencil, l' a' = l a + w Similarly, m' b' = m b + w and n' c' = n c + w. Subtracting, l' a' - m' b' = l a - m b The left hand side is a line thru C', the right hand side is a line thru C. Hence this is the line CC': l a - m b. Similarly the line thru BB' is n c - l a, and the line thru AA' is m b - n c. Since these three equations are not independent, the three lines are concurrent. The proof of the converse statement is the dual of this proof.
A homology is defined by a point W (center) and a line w (axis). It preserves the incidence relation: intersection of lines and collineation of points. If the center is improper the homology is called "affine". If it is also the direction orthogonal to the axis, the homology is orthogonal. If the axis is the ideal line, the homology is called homothetic.
Given a point A, let M be the intersection of the line AW with w, and A' the point corresponding to A, (W,M,A,A') is the invariant of the homology. If it has value -1 the homology is called harmonic (and it is an involution). A harmonic affine homology is a symmetry with respect to a line. A harmonic homothetic is a symmetry with respect to a point.
A homology is determined by
While i am at this i mention a few other interesting theorems of the plane geometry.
Pappo-Pascal theorem
Given six point P1, P2, P3 on a line p, and Q1, Q2, Q3 on a line q, consider the follwing intersections of the lines
R3 intersection of P1-Q2 and Q1-P2 R2 intersection of P1-Q3 and Q1-P3 R1 intersection of P2-Q3 and Q2-P3
Then the three point R1, R2, R3 are collinear (their line is called Pascal line).
Remark. This is a configuration of 9 points and 9 lines (every line can be considered as the Pascal line).
Brianchon theorem
This is the dual of Pappo-Pascal theorem. Given six lines p1, p2, p3 concurrent into the point P, and q1, q2, q3 concurrent into the point Q, consider the following lines,
r1 joining the point p2-q3 and the point q2-p3 r2 joining the point p3-q1 and the point q3-p1 r3 joining the point p1-q2 and the point q1-p2
The three lines r1,r2,r3 are concurrent (into the Brianchon point).
The three lines r1, r2, r3, are the "diagonal" of the hexagon
(red in the figure)
with vertices the six intersection points, denoted 1, ..., 6
in the figure.
Theorem of the quadriangle
Given four points, H,K,L,M consider the intersections A of the lines HM and KL, and B of the lines ML and HK. The line r joining A and B, and P intersection of HL and r, and Q intersection of KM and r. Then the four points A,B,P,Q are harmonic, C(A,B,P,Q) = -1
Theorem of the quadrilater (dual).
Given four lines h,k,l,m and their points of intersection, U intersection of h and k, V intersection of k and l, S intersection of l and m, T intersection of m and h, W intersection of l and h, Y intersection of m and k, and the lines b joining U and S and a joining T and V. Let R be the point of intersection of a and b and the lines p joining W and R and q joining Y and R. Then the four lines a,b,p,q are harmonic, C(a,b,p,q) = -1
Projective Space P3
Point p = (X, Y, Z, W) = (p1, p2, p3, p4) = (px, py, pz, pw) The dual object is the plane n = (n1, n2, n3, n4).
The line passing through two points p1 and p2 is an antisymmetric tensor,
A point p belongs to the line iff uijpj = 0. The matrix uij has rank 2 : a line is the intersection of two planes. from this expression it is clear that p1 and p2 lie on the line: uij p1j = 0, and uij p2j = 0. This matrix has six independent coords. because it takes (n+1 choose k) entries to describe a k-dim element in a projective n+1 dim space.
The Plucker coords. of the line are
u`ij = e`ijkl u.kl u = (u`41, u`42, u`43, u`23, u`31, u`12 ) u`ij = p1`i p2`j - p1`j p2`i
If the points are finite they are determined by the two 3-dim vectors (u41, u42, u43) = P1 - P2, direction of the line, and (u23, u31, u12) = P1 X P2, plane containing the line and the origin and distance of the line from the origin, where P1 and P2 are the euclidean 3-dim points, eg, P1 = (p1x/p1w, p1y/p1w, p1z/p1w).
In vector notation, let z = p1 - p2 and w = p1 x p2, then the contravariant form of uij is
[w]x | zt |
-z | 0 |
and the covariant form uij is
[z]x | wt |
-w | 0 |
Three points define a plane in P3,
By duality the same formula applies to the point intersection of three planes
A point p lies on the line u, join of p1 and p2, iff the three points are collinear. In formula this is
ie, the vanishing of the four 3x3 minors of the matrix
| p`x p1`x p2`x | | p`y p1`y p2`y | | p`z p1`z p2`z | | p`w p1`w p2`w |
The intersection of two planes is a (contravariant) line
Two lines u v intersects if the system of equations
has a non-trivial solution. Therefore if it has rank 3: eabcd uab vcd = 0, which written in Plucker coords is (u41 v23 + u23 v41) + (u42 v31 + u31 v42) + (u43 v12 + u12 v43) = 0
If p1, p2 are two points that define u, and q1, q2 two points that define v, this is eabcd eabkl p1k p2l ecdmn q1m q2n = 0 ie eklmn p1k p2l q1m q2n = 0 The vanishing of the determinant of the four points means that they are coplanar.
