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159 Linear Algebra B • Lecture 7 Properties of Euclidean spaces1 Theodore Th. Voronov MSS/P5 • X63682 [email protected] www.ma.umist.ac.uk/tv/159.html Orthogonal complements Let W be a subspace of an inner product space V . A vector u in V is said to be orthogonal to W if it is orthogonal to every vector in W , and the set W ⊥ of all vectors in V that are orthogonal to W is called the orthogonal complement of W . Theorem Lectures: Monday 14:00 MSS/C19 Tutorials: Tuesday 11:00 MSS/C19 Properties of lengths and distances Theorem 6.2.2. Properties of Length. If u and v are vectors in a Euclidean space V , and if k is any scalar, then: (a) kuk ≥ 0 (b) kuk = 0 if and only if u = 0 (c) kkuk = |k| kuk (d) ku + vk 6 kuk + kvk (Triangle inequality) Theorem 6.2.3. Properties of Distance. If u, v and w are vectors in an inner product space V , and if k is any scalar, then: (a) d(u, v) ≥ 0 (b) d(u, v) = 0 if and only if u = v (c) d(u, v) = d(v, u) (d) d(u, v) 6 d(u, w) + d(w, v) (Triangle inequality) We shall prove part (d) of Theorem 6.2.2 and leave the remaining parts of Theorems 6.2.2 and 6.2.3 as exercises. Proof of Theorem 6.2.2d. By definition ku + vk2 = hu + v, u + vi = hu, ui + 2hu, vi + hv, vi 6 hu, ui + 2|hu, vi| + hv, vi 6 hu, ui + 2kuk kvk + hv, vi [Property of absolute value] [By Cauchy–Bunyakovsky] = kuk2 + 2kuk kvk + kvk2 = (kuk + kvk)2 . Taking square roots gives 6.2.5. Properties of Orthogonal Complements. If W is a subspace of a finite-dimensional Euclidean vector space V , then: (a) The set W ⊥ is also a subspace of V . (b) The only vector common to W and W ⊥ is 0. (c) The orthogonal complement of W ⊥ is W ; that is, (W ⊥ )⊥ = W . We shall prove parts (a) and (b). Proof (a). Note first that h0, wi = 0 for every vector w in W , so W ⊥ contains at least the zero vector. We want to show that W ⊥ is closed under addition and multiplication by scalars; that is, we want to show that the sum of two vectors in W ⊥ is orthogonal to every vector in W and that every scalar multiple of a vector in W ⊥ is orthogonal to every vector in W . Let u and v be any vectors in W ⊥ , let k be any scalar , and let w be any vector in W . Then from the definition of W ⊥ we have hu, wi = 0 and hv, wi = 0. Using basic properties of the scalar product we have hu + v, wi = hu, wi + hv, wi = 0 + 0 = 0 hku, wi = khu, wi = k(0) = 0 which proves that u + v and ku are in W ⊥ . Proof (b). If v is common to W and W ⊥ , then hv, vi = 0, which implies that v = 0 by the Positivity Axiom (Axiom 4) for scalar products. A proof for (c) will be omitted. Clearly W ⊂ (W ⊥ )⊥ , since hu, wi = 0 for all w ∈ W and u ∈ W ⊥ . Note that the proof that W = (W ⊥ )⊥ relies on the finitedimensionality of V , which was not used in parts (a) and (b). Remark. If dim V = n, dim W = k, then dim W ⊥ = n − k. Example. Consider in R3 the subspace spanned by the the last standard basis vector: W = span(e3 ). Then W ⊥ = {u ∈ R3 | hu, e3 i = 0} = {u ∈ R3 | u1 0 + u2 0 + u3 1 = 0} = {u ∈ R3 | u3 = 0} = {(u1 , u2 , 0) ∈ R3 } is the coordinate plane of the first two coordinates. ku + vk 6 kuk + kvk. Theorem 6.2.6. (Application to linear equations.) If A is an m × n matrix, then: 1 Definitions and theorems numbered as ‘6.2.2’ are taken from the book: H. Anton, Elementary Linear Algebra, John Wiley, 2000. (a) The nullspace of A and the row space of A are orthogonal complements in Rn with respect to the standard Euclidean scalar product. 