Projective GeometryEric Bainville - Oct 2007
Antisymmetric 4x4 matrices
Let's consider a general non-zero 4x4 anti-symmetric matrix A:
The characteristic polynomial of A is X4+(a2+b2+c2+d2+e2+f2)*X2+(af-be+cd)2. A2 is a symmetric matrix, and its characteristic polynomial is (X2+(a2+b2+c2+d2+e2+f2)*X+(af-be+cd)2)2. When (af-be+cd) ≠ 0, A is invertible. When (af-be+cd)=0, A is singular; it has rank 2, and its kernel has dimension 2. Knowing this, we will define in the next chapter a line as the kernel of a non-zero antisymmetric matrix of rank 2. Before doing this, let's study these matrices a little further.
Let's define as follows the 4x4 anti-symmetric matrix P(a,b) from two vectors a,b in R3 with the properties (a,b) ≠ (0,0) (i.e. the matrix is not 0), and <a,b> = 0. The set of all such matrices will be denoted L.
If A=P(a,b), then we denote by A' the matrix P(b,a). We have (A')' = A, and we can check that A'A = AA' = 0. It means the image of A is the kernel of A', and the image of A' is the kernel of A, all these vector spaces having dimension 2.
Since A is antisymmetric, <Ax,x> = 0 for any vector x in R4: a vector and its image are always orthogonal. If we take x in E = R4/Ker(A) (i.e. not in the kernel of A), then A2x = -k2 x, with k2 = |u|2+|v|2: Ax is in E too. (x,Ax) is then an orthogonal basis of E, and so is (Ax,A2x). Now apply A' to these two orthogonal vectors: A'Ax = 0 and A'A2x = 0. It means R4/Ker(A) = Ker(A'). The same way, we have R4/Ker(A') = Ker(A). Since the symmetric matrix A2 can be diagonalized in an orthogonal basis, we can now conclude.
Ker(A) ⊗ Ker(A') is an orthogonal decomposition of R4. Both subspaces have dimension 2. A restricted to Ker(A') is a bijection mapping Ker(A') to itself, each vector being mapped to an orthogonal vector. So is A' restricted to Ker(A). A2 restricted to Ker(A') is -|A|2 Id (|A| is the L2 norm of A), and so is A'2 restricted to Ker(A).
A matrix of L can be constructed from two independent vectors u,v of R4. Let's consider the matrix A = v uT - u vT. A is a 4x4 matrix. AT = u vT - v uT = -A, so A is antisymmetric. For a vector x in R4, Ax = v <u,x> - u <v,x>, meaning the image of A is a linear combination of u and v. Since u and v are independent, the image of A is exactly the vector space generated by (u,v).
|Projective Geometry : Lines I||Top of Page||Projective Geometry : Lines II|