Linear Algebra

Section VI. Projection 253that are orthogonal and so are not-interacting.suggestion.We will now develop that2.1 Definition Vectors ⃗v 1 , . . . ,⃗v k ∈ R n are mutually orthogonal when any twoare 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 orthogonaland nonzero then that set is linearly independent.Proof Consider a linear relationship c 1 ⃗v 1 +c 2 ⃗v 2 +· · ·+c k ⃗v k = ⃗0. If i ∈ {1, .. , k}then taking the dot product of ⃗v i with both sides of the equation⃗v i • (c 1 ⃗v 1 + c 2 ⃗v 2 + · · · + c k ⃗v k ) = ⃗v i • ⃗0c i · (⃗v i • ⃗v i ) = 0shows, since ⃗v i ≠ ⃗0, that c i = 0.QED2.3 Corollary In a k dimensional vector space, if the vectors in a size k set aremutually 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 abasis.QEDOf course, the converse of Corollary 2.3 does not hold — not every basis ofevery subspace of R n has mutually orthogonal vectors. However, we can getthe partial converse that for every subspace of R n there is at least one basisconsisting of mutually orthogonal vectors.2.4 Example The members ⃗β 1 and ⃗β 2 of this basis for R 2 are not orthogonal.( (⃗β 24 1B = 〈 , 〉 ⃗β 12)3)However, we can derive from B a new basis for the same space that does havemutually orthogonal members. For the first member of the new basis we simplyuse ⃗β 1 .( )4⃗κ 1 =2For the second member of the new basis, we subtract from ⃗β 2 the part in thedirection of ⃗κ 1 . This leaves the part of ⃗β 2 that is orthogonal to ⃗κ 1 .( ( ( ( ( ⃗κ 1 1 1 2 −1 2⃗κ 2 = − proj3)[⃗κ1 ]( ) = − =3)3)1)2)