The point of intersection is the non-trivial solution to the above system of equations. The determinant of a 4x4 matrix uij is
The inverse of an antisymmetric matrix uij is
The matrix of a line has rank 2, therefore it is not invertible (the determinant is zero) and this formula does not apply. The points on the line U can be written,
| u.23 -u.24 u.34 0 | | a | p`j = | -u.13 u.14 0 -u.34 | | b | | u.12 0 -u.14 u.24 | | c | | 0 -u.12 u.13 -u.23 | | d |
where a,b,c,d are arbitrary numbers. Let U~ denote this matrix. U~ has rank 2. Infact the points on a line span a 2-dim linear subspace of R4, ie, a 1-dim subspace of Pi3.
The intersection point of u and v is the non-trivial solution of the system
Letting S = - v12 u34 + v13 u24 - v14 u23 and [i]jk = vij uik - vik uij the matrix M(U,V) = V U~ can be written
| -[1]23 [1]24 -[1]34 S | M = | [2]13 -[2]14 S [2]34 | | -[3]12 S [3]14 -[3]24 | | S [4]12 -[4]13 [4]23 |
M has rank 1 (except for the degenerate case when the two lines coincide). The non-trivial solution of M p = 0 is the intersection point of the two lines. If the two lines do not intersect there should be no (real) solution of M p = 0.
Remark: for a matrix of a line S(U,U) = - u12 u34 + u13 u24 - u14 u23 = 0 and the symbol [i]jk for (U,U) vanishes, hence U U~ = 0 (zero matrix). If two lines u, v intersect they determine a plane. The plane is defined by one line and a point lying on the other line (thus a point pj with the expression given above).
The intersection of a line u.kl and a plane n.j is
The plane containing a line u.kl and a point p`j is
Subspaces (flats)
Subspaces are sets of points p = c1 p1 + c2 p2 + ... + cr pr where r is the dimension of the subspace (or rank of the flat) and p1, p2, ..., pr is a basis of linearly independent points.
A null subspace has dimension 0. A point has dimension 1. The full space Pn has dimension n.
Intersection (meet) and union (join) of subspaces: S1 ^ S2 and S1 v S2. rank(S1) + rank(S2) = rank(S1 v S2) + rank(S1 ^ S2)
Barycentric Coordinates (August Ferdinand Moebius)
Given a triangle ABC, its area is half the determinant
| A`x A`y 1 | 2 * Area(ABC) = [ABC] = det | B`x B`y 1 | | C`x C`y 1 |
Fron a vector point of view this is A . B x C. Given a point P (inside or outside the triangle), its barycentric coordinates with respect to the triangle are
These three coordinates are not independent, in fact Pa + Pb + Pc = 1. Up to a scale factor the barycentric coords can be written
Using the barycentric coordinates, the point P is
It can be shown the a line in barycentric coordinates is experssed by a linear equation, and if three points are aligned their simple ratio has the same value in cartesian and in barycentric coordinates.
Remark. The vertices have barycentric coords, A = (1,0,0) etc. The sides of the triangles are, for example, AB: c=0, etc. a=b is the line thru C and the midpoint of the side AB.
Physical interpretation: if we place three masses, a,b,c at the vertices of the triangle, the center of mass is the point with barycentric coords (a,b,c). Infact eabc Pam Pbj Pck = [ABC] emjk therefore Pam Pa = [ABC] emjk eijk Pi = 2 [ABC] Pm
Binary representation of barycentric coordinates is related to Serpinski gasket (a fractal figure). Since this is a good point for an excursus in elementary geometry let me cite a few interesting theorems.
Menelaus theorem
Let three points D, E, F, on the sides BC, CA, AB of a triangle, they are collinear iff AE/EC + CD/DB + BF/FA = -1
Ceva theorem (Giovanni Ceva)
If three lines AD, BE, CF from the vertices A,B,C, of a triangle to points on the opposite sides, meet at one point iff AE/EC + CD/DB + BF/FA = 1
Ceva theorem has several corollaries:
Van Obel theorem
If three lines AD, BE, CF from the vertices A,B,C, of a triangle to points on the opposite sides, meet at one point K, then AK / KD = AE/EC + AF/FB
Napoleon theorem
(Here is a more general formulation then the original by Napoleon). Given a triangle ABC construct on its sides similar triangles ABF, BCD, CAE, with
AB ~ CD ~ EC BF ~ DB ~ CA FA ~ BC ~ AE
Then the triangle with vertices "corresponding" (eg. centroid) points of the three triangles is similar to them.
Napoleon theorem is related to homothetic triangles.|
Projective Basis
In order to introduce the projective basis consider two equations in P3
a=0 and b=0 are two planes, therefore they define a pencil of planes: any plane of the pencil can be written m a - n b = 0 Since the coefficients of the planes are defined up to a scale factor, a specific plane of the pencil can always be written as a - b = 0.
