A First Course in Linear Algebra, 2017a

4.11. Orthogonality and the Gram Schmidt Process 237
Solution. First we examine the product AB.
(AB)(B T A T )=A(BB T )A T = AA T = I
Since AB is square, B T A T =(AB) T is the inverse of AB, soAB is invertible, and (AB) −1 =(AB) T Therefore,
AB is orthogonal.
Next we show that A −1 = A T is also orthogonal.
(A −1 ) −1 = A =(A T ) T =(A −1 ) T
Therefore A −1 is also orthogonal.
♠
4.11.3 Gram-Schmidt Process
The Gram-Schmidt process is an algorithm to transform a set of vectors into an orthonormal set spanning
the same subspace, that is generating the same collection of linear combinations (see Definition 9.12).
The goal of the Gram-Schmidt process is to take a linearly independent set of vectors and transform it
into an orthonormal set with the same span. The first objective is to construct an orthogonal set of vectors
with the same span, since from there an orthonormal set can be obtained by simply dividing each vector
by its length.
Algorithm 4.135: Gram-Schmidt Process
Let {⃗u 1 ,···,⃗u n } be a set of linearly independent vectors in R n .
I: Construct a new set of vectors {⃗v 1 ,···,⃗v n } as follows:
⃗v 1 =⃗u 1 ( ) ⃗u2 •⃗v 1
⃗v 2 =⃗u 2 −
‖⃗v 1 ‖ 2 ⃗v 1
( ) ( )
⃗u3 •⃗v 1 ⃗u3 •⃗v 2
⃗v 3 =⃗u 3 −
‖⃗v 1 ‖ 2 ⃗v 1 −
‖⃗v 2 ‖ 2 ⃗v 2
... ( ) ( ) ( )
⃗un •⃗v 1 ⃗un •⃗v 2
⃗un •⃗v n−1
⃗v n =⃗u n −
‖⃗v 1 ‖ 2 ⃗v 1 −
‖⃗v 2 ‖ 2 ⃗v 2 −···−
‖⃗v n−1 ‖ 2 ⃗v n−1
II: Now let ⃗w i = ⃗v i
for i = 1,···,n.
‖⃗v i ‖
Then
1. {⃗v 1 ,···,⃗v n } is an orthogonal set.
2. {⃗w 1 ,···,⃗w n } is an orthonormal set.
3. span{⃗u 1 ,···,⃗u n } = span{⃗v 1 ,···,⃗v n } = span{⃗w 1 ,···,⃗w n }.
Proof. The full proof of this algorithm is beyond this material, however here is an indication of the arguments.