# Projective Geometry

Eric Bainville - Oct 2007## Antisymmetric 4x4 matrices

Let's consider a general non-zero 4x4 anti-symmetric matrix A:

0 | a | b | c |

-a | 0 | d | e |

-b | -d | 0 | f |

-c | -e | -f | 0 |

The characteristic polynomial of A is X^{4}+(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})*X^{2}+(af-be+cd)^{2}.
A^{2} is a symmetric matrix, and its characteristic polynomial is (X^{2}+(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})*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 R^{3}
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.

0 | -a_{3} | a_{2} | b_{1} |

a_{3} | 0 | -a_{1} | b_{2} |

-a_{2} | a_{1} | 0 | b_{3} |

-b_{1} | -b_{2} | -b_{3} | 0 |

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 R^{4}: a vector and its image are always orthogonal.
If we take x in E = R^{4}/Ker(A) (i.e. not in the kernel of A),
then A^{2}x = -k^{2} x, with k^{2} = |u|^{2}+|v|^{2}: Ax is in E too. (x,Ax) is then an orthogonal basis of E, and so is (Ax,A^{2}x).
Now apply A' to these two orthogonal vectors: A'Ax = 0 and A'A^{2}x = 0.
It means R^{4}/Ker(A) = Ker(A'). The same way, we have R^{4}/Ker(A') = Ker(A). Since the symmetric matrix A^{2} can be
diagonalized in an orthogonal basis, we can now conclude.

Ker(A) ⊗ Ker(A') is an orthogonal decomposition of R^{4}. 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).
A^{2} restricted to Ker(A') is -|A|^{2} Id (|A| is the L_{2} 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 R^{4}. Let's consider the matrix A = v u^{T} - u v^{T}. A
is a 4x4 matrix. A^{T} = u v^{T} - v u^{T} = -A, so A is antisymmetric.
For a vector x in R^{4}, 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 |