1. Chapter 7 Inner Product Spaces 大葉大學 資訊工程系 黃鈴玲 Linear Algebra

2. Ch7_2 Inner Product Spaces In this chapter, we extend those concepts of R n such as: dot product of two vectors, norm of a vector, angle between vectors, and distance between points, to general vector space. This will enable us to talk about the magnitudes of functions and orthogonal functions. This concepts are used to approximate functions by polynomials – a technique that is used to implement functions on calculators and computers. We will no longer be restricted to Euclidean Geometry, we will be able to create our own geometries on R n.

3. Ch7_3 7.1 Inner Product Spaces The dot product was a key concept on R n that led to definitions of norm, angle, and distance. Our approach will be to generalize the dot product of R n to a general vector space with a mathematical structure called an inner product. This in turn will be used to define norm, angle, and distance for a general vector space.

4. Ch7_4 Definition An inner product on a real spaces V is a function that associates a number, denoted 〈 u, v 〉, with each pair of vectors u and v of V. This function has to satisfy the following conditions for vectors u, v, and w, and scalar c. 1. 〈 u, v 〉 = 〈 v, u 〉 (symmetry axiom) 2. 〈 u + v, w 〉 = 〈 u, w 〉 + 〈 v, w 〉 (additive axiom) 3. 〈 cu, v 〉 = c 〈 u, v 〉 (homogeneity axiom) 4. 〈 u, u 〉  0, and 〈 u, u 〉 = 0 if and only if u = 0 (position definite axiom) A vector space V on which an inner product is defined is called an inner product space. Any function on a vector space that satisfies the axioms of an inner product defines an inner product on the space. There can be many inner products on a given vector space.

5. Ch7_5 Example 1 Let u = (x 1, x 2 ), v = (y 1, y 2 ), and w = (z 1, z 2 ) be arbitrary vectors in R 2. Prove that 〈 u, v 〉, defined as follows, is an inner product on R 2. 〈 u, v 〉 = x 1 y 1 + 4x 2 y 2 Determine the inner product of the vectors (  2, 5), (3, 1) under this inner product. Solution Axiom 1: 〈 u, v 〉 = x 1 y 1 + 4x 2 y 2 = y 1 x 1 + 4y 2 x 2 = 〈 v, u 〉 Axiom 2: 〈 u + v, w 〉 = 〈 (x 1, x 2 ) + (y 1, y 2 ), (z 1, z 2 ) 〉 = 〈 (x 1 + y 1, x 2 + y 2 ), (z 1, z 2 ) 〉 = (x 1 + y 1 ) z 1 + 4(x 2 + y 2 )z 2 = x 1 z 1 + 4x 2 z 2 + y 1 z 1 + 4 y 2 z 2 = 〈 (x 1, x 2 ), (z 1, z 2 ) 〉 + 〈 (y 1, y 2 ), (z 1, z 2 ) 〉 = 〈 u, w 〉 + 〈 v, w 〉

6. Ch7_6 Axiom 3: 〈 cu, v 〉 = 〈 c(x 1, x 2 ), (y 1, y 2 ) 〉 = 〈 (cx 1, cx 2 ), (y 1, y 2 ) 〉 = cx 1 y 1 + 4cx 2 y 2 = c(x 1 y 1 + 4x 2 y 2 ) = c 〈 u, v 〉 Axiom 4: 〈 u, u 〉 = 〈 (x 1, x 2 ), (x 1, x 2 ) 〉 = Further, if and only if x 1 = 0 and x 2 = 0. That is u = 0. Thus 〈 u, u 〉  0, and 〈 u, u 〉 = 0 if and only if u = 0. The four inner product axioms are satisfied, 〈 u, v 〉 = x 1 y 1 + 4x 2 y 2 is an inner product on R 2. The inner product of the vectors (  2, 5), (3, 1) is 〈 (  2, 5), (3, 1) 〉 = (  2  3) + 4(5  1) = 14

