Linear Algebra

264 Chapter 3. Maps Between Spaces
P ⊥ consists of the vectors that satisfy these two conditions.
⎛
⎝ 1
⎞
0⎠
⎛
⎝
3
v1
⎞
v2⎠
=0
⎛
⎝ 0
⎞
1⎠
⎛
⎝
2
v1
⎞
v2⎠
=0
v3
We can express those conditions more compactly as a linear system.
P ⊥ ⎛
= { ⎝ v1
⎞
v2⎠
� �
� 1
0
0
1
�
3
2
⎛
⎝ v1
⎞
� �
v2⎠
0
= }
0
v3
v3
We are thus left with finding the nullspace of the map represented by the matrix,
that is, with calculating the solution set of a homogeneous linear system.
P ⊥ ⎛ ⎞
v1
= { ⎝v2⎠
� � v1
⎛ ⎞
−3
+3v3 =0
} = {k ⎝−2⎠
v2 +2v3 =0
1
� � k ∈ R}
3.6 Example Where M is the xy-plane subspace of R 3 , what is M ⊥ ? A
common first reaction is that M ⊥ is the yz-plane, but that’s not right. Some
vectors from the yz-plane are not perpendicular to every vector in the xy-plane.
� �
1
1 �⊥
0
� �
0
3
2
cos θ =
v3
v3
1 · 0+1· 3+0· 2
√ 1+1+0· √ 0+9+4 so θ ≈ 0.94 rad
Instead M ⊥ is the z-axis, since proceeding as in the prior example and taking
the natural basis for the xy-plane gives this.
M ⊥ ⎛ ⎞
x
= { ⎝y⎠
z
� � �
� 1 0 0
0 1 0
⎛ ⎞
⎛ ⎞
x � � x
⎝y⎠
0
= } = { ⎝y⎠
0
z
z
� � x =0andy =0}
The two examples that we’ve seen since Definition 3.4 illustrate the first
sentence in that definition. The next result justifies the second sentence.
3.7 Lemma Let M be a subspace of R n . The orthogonal complement of M is
also a subspace. The space is the direct sum of the two R n = M ⊕ M ⊥ . And,
for any �v ∈ R n , the vector �v − proj M (�v ) is perpendicular to every vector in M.
Proof. First, the orthogonal complement M ⊥ is a subspace of R n because, as
noted in the prior two examples, it is a nullspace.
Next, we can start with any basis BM = 〈�µ1,...,�µk〉 for M and expand it to
a basis for the entire space. Apply the Gram-Schmidt process to get an orthogonal
basis K = 〈�κ1,...,�κn〉 for R n . This K is the concatenation of two bases
〈�κ1,...,�κk〉 (with the same number of members as BM) and〈�κk+1,...,�κn〉.
ThefirstisabasisforM, so if we show that the second is a basis for M ⊥ then
we will have that the entire space is the direct sum of the two subspaces.