2 Th. Th. Voronov • 159 Linear Algebra B • Lecture 7 • Properties of Euclidean spaces (b) The nullspace of AT and the column space of A are orthogonal complements in Rm with respect to the standard Euclidean scalar product. Proof (a). We want to show that the orthogonal complement of the row space of A is the nullspace of A. To do this we must show that if a vector v is orthogonal to every vector in the row space, then Av = 0, and conversely, if Av = 0, then v is orthogonal to every vector in the row space. Assume first that v is orthogonal to every vector in the row space of A. Then in particular v is orthogonal to the row vectors r1 , r2 , . . . , rn of A; that is r1 ·v = r2 ·v = · · · = rn ·v = 0 (1) This is precisely Av = 0, since these dot products are just the components of the column vector Av. Hence if v is in the orthogonal complement of the row space, it belongs to the nullspace of A. Assume now that, conversely, v is in the nullspace. Then the dot products r1 ·v, r2 ·v, . . . , rn ·v vanish. It implies that v is orthogonal to all vectors in the row space, since the vectors r1 , r2 , . . . , rn span the row space by the definition. Proof (b). Since the column space of A is the row space of AT (except for a difference in notation), the proof follows by applying the result in part (a) to the matrix AT . Isomorphism of V and V ∗ For a finite-dimensional Euclidean space V there is a natural isomorphism between V and its dual space V ∗ . There is a simple but important statement: Proposition. In a Euclidean vector space, if a vector is orthogonal to all vectors, it is a zero vector. Proof. Indeed, let u ∈ V be orthogonal to all vectors in V . Then, in particular, hu, ui = 0, which implies u = 0. the first argument). Consider the kernel of T . By definition, Ker(T ) = {u | hu, vi = 0 for all v ∈ V }. Hence Ker(T ) = {0}, by the proposition above. Therefore, the linear transformation T : V → V ∗ is injective. Hence (here we use the condition that dim V < ∞) it is an isomorphism. Notice that both spaces V and V ∗ , both being of dimension n, are of course isomorphic, as are all vector spaces of the same dimension. Such an isomorphism can be constructed by a random choice of bases in V and V ∗ and identifying both spaces with Rn . It is does not require any Euclidean structure in V . There are many such isomorphisms. However, the isomorphism constructed above using a scalar product in V is ‘natural’ in the sense that it does not depend on any random choices. Orthonormal bases A set of vectors in an inner product space is called an orthogonal set if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is called orthonormal. We shall be particularly interested in orthogonal and orthonormal bases. First of all, orthogonality automatically implies linear independence: Theorem 6.3.3 If S = {v1 , v2 , . . . , vn } is an orthogonal set of nonzero vectors in a Euclidean vector space, then S is linearly independent. Proof. Assume that k1 v1 + k2 v2 + . . . + kn vn = 0 (2) To demonstrate that S = {v1 , v2 , . . . , vn } is linearly independent, we must prove that k1 = k2 = · · · = kn = 0. For each vi in S, it follows from (2) that Theorem. Let V be a finite-dimensional Euclidean space. Then the map T : V → V ∗ , T : u 7→ T (u) = ξu , where the linear function ξu is defined by ξu (v) = hu, vi for all v ∈ V , is an isomorphism of vector spaces. Proof. Notice first that ξu for any u ∈ V is indeed a linear function V → R. This follows from the properties of a scalar product (additivity and homogeneity w.r.t. the second argument). Next, the map T : u 7→ T (u) = ξu is linear for the same reason (additivity and homogeneity of the scalar product w.r.t. hk1 v1 + k2 v2 + . . . + kn vn , vi i = h0, vi i = 0 or equivalently, k1 hv1 , vi i + k2 hv2 , vi i + · · · + kn hvn , vi i = 0 From the orthogonality of S it follows that hvj , vi i = 0 when j 6= i, so that this equation reduces to ki hvi , vi i = 0 Since the vectors in S are assumed to be nonzero, hvi , vi i 6= 0 by the positivity axiom for inner products. Therefore, ki = 0. Since the subscript i is arbitrary, we have k1 = k2 = · · · = kn = 0; 3 Th. Th. Voronov • 159 Linear Algebra B • Lecture 7 • Properties of Euclidean spaces thus, S is linearly independent. Hence for an orthogonal (moreover, orthonormal) set of nonzero vectors S to be a basis in V , is equivalent to the condition that S spans V . It is very easy to find the coordinates of vectors in an orthonormal basis: Theorem 6.3.1 If S = {e1 , e2 , . . . , en } is an orthonormal basis for a Euclidean vector space V , and u is any vector in V , then u = hu, e1 ie1 + hu, e2 ie2 + · · · + hu, en ien The matrix G is known as the Gram matrix of the basis B. It is defined by the properties of a basis: the lengths of the basis vectors and angles between them. Now it is clear that a basis B is orthonormal if its Gram matrix is the identity matrix: ( 1 if i = j gij = δij = 0 if i 6= j (the symbol δij is known as the Kronecker symbol or Kronecker delta). It follows that in an orthonormal basis B we have the following simple formulae. For scalar product: X hu, vi = u i vi = u 1 v1 + . . . + u n vn . Proof. Since S = {e1 , e2 , . . . , en } i is a basis, a vector u can be expressed in the form For norm: u = k1 e1 + k2 e2 + · · · + kn en kuk2 = We shall complete the proof by showing that ki = hu, ei i for i = 1, 2, . . . , n. For each vector ei in S we have = k1 he1 , ei i + k2 he2 , ei i + · · · + kn hen , ei i Since S = {e1 , e2 , . . . , en } is an orthonormal set, we have hei , ei i = ||ei ||2 = 1 and hej , ei i = 0 if j 6= i. Therefore, the above expression for hu, ei i simplifies to u2i = u21 + . . . + u2n . i For distance: d(u, v) = hu, ei i = hk1 e1 + k2 e2 + . . . + kn en , ei i X p (u1 − v1 )2 + . . . + (un − vn )2 . Remark. The formulae above, in particular, mean that in an orthonormal basis B the scalar product of two vectors in V equals to the standard scalar product in Rn of their coordinate column (or row) vectors: hu, vi = [u]B · [v]B . hu, ei i = ki . We shall use this in the sequel. The main point of considering orthonormal bases is that all calculations are greatly simplified. To see this, let B be an arbitrary basis. How one can calculate the scalar product hu, vi? We have u = u 1 e1 + . . . + u n en and v = v1 e1 + . . . + vn en . Hence, by the properties of scalar products, we obtain hu, vi = hu1 e1 + . . . + un en , v1 e1 + . . . + vn en i = DX E X X ui ei , vj ej = ui vj hei , ej i , i j i,j or, introducing a matrix G = (gij ) where gij = hei , ej i , we finally can write hu, vi = X i,j ui vj gij . In an orthonormal basis, the natural isomorphism V → V ∗ of a Euclidean vector space with its dual, looks especially simple. If T : V → V ∗ , u 7→ T (u) = ξu , where ξu (v) = hu, vi as described above, then ξu = u1 e∗1 + . . . + un e∗n . Here e∗i ∈ V ∗ denote the covectors of the dual basis for B, if B = {e1 , . . . , en }. That means that if a basis is orthonormal, then a vector and the corresponding covector have the same components. In particular, the basis vectors e1 , . . . , en are mapped precisely to the covectors of the dual basis e∗1 , . . . , e∗n , respectively. Remark. In an arbitrary basis in a Euclidean space V , the natural isomorphism V → V ∗ will have the appearance X u 7→ ξu = ui gij e∗j ij where (gij ) is the Gram matrix.