OrthogonalityThe inner product and d istancesDefinition 1. The inner product (or dot product) of v, w in Rn:v ·w = vTw = v1w1++ vnwn.Example 2. For instance,123·1−1−2=Definition 3.•The norm (or length) of a vector v in Rniskvk = v ·v√= v12++ vn2p.This is the distance to the origin.• The distance between points v a nd w in Rnisdist(v, w) = kv −wk.vwv − wExample 4. For instance, in R2,distx1y1,x2y2=Or thogonal vectorsDefinition 5. v and w in Rnare orthogonal ifv ·w = 0.Armin [email protected] is this related to ou r understand ing of right angl es?Example 6. Are the following vectors orthogona l?(a)12,−21(b)121,−211Theorem 7. Suppose tha t v1,, vnare nonzero and pairwise orthogonal. Then v1,,vnare independent.Proof. Su ppose tha tc1v1++ cnvn= 0. Example 8. Let us consider A =1 22 43 6.FindNul(A) and Col(AT). Observe!Solution.Armin [email protected] 9. Repeat for A =1 2 12 4 03 6 0.Solution.The fundamental t heorem, s econ d actDefinition 10. Let W be a subspace of Rn, a n d v in Rn.• v is ortho gonal to W , if v ·w = 0 for all w in W .• Another subspace V is orthogonal to W , if every vector in V is orthogonal to W .• The orth ogonal complement of W is the space W⊥of all vectors that are orthog-onal to W .Exercise: show that the orthogonal complement is indeed a vector space.Example 11. In the previous example, A =1 2 12 4 03 6 0.We found thatNul(A) = span−210, Col(AT) = span123,100are orthogonal subspaces.Theore m 12. (Fundamental Theorem of Lin ear Algebra, Part I)Let A be a n m ×n matrix of rank r.• dim Col(A) = r (subspa ce of Rm)• dim Col(AT) = r (subspace of Rn)• dim Nul(A) = n −r (subspace of Rn)• dim Nul(AT) = m −r (subspace of Rm)Armin [email protected] m 13. (Fundamental Theorem of Lin ear Algebra, Part II)•Nul(A) is orthogonal to C ol(AT). (both subspaces of Rn)Note thatdimNul(A) + dimCol(AT) = n.Hence, the two spaces are orthogonal complements.• Nul(AT) is orthog onal to Col(A).Again, the two spaces are orthogonal complements.Why?Corollary 14. Ax = b is solvableyTb = 0 whenever yTA = 0Proof. Mo tivationExample 15. Not all linear systems h ave soluti ons.In fact, for many applications, data needs to be fi tted and there isno hope for a perfect match.For instance,Ax = b with1 22 4x =−12has no solution:•−12is not i n Col(A) = spann12o• Instead of giving up, we want the x which makes Ax and b asclose a s possibl e.• Such x is c haracterized byAxbArmin
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