7. Ch7_7 Example 2 Consider the vector space M 22 of 2  2 matrices. Let u and v defined as follows be arbitrary 2  2 matrices. Prove that the following function is an inner product on M 22. 〈 u, v 〉 = ae + bf + cg + dh Determine the inner product of the matrices. Solution Axiom 1: 〈 u, v 〉 = ae + bf + cg + dh = ea + fb + gc + hd = 〈 v, u 〉 Axiom 3: Let k be a scalar. Then 〈 ku, v 〉 = kae + kbf + kcg + kdh = k(ae + bf + cg + dh) = k 〈 u, v 〉

8. Ch7_8 Example 3 Consider the vector space P n of polynomials of degree  n. Let f and g be elements of P n. Prove that the following function defines an inner product of P n. Determine the inner product of polynomials f(x) = x 2 + 2x – 1 and g(x) = 4x + 1 Solution Axiom 1: Axiom 2:

9. Ch7_9 We now find the inner product of the functions f(x) = x 2 + 2x – 1 and g(x) = 4x + 1

10. Ch7_10 Norm of a Vector Definition Let V be an inner product space. The norm of a vector v is denoted ||v|| and it defined by The norm of a vector in R n can be expressed in terms of the dot product as follows Generalize this definition: The norms in general vector space do not necessary have geometric interpretations, but are often important in numerical work.

11. Ch7_11 Example 4 Consider the vector space P n of polynomials with inner product The norm of the function f generated by this inner product is Determine the norm of the function f(x) = 5x 2 + 1. Solution Using the above definition of norm, we get The norm of the function f(x) = 5x 2 + 1 is

12. Ch7_12 Example 2’ ( 補充 ) Consider the vector space M 22 of 2  2 matrices. Let u and v defined as follows be arbitrary 2  2 matrices. It is known that the function 〈 u, v 〉 = ae + bf + cg + dh is an inner product on M 22 by Example 2. The norm of the matrix is

13. Ch7_13 Definition Let V be an inner product space. The angle  between two nonzero vectors u and v in V is given by The dot product in R n was used to define angle between vectors. The angle  between vectors u and v in R n is defined by Angle between two vectors

14. Ch7_14 Example 5 Consider the inner product space P n of polynomials with inner product The angle between two nonzero functions f and g is given by Determine the cosine of the angle between the functions f(x) = 5x 2 and g(x) = 3x Solution We first compute ||f || and ||g||. Thus

15. Ch7_15 Example 2” ( 補充 ) Consider the vector space M 22 of 2  2 matrices. Let u and v defined as follows be arbitrary 2  2 matrices. It is known that the function 〈 u, v 〉 = ae + bf + cg + dh is an inner product on M 22 by Example 2. The norm of the matrix is The angle between u and v is

16. Ch7_16 Orthogonal Vectors Def. Let V be an inner product space. Two nonzero vectors u and v in V are said to be orthogonal if Example 6 Show that the functions f(x) = 3x – 2 and g(x) = x are orthogonal in P n with inner product Solution Thus the functions f and g are orthogonal in this inner product Space.

17. Ch7_17 Distance Definition Let V be an inner product space with vector norm defined by The distance between two vectors (points) u and v is defined d(u,v) and is defined by As for norm, the concept of distance will not have direct geometrical interpretation. It is however, useful in numerical mathematics to be able to discuss how far apart various functions are.

18. Ch7_18 Example 7 Consider the inner product space P n of polynomials discussed earlier. Determine which of the functions g(x) = x 2 – 3x + 5 or h(x) = x 2 + 4 is closed to f(x) = x 2. Solution Thus The distance between f and h is 4, as we might suspect, g is closer than h to f.

19. Ch7_19 Inner Product on C n For a complex vector space, the first axiom of inner product is modified to read. An inner product can then be used to define norm, orthogonality, and distance, as far a real vector space. Let u = (x 1, …, x n ) and v = (y 1, …, y n ) be element of C n. The most useful inner product for C n is 

20. Ch7_20 Example 8 Consider the vectors u = (2 + 3i,  1 + 5i), v = (1 + i,  i) in C 2. Compute (a) 〈 u, v 〉, and show that u and v are orthogonal. (b) ||u|| and ||v|| (c) d(u, v) Solution

