Page 267 - 35Linear Algebra

14.6 Orthogonal Complements 267

vectors. To achieve that we divide each column of the first matrix by its length and
multiply the corresponding row of the second matrix by the same amount:
 √ √ √   √ √ 
2 5 − 7 30 − 6 5 5 0
5 90 18 5
 √ √ √   √ √ 

 5 7 30 6   3 30 30  = QR .
5 45 9 5 2
M =    0 −
√ √ √
   
0 30 − 7 6 0 0 6
18 18 2
Geometrically what has happened here is easy to see. We started with three vectors
given by the columns of M and rotated them such that the first lies along the x-
axis, the second in the xy-plane and the third in some other generic direction (here it
happens to be in the yz-plane).
A nice check of the above result is to verify that entry (i, j) of the matrix R equals
the dot product of the i-th column of Q with the j-th column of M. (Some people
memorize this fact and use it as a recipe for computing QR decompositions.) A good
test of your own understanding is to work out why this is true!

Another QR decomposition example

14.6 Orthogonal Complements

Let U and V be subspaces of a vector space W. In Review Exercise 2,
Chapter 9, you are asked to show that U ∩ V is a subspace of W, and that
U ∪ V is not a subspace. However, span(U ∪ V ) is certainly a subspace,
since the span of any subset of a vector space is a subspace. Notice that all
elements of span(U ∪ V ) take the form u + v with u ∈ U and v ∈ V . We call
the subspace

U + V := span(U ∪ V ) = {u + v | u ∈ U, v ∈ V }

the sum of U and V . Here, we are not adding vectors, but vector spaces to
produce a new vector space.

Example 137
             
 1  0  1 0

 0   0   0 


1 1 1 0 1 1 0
     
             
,
,
,
,
span     + span     = span       .
 0  1  0 1
     
   
   

 1   1   1 


0 0 0 1 0 0 1
     
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