COURSE OF PROJECTIVE GEOMETRY

§ 14:

*O55* Suppose A = ((a,b),(c,d)) meets the requirements. A maps (4,-5) to ρ(1,2) and (1,2) to σ(4,-5). We get A = μ((2.1),(-1,-2)).

*O56* Suppose A = ((a,b),(c,d)) meets the requirements. Using A^{2} = λE, we find a^{2} = d^{2} and c(a+d) = b(a+d) = 0.

If a=d≠0 then b=c=0, but then the projectivity is trivial. So a=-d.

We find the eigenvalues by equating the characteristic polynomial λ^{2} - a^{2} - bc to 0. So hyperbolic if a^{2} + bc is positive,
elliptic if a^{2} + bc is negative. If a^{2} + bc = 0, then the matrix is singular.

*O57* Rename *X*_{3}=*Y*_{1} and prove: *Y*_{3}=*X*_{1} ↔ *P* on *p*.

(*Y*_{3}=*X*_{1} ↔ *PY*_{3}=*PX*_{1} ↔ *Y*_{2}*X*_{1}=*PX*_{1})

"→" : If *Y*_{3}=*X*_{1} then *Y*_{2}*X*_{1}=*PX*_{1} and
*Y*_{2}*X*_{1}.*Y*_{1}*X*_{2} = *PX*_{1}.*PX*_{2} = *P*. So *P* on *p*.

"→" : If *P* on *p* then *P* = *p*.*Y*_{1}*X*_{2} so *P* op *Y*_{2}*X*_{1}. Then
*Y*_{2}*X*_{1}=*PX*_{1} so *Y*_{3}=*X*_{1}.

*O58* From the data it follows that (*P*, *Q*; *A*, *A'*) = (*P*, *Q*; *B*, *B'*) = -1 = (*Q*, *P*; *B*, *B'*).

Hence, the projectivity that maps *P* to *Q*, and *Q* to *P*, and *A* to *B*, maps *A'* to *B'*.

(It is an involution because it interchanges *P* and *Q*.)