Page 270 - 35Linear Algebra

270 Orthonormal Bases and Complements

Possibly by now you are feeling overwhelmed, it may help to watch this quick
overview video.

Overview

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Example 138 Consider any plane P through the origin in R . Then P is a subspace,
and P ⊥ is the line through the origin orthogonal to P. For example, if P is the
xy-plane, then

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R = P ⊕ P ⊥ = {(x, y, 0)|x, y ∈ R} ⊕ {(0, 0, z)|z ∈ R}.

Theorem 14.6.2. Let U be a subspace of a finite-dimensional vector space W.
⊥
Then the set U ⊥ is a subspace of W, and W = U ⊕ U .
Proof. First, to see that U ⊥ is a subspace, we only need to check closure,
⊥
which requires a simple check: Suppose v, w ∈ U , then we know
v u = 0 = w u (∀u ∈ U) .

Hence
⇒ u (αv + βw) = αu v + βu w = 0 (∀u ∈ U) ,
⊥
and so αv + βw ∈ U .
Next, to form a direct sum between U and U ⊥ we need to show that
⊥
U ∩ U = {0}. This holds because if u ∈ U and u ∈ U ⊥ it follows that
u u = 0 ⇔ u = 0.

⊥
Finally, we show that any vector w ∈ W is in U ⊕ U . (This is where
we use the assumption that W is finite-dimensional.) Let e 1 , . . . , e n be an
orthonormal basis for U. Set:

u = (w e 1 )e 1 + · · · + (w e n )e n ∈ U ,
u ⊥ = w − u .

⊥
It is easy to check that u ∈ U ⊥ (see the Gram-Schmidt procedure). Then
⊥
⊥
w = u + u , so w ∈ U ⊕ U , and we are done.

Reading homework: problem 4

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