Page 436 - Matrix Analysis & Applied Linear Algebra

432 Chapter 5 Norms, Inner Products, and Orthogonality

x 3
x 1
x 2

x 1
x 2
(I − P 3 )x 3
(I − P 2 )x 2
x 1
Figure 5.13.3
Problem: Determine the volumes of a two-dimensional and a three-dimensional
parallelepiped, and then make the natural extension to define the volume of an
n-dimensional parallelepiped.
Solution: In the two-dimensional case, volume is area, and it’s evident from
Figure 5.13.3 that the area of the shaded parallelogram is the same as the area
of the dotted rectangle. The width of the dotted rectangle is ν 1 = x 1 , and
2
the height is ν 2 = (I − P 2 )x 2 , where P 2 is the orthogonal projector onto
2
the space (line) spanned by x 1 , and I − P 2 is the orthogonal projector onto
⊥
span {x 1 } . In other words, the area, V 2 , of the parallelogram is the length of
its base times its projected height, ν 2 , so
V 2 = x 1 (I − P 2 )x 2 = ν 1 ν 2 .
2 2
Similarly, the volume of a three-dimensional parallelepiped is the area of its
base times its projected height. The area of the base was just determined to be
V 2 = x 1 (I − P 2 )x 2 = ν 1 ν 2 , and it’s evident from Figure 5.13.3 that the
2 2
projected height is ν 3 = (I − P 3 )x 3 , where P 3 is the orthogonal projector
2
onto span {x 1 , x 2 } . Therefore, the volume of the parallelepiped generated by
{x 1 , x 2 , x 3 } is
V 3 = x 1 (I − P 2 )x 2 (I − P 3 )x 3 = ν 1 ν 2 ν 3 .
2 2 2
It’s now clear how to inductively define V 4 ,V 5 , etc. In general, the volume of
the parallelepiped generated by a linearly independent set {x 1 , x 2 ,..., x n } is

V n = x 1 (I − P 2 )x 2 (I − P 3 )x 3 ··· (I − P n )x n = ν 1 ν 2 ··· ν n ,
2 2 2 2
where P k is the orthogonal projector onto span {x 1 , x 2 ,..., x k−1 } , and where

ν 1 = x 1 and ν k = (I − P k )x k for k> 1. (5.13.7)
2 2
Note that if {x 1 , x 2 ,..., x n } is an orthogonal set, V n = x 1 x 2 ··· x n ,
2 2 2
which is what we would expect.