Page 425 - 35Linear Algebra

G.13 Orthonormal Bases and Complements 425

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Taking a direct sum we again get the whole space, U ⊕ V = R .
Now we come to an orthogonal complement. There is not really a notion of
subtraction for subspaces but the orthogonal complement comes close. Given U
it provides a space U ⊥ such that the direct sum returns the whole space:
3
U ⊕ U ⊥ = R .
The orthogonal complement U ⊥ is the subspace made from all vectors perpen-
dicular to any vector in U. Here, we need to just tilt the line V above until
it hits U at a right angle:

Notice, we can apply the same operation to U ⊥ and just get U back again, i.e.
⊥ ⊥

U = U .
Hint for Review Question 2
You are asked to consider an orthogonal basis {v 1 , v 2 , . . . v n }. Because this is a
basis any v ∈ V can be uniquely expressed as
2
n
1
v = c v 1 + c v 2 + · · · + v c n ,
and the number n = dim V . Since this is an orthogonal basis
v i v j = 0 , i 6= j .
So different vectors in the basis are orthogonal:

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