# Inner product. Definition of inner product

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1 Math 20F Linear Algebra Lecture 25 1 Inner product Review: Definition of inner product. Slide 1 Norm and distance. Orthogonal vectors. Orthogonal complement. Orthogonal basis. Definition of inner product Slide 2 Definition 1 (Inner product) Let V be a vector space over IR. An inner product (, ) is a function V V IR with the following properties 1. u V, (u, u) 0, and (u, u) = 0 u = 0; 2. u, v V, holds (u, v) = (v, u); 3. u, v, w V, and a, b IR holds (au + bv, w) = a(u, w) + b(v, w). Notation: V together with (, ) is called an inner product space.

2 Math 20F Linear Algebra Lecture 25 2 Examples The Euclidean inner product in IR 2. Let V = IR 2, and {e 1, e 2 } be the standard basis. Given two arbitrary vectors x = x 1 e 1 + x 2 e 2 and y = y 1 e 1 + y 2 e 2, then (x, y) = x 1 y 1 + x 2 y 2. Slide 3 Notice that (e 1, e 1 ) = 1, (e 2, e 2 ) = 1, and (e 1, e 2 ) = 0. It is also called dot product, and denoted as x y. The Euclidean inner product in IR n. Let V = IR n, and {e i } n i=1 be the standard basis. Given two arbitrary vectors x = n i=1 x ie i and y = n i=1 y ie i, then n (x, y) = x i y i. i=1 Notice that (e i, e j ) = I ij Examples An inner product in the vector space of continuous functions in [0, 1], denoted as V = C([0, 1]), is defined as follows. Given two arbitrary vectors f(x) and g(x), introduce the inner product Slide 4 (f, g) = 1 0 f(x)g(x) dx. An inner product in the vector space of functions with one continuous first derivative in [0, 1], denoted as V = C 1 ([0, 1]), is defined as follows. Given two arbitrary vectors f(x) and g(x), then (f, g) = 1 0 [f(x)g(x) + f (x)g (x)] dx.

3 Math 20F Linear Algebra Lecture 25 3 Norm An inner product space induces a norm, that is, a notion of length of a vector. Slide 5 Definition 2 (Norm) Let V, (, ) be a inner product space. The norm function, or length, is a function V IR denoted as, and defined as u = (u, u). Example: The Euclidean norm in IR 2 is given by u = (x, x) = (x 1 ) 2 + (x 2 ) 2. Examples The Euclidean norm in IR n is given by u = (x, x) = (x 1 ) (x n ) 2. Slide 6 A norm in the space of continuous functions V = C([0, 1]) is given by 1 f = (f, f) = 0 [f(x)] 2 dx. For example, one can check that the length of f(x) = 3x is 1.

4 Math 20F Linear Algebra Lecture 25 4 Distance A norm in a vector space, in turns, induces a notion of distance between two vectors, defined as the length of their difference. Slide 7 Definition 3 (Distance) Let V, (, ) be a inner product space, and be its associated norm. The distance between u and v V is given by dist(u, v) = u v. Example: The Euclidean distance between to points x and y IR 3 is x y = (x 1 y 1 ) 2 + (x 2 y 2 ) 2 + (x 3 y 3 ) 2. Orthogonal vectors Theorem 1 Let V be a vector space and u, v V. Then, u + v = u v (u, v) = 0. Slide 8 Proof: then, u + v 2 = (u + v, u + v) = u 2 + v 2 + 2(u, v). u v 2 = (u v, u v) = u 2 + v 2 2(u, v). u + v 2 u v 2 = 4(u, v).