21. Ch7_21 Homework Exercise 7.1: 1, 4, 8(a), 9(a), 10, 12, 13, 15, 17(a), 19, 20(a)

22. Ch7_22 7.2 Non-Euclidean Geometry and Special Relativity Different inner products on R n lead to different measures of vector norm, angle, and distance – that is, to different geometries. dot product  Euclidean geometry other inner products  non-Euclidean geometries Example Let u = (x 1, x 2 ), v = (y 1, y 2 ) be arbitrary vectors in R 2. It is proved that 〈 u, v 〉, defined as follows, is an inner product on R 2. 〈 u, v 〉 = x 1 y 1 + 4x 2 y 2 The inner product differs from the dot product in the appearance of a 4. Consider the vector (0, 1) in this space. The norm of this vector is

23. Ch7_23 Figure 7.1 The norm of this vector in Euclidean geometry is 1; in our new geometry, however, the norm is 2.

24. Ch7_24 Consider the vectors (1, 1) and (  4, 1). The inner product of these vectors is Figure 7.2 Thus these two vectors are orthogonal.

25. Ch7_25 Let us use the definition of distance based on this inner product to compute the distance between the points(1, 0) and (0, 1). We have that Figure 7.3

26. Ch7_26 7.4 Least-Squares Curves To find a polynomial that best fits given data points. Ax = y : (1) if n equations, n variables, and A  1 exists  x = A  1 y (2) if n equations, m variables with n > m  overdetermined How to solve it? We will introduce a matrix called the pseudoinverse of A, denoted pinv(A), that leads to a least-squares solution x = pinv(A)y for an overdetermined system.

27. Ch7_27 Definition Let A be a matrix. The matrix (A t A)  1 A t is called the pseudoinverse of A and is denoted pinv(A). Example 1 Find the pseudoinverse of A = Solution

28. Ch7_28 Ax = yx = pinv(A)y system least-squares solution If the system Ax=y has a unique solution, the least-squares solution is that unique solution. If the system is overdetermined, the least-squares solution is the closest we can get to a true solution. The system cannot have many solutions. Let Ax = y be a system of n linear equations in m variables with n > m, where A is of rank m. Ax=y  A t Ax=A t y  x = (A t A)  1 A t y A t A is invertible

29. Ch7_29 Example 2 Find the least-squares solution of the following overdetermined system of equations. Sketch the solution. Solution The matrix of coefficients is rank(A)=2 

30. Ch7_30 The least-squares solution is The least-squares solution is the point Figure 7.9

31. Ch7_31 Least-Square Curves Figure 7.10 Least-squares line or curve minimizes

32. Ch7_32 Example 3 Find the least-squares line for the following data points. (1, 1), (2, 4), (3, 2), (4, 4) Solution Let the equation of the line by y = a + bx. Substituting for these points into the equation of the line, we get the overdetermined system We find the least squares solution. The matrix of coefficients A and column vector y are as follows. It can be shown that

33. Ch7_33 The least squares solution is Thus a = 1, b = 0.7. The equation of the least-squares line for this data is y = 1 + 0.7x Figure 7.11

34. Ch7_34 Example 4 Find the least-squares parabola for the following data points. (1, 7), (2, 2), (3, 1), (4, 3) Solution Let the equation of the parabola be y = a + bx + cx 2. Substituting for these points into the equation of the parabola, we get the system We find the least squares solution. The matrix of coefficients A and column vector y are as follows. It can be shown that

35. Ch7_35 The least squares solution is Thus a = 15.25, b = -10.05, c = 1.75. The equation of the least-squares parabola for these data points is y = 15.25 – 10.05x + 1.75x 2 Figure 7.12

36. Ch7_36 Let (x 1, y 1 ), …, (x n, y n ) be a set of n data points. Let y = a 0 + … + a m x m be a polynomial of degree m (n > m) that is to be fitted to these points. Substituting these points into the polynomial leads to a system Ax = y of n linear equations in the m variables a 0, …, a m, where The least-squares solution to this system gives the coefficients of the least-squares polynomial for these data points. Theorem 7.1 Figure 7.13 y’ is the projection of y onto range(A)

37. Ch7_37 Homework Exercise 7.4 3, 11, 21