Algebra 413(p+q)A^2 B^2 =ABand(p+q)B^2 A^2 =BA. We have seen thatA^2 andB^2 commute, and
so we find thatAandBcommute as well, which contradicts the hypothesis. Therefore,
p=q.
(V. Vornicu)
203.For any numbert,
(
1 t
01)(
1 −t
01)
=
(
1 −t
01)(
1 t
01)
=
(
10
01
)
.
The equality from the statement can be rewritten
(
1 u
01)(
ab
cd)(
1 v
01)
=
(
10
c 1)
.
This translates to
(
a+uc v(a+uc)+b+ud
ccv+d
)
=
(
10
c 1)
.
Becausec =0 we can chooseusuch thata+uc=1. Then choosev=−(b+ud).
Theresulting matrix has 1 in the upper left corner and 0 in the upper right corner. In the
lower right corner it has
cv+d=−c(b+ud)+d=−bc−cud+d= 1 −ad−ucd+d
= 1 −(a+uc)d+d= 1.This also follows from the fact that the determinant of the matrix is 1. The numbersu
andvthat we have constructed satisfy the required identity.
Remark.This factorization appears in Gaussian optics. The matrices
(
1 ±u
01
)
and(
1 ±v
01)
model a ray of light that travels on a straight line through a homogeneous medium, while
the matrix
(
10
c 1
)
models refraction between two regions of different refracting indices. The result we have
just proved shows that any SL( 2 ,R)matrix with nonzero lower left corner is an optical
matrix.