Review• v ·w = vTw = v1w1++ vnwn, the inner product of v, w in Rn◦ Length of v: kv k= v ·v√= v12++ vn2p◦ Distance between points v and w : kv −w k• v and w in Rnare orthogonal if v ·w = 0.◦ This simple criterion is equivalent to Pythagoras theorem.Example 1. The vectors100,010,001• are orthogonal to each other, and• have length 1.We are going to call such a basis orthonormal soon.Theorem 2. Suppose that v1,, vnare nonzero and (pairwise) orthogonal. Then v1,,vnare independent.Proof. Suppose thatc1v1++ cnvn= 0.Take t he dot product of v1with both sides:0 = v1·(c1v1++ cnvn)= c1v1·v1+ c2v1·v2++ cnv1·vn= c1v1·v1= c1kv1k2But kv1k0 and hence c1= 0.Likewise, we findc2= 0,, cn= 0. Hence, the vectors are independent. Example 3. Let us consider A =1 22 43 6.FindNul(A) and C ol(AT). Observe!Solution.Nul(A) = spann−21ocan you see it?if not, do it!Col(AT) = spann12oThe two basis vectors are orthogonal!−21·12= 0Armin [email protected] 4. Repeat for A =1 2 12 4 03 6 0.Solution.1 2 12 4 03 6 0 1 2 10 0 −20 0 −3>RREF1 2 00 0 10 0 0Nul(A) = span(−210)Col(AT) = span(120,001)The 2 vectors form a basis.Again, the vectors are orthogona l!−210·120= 0,−210·001= 0.Note: Because−210is orthogonal to both basis vect ors, it is orthogonal to every vectorin the row space.Vectors in Nul(A) are orthogonal to vectors in Col(AT).The fundamental theorem, second actDefinition 5 . Let W be a subspace of Rn, and v in Rn.• v is orthogonal to W , if v ·w = 0 for all w in W .(v is orthogonal to each vector in a basis of W )• Another subspace V is orthogonal to W , if every vector in V is orthogon al to W .• The orthogonal complement of W is the space W⊥of all vectors that are orthog-onal to W .Exercise: show that the orthogo nal complement is indeed a vector space.Example 6. In the previous example, A =1 2 12 4 03 6 0.We found thatNul(A) = span−210, C ol(AT) = span120,001Armin [email protected] orthogonal subspaces.Indeed,Nul(A) and C ol(AT) are orthogonal complements.Why? Because−210,120,001are orthogonal, hence independent, and hence a ba sis of all of R3.Remark 7. Reca ll that, for an m ×n matrix A, Nul(A) lives in Rnand Col(A) livesin Rm. Hence, they cannot be related in a similar way.In the previou s example, they happen to be both subspaces ofR3:Nul(A) = span−210, C ol(A) = span123,100But these spaces are not orthogonal:−210·1000Theorem 8. (Fundamental Theorem of Linear Algebra, Par t I)Let A be an m ×n matrix of rank r.• dim C o l(A) = r (subspace of Rm)• dim C o l(AT) = r (subspace of Rn)• dim N u l(A) = n −r (subspace of Rn)• dim N u l(AT) = m −r (subspace of Rm)Theorem 9. (Fundamental Theorem of Linear Algebra, Par t II)• Nul(A) is orthogonal to C ol(AT). (both subspaces of Rn)Note tha tdim Nul(A) + dim Col(AT) = n.Hence, the two spaces are orthogonal complements inRn.• Nul(AT) is orthogonal to Col(A).Again, the two spaces are orthogonal complements.Armin
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