Page 122 - Linear Algebra Done Right

Orthonormal Bases
Let 1 ≤ k<j. Then
e j ,e k = v j −Pv j ,e 1 e 1 − ··· − v j ,e j−1 e j−1 ,e k 109
v j −Pv j ,e 1 e 1 − ··· − v j ,e j−1 e j−1
v j ,e k −Pv j ,e k
=
v j −Pv j ,e 1 e 1 −· · ·−Pv j ,e j−1 e j−1
= 0.
Thus (e 1 ,...,e j ) is an orthonormal list.
From 6.23, we see that v j ∈ span(e 1 ,...,e j ). Combining this infor-
mation with 6.22 shows that

span(v 1 ,...,v j ) ⊂ span(e 1 ,...,e j ).

Both lists above are linearly independent (the v’s by hypothesis, the e’s
by orthonormality and 6.16). Thus both subspaces above have dimen-
sion j, and hence they must be equal, completing the proof.

Now we can settle the question of the existence of orthonormal
bases.

6.24 Corollary: Every finite-dimensional inner-product space has an Until this corollary,
orthonormal basis. nothing we had done
with inner-product
Proof: Choose a basis of V. Apply the Gram-Schmidt procedure spaces required our
(6.20) to it, producing an orthonormal list. This orthonormal list is standing assumption
linearly independent (by 6.16) and its span equals V. Thus it is an that V is finite
orthonormal basis of V. dimensional.

As we will soon see, sometimes we need to know not only that an
orthonormal basis exists, but also that any orthonormal list can be
extended to an orthonormal basis. In the next corollary, the Gram-
Schmidt procedure shows that such an extension is always possible.

6.25 Corollary: Every orthonormal list of vectors in V can be ex-
tended to an orthonormal basis of V.

Proof: Suppose (e 1 ,...,e m ) is an orthonormal list of vectors in V.
Then (e 1 ,...,e m ) is linearly independent (by 6.16), and hence it can be
extended to a basis (e 1 ,...,e m ,v 1 ,...,v n ) of V (see 2.12). Now apply