Linear Algebra, 2020a

280 Chapter Three. Maps Between Spaces
VI.2
Gram-Schmidt Orthogonalization
The prior subsection suggests that projecting ⃗v into the line spanned by ⃗s
decomposes that vector into two parts
⃗v
⃗v − proj [⃗s] (⃗p)
proj [⃗s] (⃗p)
⃗v = proj [⃗s ] (⃗v) + ( ⃗v − proj [⃗s ] (⃗v) )
that are orthogonal and so are “non-interacting.” We now develop that suggestion.
2.1 Definition Vectors ⃗v 1 ,...,⃗v k ∈ R n are mutually orthogonal when any two
are orthogonal: if i ≠ j then the dot product ⃗v i • ⃗v j is zero.
2.2 Theorem If the vectors in a set {⃗v 1 ,...,⃗v k } ⊂ R n are mutually orthogonal
and nonzero then that set is linearly independent.
Proof Consider ⃗0 = c 1 ⃗v 1 + c 2 ⃗v 2 + ···+ c k ⃗v k .Fori ∈ {1,..,k}, taking the dot
product of ⃗v i with both sides of the equation ⃗v i •(c 1 ⃗v 1 +c 2 ⃗v 2 +···+c k ⃗v k )=⃗v i •⃗0,
which gives c i · (⃗v i • ⃗v i )=0, shows that c i = 0 since ⃗v i ≠ ⃗0.
QED
2.3 Corollary In a k dimensional vector space, if the vectors in a size k set are
mutually orthogonal and nonzero then that set is a basis for the space.
Proof Any linearly independent size k subset of a k dimensional space is a
basis.
QED
Of course, the converse of Corollary 2.3 does not hold — not every basis of
every subspace of R n has mutually orthogonal vectors. However, we can get
the partial converse that for every subspace of R n there is at least one basis
consisting of mutually orthogonal vectors.
2.4 Example The members ⃗β 1 and ⃗β 2 of this basis for R 2 are not orthogonal.
( (
⃗β 2
4 1
B = 〈 , 〉 ⃗β 1
2)
3)
We will derive from B a new basis for the space 〈⃗κ 1 ,⃗κ 2 〉 consisting of mutually
orthogonal vectors. The first member of the new basis is just ⃗β 1 .
( )
4
⃗κ 1 =
2