Page 272 - 35Linear Algebra
P. 272
272 Orthonormal Bases and Complements
So the set
1 1
1
2
3
−1 1 1
,
,
2 3
0 −1 1
3
0 0 −1
⊥
is an orthogonal basis for L . Dividing each basis vector by its length yields
√
1 1 3
√ √
6 √ 6
2
− √
1 √ 1 3
2 6 , √ 6 ,
, 2
3
0 − √
6 6
√
0 0 − 3
2
⊥
and orthonormal basis for L . Moreover, we have
c x
c
y
⊥
4
R = L ⊕ L = c ∈ R ⊕ ∈ R x + y + z + w = 0 ,
4
z
c
c w
4
a decomposition of R into a line and its three dimensional orthogonal complement.
⊥ ⊥
Notice that for any subspace U, the subspace (U ) is just U again. As
such, ⊥ is an involution on the set of subspaces of a vector space. (An invo-
lution is any mathematical operation which performed twice does nothing.)
14.7 Review Problems
Reading Problems 1 , 2 , 3 , 4
Gram–Schmidt 5
Webwork:
Orthogonal eigenbasis 6, 7
Orthogonal complement 8
0
λ 1
1. Let D = .
0 λ 2
(a) Write D in terms of the vectors e 1 and e 2 , and their transposes.
a b
(b) Suppose P = is invertible. Show that D is similar to
c d
1 λ 1 ad − λ 2 bc −(λ 1 − λ 2 )ab
M = .
ad − bc (λ 1 − λ 2 )cd −λ 1 bc + λ 2 ad
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