5 Math 20F Linear Algebra Lecture 26 5 Orthogonal vectors Definition 4 (Orthogonal vectors) Let V, (, ) be an inner product space. Two vectors u, v V are orthogonal, or perpendicular, if and only if Slide 9 (u, v) = 0. We call them orthogonal, because the diagonal of the parallelogram formed by u and v have the same length. Theorem 2 Let V be a vector space and u, v V be orthogonal vectors. Then u + v 2 = u 2 + v 2. Example Vectors u = [1, 2] T and v = [2, 1] T in IR 2 are orthogonal with the inner product (u, v) = u 1 v 1 + u 2 v 2, because, Slide 10 (u, v) = 2 2 = 0. The vectors cos(x), sin(x) C([0, 2π]) are orthogonal, with the inner product (f, g) = 2π 0 fg dx, because 2π 2π (cos(x), sin(x)) = sin(x) cos(x) dx = 1 sin(2x) dx, (cos(x), sin(x)) = 1 ( ) cos(2x) 2π 0 = 0. 4

6 Math 20F Linear Algebra Lecture 26 6 Review: Orthogonal vectors. Orthogonal vectors Slide 11 Orthogonal projection along a vector. Orthogonal bases. Orthogonal projection onto a subspace. Gram-Schmidt orthogonalization process. Review or orthogonal vectors Slide 12 Definition 5 (Orthogonal vectors) Let V, (, ) be an inner product vector space. Two vectors u, v V are orthogonal, or perpendicular, if and only if (u, v) = 0. Theorem 3 Let V, (, ) be an inner product vector space. u, v V are orthogonal u + v = u v, u + v 2 = u 2 + v 2.

7 Math 20F Linear Algebra Lecture 26 7 Orthogonal projection along a vector Fix V, (, ), and u V, with u 0. Can any vector x V be decomposed in orthogonal parts with respect to u, that is, x = ˆx + x with (ˆx, x ) = 0 and ˆx = cu? Is this decomposition unique? Slide 13 Theorem 4 (Orthogonal decomposition along a vector) V, (, ), an inner product vector space, and u V, with u 0. Then, any vector x V can be uniquely decomposed as where ˆx = x = ˆx + x, (x, u) u 2 u, x = x ˆx. Therefore, ˆx is proportional to u, and (ˆx, x ) = 0. Orthogonal projection along a vector Proof: Introduce ˆx = cu, and then write x = cu + x. The condition (ˆx, x ) = 0 implies that (u, x ) = 0, then (x, u) = c(u, u), c = (x, u) u 2, Slide 14 then ˆx = (x, u) u 2 u, x = x ˆx. This decomposition is unique, because, given a second decomposition x = ŷ + y with y = du, and (ŷ, y ) = 0, then, (u, y ) = 0 and cu + x = du + y c = d, from a multiplication by u, and then, x = y.

8 Math 20F Linear Algebra Lecture 26 8 Orthogonal bases Slide 15 Definition 6 Let V, (, ) be an n dimensional inner product vector space, and {u 1,, u n } be a basis of V. The basis is orthogonal (u i, u j ) = 0, for all i j. The basis is orthonormal it is orthogonal, and in addition, u i = 1, for all i, where i, j = 1,, n. Orthogonal bases Slide 16 Theorem 5 Let V, (, ) be an n dimensional inner product vector space, and {u 1,, u n } be an orthogonal basis. Then, any x V can be written as x = c 1 u c n u n, with the coefficients have the form c i = (x, u i), i = 1,, n. u i 2 Proof: The set {u 1,, u n} is a basis, so there exist coefficients c i such that x = c 1u c nu n. The basis is orthogonal, so multiplying the expression of x by u i, and recalling (u i, u j) = 0 for all i j, one gets, (x, u i) = c i(u i, u i). The u i are nonzero, so (u i, u i) = u i 2 0, so c i = (x, u i)/ u i 2.

9 Math 20F Linear Algebra Lecture 26 9 Slide 17 Notice: Orthogonal projections onto subspaces To write x in an orthogonal basis means to do an orthogonal decomposition of x along each basis vector. (All this holds for vector spaces of functions.) Theorem 6 Let V, (, ) be an n dimensional inner product vector space, and W V be a p dimensional subspace. Let {u 1,, u p } be an orthogonal basis of W. Then, any x V can be decomposed as x = ˆx + x with (ˆx, x ) = 0 and ˆx = c 1 u c p u p, where the coefficients c i are given by c i = (x, u i), i = 1,, p. u i 2 Slide 18 Gram-Schmidt Orthogonalization process Orthogonal bases are convenient to carry out computations. Jorgen Gram and Erhard Schmidt by the year 1900 made standard a process to compute an orthogonal basis from an arbitrary basis. (They actually needed it for vector spaces of functions. Laplace, by 1800, used this process on IR n.)

