Page 125 - Linear Algebra Done Right

112
6.31
v = v, e 1 e 1 + ··· + v, e m e m + v −Pv, e 1 e 1 −· · ·−Pv, e m e m .

w
u Chapter 6. Inner-Product Spaces
Clearly u ∈ U. Because (e 1 ,...,e m ) is an orthonormal list, for each j
we have
w, e j = v, e j −Pv, e j
= 0.
Thus w is orthogonal to every vector in span(e 1 ,...,e m ). In other
words, w ∈ U . Thus we have written v = u + w, where u ∈ U
⊥
and w ∈ U , completing the proof of 6.30.
⊥
If v ∈ U ∩ U , then v (which is in U) is orthogonal to every vector
⊥
in U (including v itself), which implies that v, v = 0, which implies
that v = 0. Thus
6.32 U ∩ U ⊥ ={0}.

Now 6.30 and 6.32 imply that V = U ⊕ U ⊥ (see 1.9).

The next corollary is an important consequence of the last theorem.

6.33 Corollary: If U is a subspace of V, then

U = (U ) .
⊥ ⊥
Proof: Suppose that U is a subspace of V. First we will show that

6.34 U ⊂ (U ) .
⊥ ⊥
To do this, suppose that u ∈ U. Then u, v = 0 for every v ∈ U ⊥ (by
the definition of U ). Because u is orthogonal to every vector in U ,
⊥
⊥
we have u ∈ (U ) , completing the proof of 6.34.
⊥ ⊥
To prove the inclusion in the other direction, suppose v ∈ (U ) .
⊥ ⊥
By 6.29, we can write v = u + w, where u ∈ U and w ∈ U . We have
⊥
v − u = w ∈ U . Because v ∈ (U ) and u ∈ (U ) (from 6.34), we
⊥ ⊥
⊥
⊥ ⊥
have v −u ∈ (U ) . Thus v −u ∈ U ∩(U ) , which implies that v −u
⊥ ⊥
⊥ ⊥
⊥
is orthogonal to itself, which implies that v −u = 0, which implies that
v = u, which implies that v ∈ U. Thus (U ) ⊂ U, which along with
⊥ ⊥
6.34 completes the proof.