Math 304–504Linear AlgebraLecture 24:Orthogonal subspaces.Scalar product in RnDefinition. The scalar product of vectorsx = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) isx · y = x1y1+ x2y2+ · · · + xnyn.Properties of scalar product:x · x ≥ 0, x · x = 0 only if x = 0 (posi tivity)x · y = y · x (symmetry)z(x + y) · z = x · z + y · z (distributive law)(rx) · y = r(x · y) (homogeneity)In particular, x · y is a bilinear function (i.e. , it isboth a linear function of x and a linear function of y).OrthogonalityDefinition 1. Vectors x, y ∈ Rnare said to beorthogonal (denoted x ⊥ y) ifx · y = 0.Definition 2. A vector x ∈ Rnis said to beorthogonal to a nonempty set Y ⊂ Rn(denotedx ⊥ Y ) if x · y = 0 for any y ∈ Y .Definition 3. Nonempty sets X , Y ⊂ Rnare saidto be orthogonal (denoted X ⊥ Y ) if x · y = 0for any x ∈ X and y ∈ Y .Examples in R3. • The line x = y = 0 isorthogonal to the line y = z = 0.Indeed, if v = (0, 0, z) and w = (x, 0, 0) then v · w = 0.• The line x = y = 0 is ortho gonal to the planez = 0.Indeed, if v = (0, 0, z) and w = (x, y, 0) then v · w = 0 .• The line x = y = 0 is not ortho gonal to theplane z = 1.The vector v = (0, 0, 1) belongs to both the line and theplane, and v · v = 1 6= 0.• The plane z = 0 is not orthogonal to the planey = 0.The vector v = (1, 0, 0) belongs to both planes andv · v = 1 6= 0.Proposition 1 If X , Y ∈ Rnare orthogonal setsthen either they are disjoint or X ∩ Y = {0}.Proof: v ∈ X ∩ Y =⇒ v ⊥ v =⇒ v · v = 0 =⇒ v = 0.Proposition 2 Let V be a subspace of Rnand Sbe a spanning set for V . Then for any x ∈ Rnx ⊥ S =⇒ x ⊥ V .Proof: Any v ∈ V is represented as v = a1v1+ · · · + akvk,where vi∈ S and ai∈ R. If x ⊥ S thenx · v = a1(x · v1) + · · · + ak(x · vk) = 0 =⇒ x ⊥ v.Example. The vector v = (1, 1 , 1) is orthogonal tothe plane spanned by vectors w1= (2, −3, 1) andw2= (0, 1, −1) (because v · w1= v · w2= 0).Orthogonal complementDefinition. Let S ⊂ Rn. The orthogonalcomplement of S, deno ted S⊥, is the set of allvectors x ∈ Rnthat are ortho gonal to S. That is,S⊥is the largest subset of Rnorthogonal to S.Theorem 1 S⊥is a subspace of Rn.Note that S ⊂ (S⊥)⊥, hence Span(S) ⊂ (S⊥)⊥.Theorem 2 (S⊥)⊥= Span(S). In particular, forany subspace V we have (V⊥)⊥= V .Example. Consider a line L = {(x, 0, 0) | x ∈ R}and a plane Π = {(0, y , z) | y , z ∈ R} in R3.Then L⊥= Π and Π⊥= L.Fundamental subspacesDefinition. Given an m×n matrix A, letN(A) = {x ∈ Rn| Ax = 0},R(A) = {b ∈ Rm| b = Ax for some x ∈ Rn}.R(A) is the range of a linear m apping L : Rn→ Rm,L(x) = Ax. N(A) is the kernel of L.Also, N(A) is the nullspace of the matrix A whileR(A) is the colum n space of A. The row space ofA is R(AT).The subspaces N(A) , R(AT) ⊂ RnandR(A), N(AT) ⊂ Rmare fundamental subspacesassociated to the matrix A.Theorem N(A) = R(AT)⊥, N(AT) = R(A)⊥.That is, the nullspace of a matrix is the orthogonalcomplement of its row space.Proof: The equality Ax = 0 means that the v ector x isorthogonal to rows of the matrix A. Therefore N(A) = S⊥,where S is the set of rows of A. It remains to note thatS⊥= Span(S)⊥= R(AT)⊥.Corollary Let V be a subspace of Rn. Thendim V + dim V⊥= n.Proof: Pick a basis v1, . . . , vkfor V . Let A be the k×nmatrix whose rows are vectors v1, . . . , vk. T hen V = R(AT)and V⊥= N(A). Consequently, dim V and dim V⊥are rankand nullity of A. Therefore dim V + dim V⊥equals thenumber of columns of A, which is n.Direct sumDefinition. Let U, V be subspaces of a vectorspace W . We say that W is a direct sum of U andV (denoted W = U ⊕ V ) if any w ∈ W isuniquely represented as w = u + v, where u ∈ Uand v ∈ V .Remark. Given subspaces U, V ⊂ W , we can define a setU + V = {u + v | u ∈ U, v ∈ V }, which is also a subspace.However U ⊕ V may not be well defined.Proposition The direct sum U ⊕ V is well defi nedif and only if U ∩ V = {0}.Proof: U ⊕ V is well defined if for any u1, u2∈ U andv1, v2∈ V we have u1+v1=u2+v2=⇒ u1=u2and v1=v2.Now note that u1+v1=u2+v2⇐⇒ u1−u2=v2−v1.Theorem dim U ⊕ V = dim U + dim V .Proof: Pick a basis u1, . . . , ukfor U and a basis v1, . . . , vmfor V . Then u1, . . . , uk, v1, . . . , vmis a spanning set forU ⊕ V . Linear independence of this set follows from the factthat U ∩ V = {0}.Theorem Let V be a subspace of Rn. ThenRn= V ⊕ V⊥.Proof: V ⊥ V⊥=⇒ V ∩ V⊥= {0} =⇒ V ⊕ V⊥iswell defined. Since dim V ⊕ V⊥= dim V + dim V⊥= n, itfollows that V ⊕ V⊥is the entire space Rn.Given a vector x ∈ Rnand a subspace V ⊂ Rn,there exists a unique repr es entation x = p + o suchthat p ∈ V whi le o ⊥ V . The vector p is calledthe orthogonal projection of x onto V .Problem. Find the or tho gonal projection of thevector x = (2, 1, 0) onto the plane Π spanned byvectors v1= (1, 0, −2) and v2= (0, 1, 1).We have x = p + o, where p ∈ Π and o ⊥ Π.Then p = αv1+ βv2for some α, β ∈ R. Also,o · v1= o · v2= 0. Note thato·vi= (x−αv1−βv2)·vi= x·vi−α(v1·vi)−β(v2·vi).α(v1· v1) + β(v2· v1) = x · v1α(v1· v2) + β(v2· v2) = x · v2⇐⇒5α − 2 β = 2−2α + 2β = 1⇐⇒α = 1β = 3/2Thus p = v1+32v2= (1,32,
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