Page 408 - Matrix Analysis & Applied Linear Algebra

404 Chapter 5 Norms, Inner Products, and Orthogonality

Proof. Observe that M∩M ⊥ = 0 because if x ∈M and x ∈M , then
⊥
x must be orthogonal to itself, and x x =0 implies x =0. To prove that
⊥
M⊕M = V, suppose that B M and B M ⊥ are orthonormal bases for M
⊥ ⊥
and M , respectively. Since M and M are disjoint, B M ∪B M ⊥ is an
⊥
orthonormal basis for some subspace S = M⊕M ⊆V. If S = V, then
the basis extension technique of Example 4.4.5 followed by the Gram–Schmidt
orthogonalization procedure of §5.5 yields a nonempty set of vectors E such that
B M ∪B M ⊥ ∪E is an orthonormal basis for V. Consequently,
⊥
=⇒E ⊥ M =⇒E ⊆ M =⇒E ⊆ span (B M ⊥) .
E⊥ B M
M ⊥ ∪E is linearly independent. Therefore,
But this is impossible because B M ∪B
E is the empty set, and thus V = M⊕M . To prove statement (5.11.2),
⊥
note that N⊥ M implies N⊆ M , and coupling this with the fact that
⊥
M⊕M = V = M⊕N together with (4.4.19) insures
⊥
dim N = dim V− dim M = dim M .
⊥
Example 5.11.1

Problem: Let U m×m = U 1 | U 2 be a partitioned orthogonal matrix. Explain
why R (U 1 ) and R (U 2 )must be orthogonal complements of each other.
m
Solution: Statement (5.9.4) insures that = R (U 1 ) ⊕ R (U 2 ), and we know
that R (U 1 ) ⊥ R (U 2 )because the columns of U are an orthonormal set.
⊥
Therefore, (5.11.2) guarantees that R (U 2 )= R (U 1 ) .

Perp Operation
If M is a subspace of an n-dimensional inner-product space, then the
following statements are true.
⊥
• dim M = n − dim M. (5.11.3)
⊥
⊥
• M = M. (5.11.4)
Proof. Property (5.11.3) follows from the fact that M and M ⊥ are comple-
⊥
⊥
mentary subspaces—recall (4.4.19). To prove (5.11.4), first show that M ⊆
⊥
⊥
M. If x ∈M , then (5.11.1) implies x = m + n, where m ∈M and
n ∈M , so
⊥
0= n x = n m + n = n m + n n = n n =⇒ n =0 =⇒ x ∈M,
⊥
⊥ ⊥
and thus M ⊆M. We know from (5.11.3) that dim M = n − dim M and
⊥ ⊥
⊥ ⊥ ⊥
dim M = n−dim M , so dim M = dim M. Therefore, (4.4.6) guarantees
⊥
⊥
that M = M.