Graphics Reference
In-Depth Information
Solution.
We follow the steps outlined in Theorem 1.9.11. If the elementary
matrices E, F, and G are defined by
Ê
ˆ
100
110
001
100
010
201
101
010
Ê
ˆ
Ê
ˆ
Á
Á
Á
˜
˜
˜
Á
Á
˜
˜
Á
Á
˜
˜
E
=
,
F
=
,
and
G
=
,
1
2
Ë
¯
Ë
¯
-
0
1
Ë
¯
then
10 0
040
00 2
Ê
ˆ
Á
Á
˜
TT T
AGFEAEFG
=
=
-
˜
.
1
Ë
¯
-
Finally, define the elementary diagonal matrix H by
Ê
ˆ
10 0
Á
Á
Á
Á
Á
˜
˜
˜
˜
˜
1
2
0
0
H =
1
2
00
Ë
¯
and observe that
10 0
010
00 1
Ê
ˆ
Á
Á
˜
AHAH
T
=
=
-
˜
.
2
1
Ë
¯
-
Therefore, if M = HGFE, then MAM
T
= A
2
is the desired matrix and we are done.
It is worth pointing out one consequence of Theorem 1.9.10.
1.9.13. Corollary.
A positive definite quadratic form is nondegenerate and all of its
discriminants are positive.
Proof.
We may assume that the vector space is
R
n
and then must have s = n in
Theorem 1.9.10. The discriminant is certainly positive with the respect to the ortho-
normal basis guaranteed by the theorem. The reason that the discriminant is always
positive is that the determinant of congruent matrices differs by a square.
1.10
The Cross Product Reexamined
In Section 1.5 we observed that
R
3
has not only a dot product but also a cross product.
Note that the cross product produces another vector, whereas the dot product was a
real number. Various identities involving the dot and cross product are known. The