A fourth plane on the pencil can then be written h a - b = 0. It can be checked that h is the cross ratio of the four plane,
Given five points in P3, four at a time independent, p1, p2, p3, p4 and U, we consider the four planes,
n1 = plane of p2, p3, p4 n2 = plane of p3, p4, p1 n3 = plane of p4, p1, p2 n4 = plane of p1, p2, p3
The pencil of planes with axis the line p1-p2 is spanned by n3 and n4, n = h n4 - n3. Let n4 - n3 = 0 be the plane of this pencil passing thru U. Similarly let n4 - n1 = 0 be the plane of the pencil on p2-p3 thru U, and n4 - n2 = 0 the plane of the pencil on p1-p3 thru U.
Given a point p, consider the cross-ratio coordinates
where ni(p) is the plane of the pencil h n4 - ni = 0 passing thru P. Therfore ni(p) = zi n4 - ni.
(z1, z2, z3) are the non-homogeneous projective coordinates of p in the projective reference (p1,p2,p3,p4,U). They identify three planes passing thru the point p. Explicitly,
The homogeneous projective coords are
In general a projective basis of Pn is a set on n+1 points p0, p1, ..., pn , such that any n of them are independent. Any point can be written as linear combination of any set on n of them, for example
Example (1,0,0,0) (0,1,0,0) (0,0,1,0) (0,0,0,1) and (1,1,1,1) is a projective basis of P3, consisting of the ideal points in the three X,Y,Z directions, the origin and the unit. The points p1, p2, p3 are the ideal points on the axes; p4 is the origin and U is the unit point.
The cartesian reference is a particular case of projective reference, with the first n-1 base points at infinity in orthogonal (if the cartesian reference is orthogonal) directions.
A plane remains a linear equation when expressed in projective coords. Consider the reference planes
Any plane ni has projective coords
I leave to verify this relation (in P2) for lines. The transformation to projective coordinates preserves the incidence relations,
z`a v.a = e`abcd e.ijkl p`i p.b`j p.c`k p.d`l n.m p.a`m = n.m e.ijkl p1`m p2`j p3`k p4`l p`i = det n.m d.im p`i = det n.i p`i
Homography
In projective coordinates the relation between the coordinates in two projective frames is
with inverse
H is a 3x3 matrix and defines a homography. It has 8 independent coefficients (all relations are up to a scale factor). If det(H) != 0 the homography is non-degenerate.
A homography maps planes into planes.
infact mb yb ~ (Ht-1)ba (H)bc na zc = (H-1)ab (H)bc na zc = na za.
The equation of the invariant points of a homography is a fourth order algebraic equation, r ya = za. The homographies form a non-abelian group, with unit the identity matrix.
Theorem of the homography
A homography (in P3) is determined by five corresponding points.
Remark. A homography in P3 with five invariant elemnts is the identity.
Camera Centered Projection (P3 --> P2)
Given a point pi and a plane nj the projection of the points of P3 onto the plane nj with center pi is obtained by intersecting the plane with the line (ray) joining the point to pi:
The projection points lie on the plane ni, ie, Qi ni = 0 as is obvious by the previous equation. If we take a projective basis on the plane p1, p2, ..., p4 (thus ni p1i = ni p2i = ... = ni p4i = 0) so that we can write
and add to it the center p = p0
p`i = p0`i q`i = c_0 p0'i + c_1 p1`i + ... + c_3 p3`i
and Qi = C1 p1i + ... + C3 p3i with
Remark: the expression for Q is homogeneous in p0, p1, ..., p3 and q.
Example. Consider the image plane p1=(1,0,0,1), p2=(0,1,0,1), p3=(0,0,0,1) p0=(0,0,f,1) and q=(X,Y,Z,1). Then the plane is n=(0,0,1,0) and the line p0-q is
u.12 = f - Z u.23 = - X u.13 = Y u.24 = f X u.14 = -f Y u.34 = 0
ie, the plucker coords,
u`12 = 0 u`41 = X u`13 = -f X u`42 = Y u`23 = -f Y u`43 = Z - f
The intersection of the line and the plane is
C_1 = f X C_2 = f Y C_3 = f - Z - f X - f Y
therefore
Q = f X p1 + f Y p2 + (f - Z - f X - f Y) p3 = ( f X, f Y, 0, f - Z) = ( -f X / (Z-f), -f Y / (Z-f), 0, 1 )
Least Square Fitting
We now consider the least square fitting of projective points in P2 to find the interpolating line. The function to minimize is the sun of the Euclidean square distances of the points from the line,
By differentiating we have the system
| [XX] [XY] [X] | | u.1 | | [XY] [YY] [Y] | | u.2 | = 0 | [X] [Y] [1] | | u.3 |
If the points are not all aligned this system does not have a solution (the determinant is not zero). In general the matrix has three positive eigenvalues Li, as it is symmetric positive definite, and can be written
therefore
The eigenvector of the smallest eigenvalue can be taken as the best fitting line. An analogous problem is the best fitting point of a set of lines (ie, their intersection). In this case we normalize the lines so that uk12 + uk22 = 1. Then we minimize the function
This is the usual least square fitting problem, in fact it does not make sense to minimize Euclidean distances unless the lines are all proper (not the ideal line), and therefore the solution is a proper point.