10 Math 20F Linear Algebra Lecture Gram-Schmidt Orthogonalization process Theorem 7 Let V, (, ) be an inner product vector space, and {u 1,, u n } be an arbitrary basis of V. Then, an orthogonal basis of V is given by the vectors {v 1,, v n }, where Slide 19 v 1 = u 1, v 2 = u 2 (u 2, v 1 ) v 1 2 v 1, v 3 = u 3 (u 3, v 1 ) v 1 2 v 1 (u 3, v 2 ) v 2 2 v 2,. v n = u n. n 1 i=1 (u n, v i ) v i 2 v i. Least-squares approximation Slide 20 Review: Gram-Schmidt orthogonalization process. Least-squares approximation. Definition. Normal equation. Examples.

11 Math 20F Linear Algebra Lecture Gram-Schmidt Orthogonalization process Theorem 8 Let V, (, ) be an inner product vector space, and {u 1,, u n } be an arbitrary basis of V. Then, an orthogonal basis of V is given by the vectors {v 1,, v n }, where Slide 21 v 1 = u 1, v 2 = u 2 (u 2, v 1 ) v 1 2 v 1, v 3 = u 3 (u 3, v 1 ) v 1 2 v 1 (u 3, v 2 ) v 2 2 v 2,. v n = u n. n 1 i=1 (u n, v i ) v i 2 v i. Least-squares approximation Slide 22 Let V, W be vector spaces and A : V W be linear. Given b W then the linear equation Ax = b either has a solution x or it has no solutions. Suppose now that there is an inner product in W, say (, ) W, with associated norm W. Then there exists a notion of approximate solution, given as follows.

12 Math 20F Linear Algebra Lecture Least-squares approximation Definition 7 (Approximate solution) Let V, W be vector spaces and let (, ) W, W be an inner product and its associate norm in W. Let A : V W be linear, and b W be an arbitrary vector. An approximate solution to the linear equation Slide 23 is a vector ˆx V such that Ax = b, Aˆx b W Ax b W,, x V. Remark: ˆx is also called least-squares approximation, because ˆx makes the number Aˆx b 2 W = [(Ax) 1 b 1] [(Ax) m b m] 2 as small as possible, where m = dim(w ). Least-squares approximation Slide 24 Theorem 9 Let V, W be vector spaces and let (, ) W, W be an inner product and its associate norm in W. Let A : V W be linear, and b W be an arbitrary vector. If ˆx V satisfies that (Aˆx b) Range(A), then ˆx is a least-squares solution of Ax = b.

13 Math 20F Linear Algebra Lecture Least-squares approximation Proof: The hypothesis (Aˆx b) Range(A) implies that for all x V holds Slide 25 Ax b = Aˆx b + Ax Aˆx, = (Aˆx b) + A(x ˆx). The two terms on the right hand side are orthogonal, by hypothesis, then Pythagoras theorem holds, so Ax b 2 W = Aˆx b 2 W + A(x ˆx) 2 W Aˆx b 2 W, so ˆx is a least-squares solution. Least-squares approximation Theorem 10 Let IR n, (, ) n, and IR m, (, ) m be the Euclidean inner product spaces and A : IR n IR m be linear, identified with an m n matrix. Fix b IR m. Then, Slide 26 ˆx IR n is solution of A T Aˆx = A T b (Aˆx b) Col(A). Proof: Let ˆx such that A T Aˆx = A T b. Then, A T Aˆx = A T b, A T (Aˆx b) = 0, (a i, (Aˆx b)) m = 0, for all a i column vector of A, where we used the notation A = [a 1,, a n]. Therefore, the condition A T Aˆx = A T b is equivalent to (Aˆx b) Col(A).

14 Math 20F Linear Algebra Lecture Least-squares approximation The previous two results can be summarized in the following one: Slide 27 Theorem 11 Let IR n, (, ) n, and IR m, (, ) m be the Euclidean inner product spaces and A : IR n IR m be linear, identified with an m n matrix. Fix b IR m. If ˆx IR n is solution of A T Aˆx = A T b, then ˆx is a least squares solution of Ax = b.

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