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6.6 Orthogonal Complements and Projections 179

The vector on the left is in S and the vector on the right is in S . Therefore both sides equal the
⊥
zero vector, so
u S = U and u = U .
⊥
⊥
This completes the proof.
Notice in the theorem that, if u is actually in S, then u S = u and u = O.
⊥

The vector u S produced in the proof is called the orthogonal projection of u onto S.Itis
the sum of the projections of u onto an orthogonal basis for S.

It would appear from the way u S was formed that this orthogonal projection depends on the
orthogonal basis specified for S. In fact it does not, and any orthogonal basis for S leads to the
same orthogonal projection u S , justifying the term the orthogonal projection of u onto S.The
reason for this is that, given u, the orthogonal projection of u onto S is the unique vector in S
⊥
such that u is the sum of this projection and a vector in S .
It is therefore true that, if we write a vector u as the sum of a vector in S and a vector in
S , then necessarily the vector in S is u S and the vector in S is u − u S . In particular, u − u S is
⊥
⊥
orthogonal to every vector in S.
EXAMPLE 6.20
5
Let S be the subspace of R consisting of all < x,0, y,0, z > having zero second and fourth
components. Let
u =< 1,4,1,−1,3 >.
We will determine at u S and u . First use the orthogonal basis
⊥
V 1 =< 1,0,0,0,0 >,V 2 =< 0,0,1,0,2 >,V 3 =< 0,0,2,0,−1 >
for S. The orthogonal projection u S is
u · V 1 u · V 2 u · V 3
u S = V 1 + V 2 + V 3
V 1 · V 1 V 2 · V 2 V 3 · V 3
7 1
= V 1 + V 2 − V 3
5 5
=< 1,0,1,0,3 >,
and
u = u − u S =< 0,4,0,−1,0 >
⊥
⊥
is in the orthogonal complement of S, being orthogonal to every vector in S, and u = u S + u .
Suppose we used a different orthogonal basis for S,say
∗
∗
∗
V =< 1,0,1,0,0 >,V =< −3,0,3,0,0 >,V =< 0,0,0,0,6 >.
2
3
1
Now compute the orthogonal projection of u with respect to this basis:
u · V ∗ 1 u · V ∗ 2 u · V ∗ 3
∗
∗
V + V + V ∗ 3
1
2
∗
∗
∗
V · V ∗ 1 V · V ∗ 2 V · V ∗ 3
1
3
2
1
∗
= V + 0V + V ∗ 3
∗
2
1
2
=< 1,0,1,0,3 >,
the same as obtained using the first orthogonal basis. This illustrates the uniqueness of u S ,given
u and S.
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October 14, 2010 14:21 THM/NEIL Page-179 27410_06_ch06_p145-186