4 Some Matrix Theory 237Proof Since
A=
[
1 O
ent− 1 I][
1 en− 1
O − 2 B]
,
whereemdenotes a row ofm1’s, the matrixAhas determinant(− 2 )n−^1 detB.The
rest of the lemma is obvious.
LetAbe ann×nmatrix with entries±1. By multiplying rows and columns ofA
by−1 we can make all elements in the first row and first column equal to 1 without
altering the value of det(AtA). It follows from Lemma 9 that ifαnis the maximum
of det(AtA)among alln×nmatricesA=(αjk)withαjk∈{− 1 , 1 },andifβn− 1
is the maximum of det(BtB)among all(n− 1 )×(n− 1 )matricesB=(βjk)with
βjk∈{ 0 , 1 },then
αn= 22 n−^2 βn− 1.4 SomeMatrixTheory.........................................
In rectangular coordinates the equation of an ellipse with centre at the origin has the
form
Q:=ax^2 + 2 bxy+cy^2 =const. (∗)This is not the form in which the equation of an ellipse is often written, because of the
‘cross product’ term 2bxy. However, we can bring it to that form by rotating the axes,
so that the major axis of the ellipse lies along one coordinate axis and the minor axis
along the other. This is possible because the major and minor axes are perpendicular
to one another. These assertions will now be verified analytically.
In matrix notation,Q=ztAz,where
A=[
ab
bc]
, z=[
x
y]
.
A rotation of coordinates has the formz=Tw,where
T=[
cosθ −sinθ
sinθ cosθ]
,w=[
u
v]
.
ThenQ=wtBw,whereB=TtAT. Multiplying out, we obtain
B=
[
a′ b′
b′ c′]
,
where
b′=b(cos^2 θ−sin^2 θ)−(a−c)sinθcosθ.To eliminate the cross product term we chooseθ so thatb(cos^2 θ−sin^2 θ) =
(a−c)sinθcosθ; i.e., 2bcos 2θ=(a−c)sin 2θ,or
tan 2θ= 2 b/(a−c).