Lecture 0: 6.1 Inner Products Wei-Ta Chu 011/1/5
Definition An inner product on a real vector space V is a function that associates a real number u, v with each pair of vectors u and v in V in such a way that the following axioms are satisfied for all vectors u, v, and w in V and all scalars k. u, v = v, u u + v, w = u, w + v, w ku, v = k u, v u, u 0 and u, u = 0 if and only if u = 0 A real vector space with an inner product is called a real inner product space.
Euclidean Inner Product on R n If u = (u 1, u,, u n ) and v = (v 1, v,, v n ) are vectors in R n, then the formula v, u = u v = u 1 v 1 + u v + + u n v n defines v, u to be the Euclidean inner product on R n. Inner products can be used to define norm and distance in a general inner product space. 3
Theorem 6.1.1 If u and v are vectors in a real inner product space V, and if k is a scalar, then v 0 with equality if and only if v = 0 kv = k v d(u,v) = d(v,u) d(u,v) 0 with equality if and only if u = v 4
Euclidean Inner Product vs. Inner Product The Euclidean inner product is the most important inner product on R n. However, there are various applications in which it is desirable to modify the Euclidean inner product by weighting its terms differently. More precisely, if w 1,w,,w n are positive real numbers, and if u=(u 1,u,,u n ) and v=(v 1,v,,v n ) are vectors in R n, then it can be shown which is called weighted Euclidean inner product with weights w 1,w,,w n 5
Weighted Euclidean Inner Product Let u = (u 1, u ) and v = (v 1, v ) be vectors in R. Verify that the weighted Euclidean inner product u, v = 3u 1 v 1 + u v satisfies the four product axioms. Solution: Note first that if u and v are interchanged in this equation, the right side remains the same. Therefore, u, v = v, u. If w = (w 1, w ), then u + v, w = 3(u 1 +v 1 )w 1 + (u +v )w = (3u 1 w 1 + u w ) + (3v 1 w 1 + v w )= u, w + v, w which establishes the second axiom. 6
Weighted Euclidean Inner Product ku, v =3(ku 1 )v 1 + (ku )v = k(3u 1 v 1 + u v ) = k u, v which establishes the third axiom. v, v = 3v 1 v 1 +v v = 3v 1 + v.obviously, v, v = 3v 1 + v 0. Furthermore, v, v = 3v 1 + v = 0 if and only if v 1 = v = 0, That is, if and only if v = (v 1,v )=0. Thus, the fourth axiom is satisfied. 7
Applications of Weighted Euclidean Inner Products Suppose that some physical experiment can produce any of n possible numerical values x 1,x,,x n. We perform m repetitions of the experiment and yield these values with various frequencies; that is, x 1 occurs f 1 times, x occurs f times, and so forth. f 1 +f + +f n =m. Thus, the arithmetic average, or mean, is 8
Applications of Weighted Euclidean Inner Products If we let f=(f 1,f,,f n ), x=(x 1,x,,x n ), w 1 =w = =w n =1/m Then this equation can be expressed as the weighted inner product 9
Weighted Euclidean Inner Product The norm and distance depend on the inner product used. If the inner product is changed, then the norms and distances between vectors also change. For example, for the vectors u = (1,0) and v = (0,1) in R with the Euclidean inner product, we have u = 1 and d( u, v) = u v (1, 1) ( 1) However, if we change to the weighted Euclidean inner product u, v = 3u 1 v 1 + u v, then we obtain = = 1 = u = u, u 1/ = (3 1 1+ 0 0) 1/ = 3 d( u, v) = u v = (1, 1),(1, 1) 1/ = [3 1 1+ ( 1) ( 1)] 1/ = 5 10
Definition If V is an inner product space, then the norm (or length) of a vector u in V is denoted by u and is defined by u = u, u ½ The distance between two points (vectors) u and v is denoted by d(u,v) and is defined by d(u, v) = u v If a vector has norm 1, then we say that it is a unit vector. 11
Unit Circles and Spheres in Inner Product Space If V is an inner product space, then the set of points in V that satisfy u = 1 is called the unite sphere or sometimes the unit circle in V. In R and R 3 these are the points that lie 1 unit away form the origin. 1
Unit Circles in R Sketch the unit circle in an xy-coordinate system in R using the Euclidean inner product u, v = u 1 v 1 + u v Sketch the unit circle in an xy-coordinate system in R using the Euclidean inner product u, v = 1/9u 1 v 1 + 1/4u v Solution If u = (x,y), then u = u, u ½ = (x + y ) ½, so the equation of the unit circle is x + y = 1. If u = (x,y), then u = u, u ½ = (1/9x + 1/4y ) ½, so the equation of the unit circle is x /9 + y /4 = 1. 13
Inner Products Generated by Matrices u1 v1 u v Let u = and v= be vectors in R n (expressed as n 1 : : un vn matrices), and let A be an invertible n n matrix. If u v is the Euclidean inner product on R n, then the formula u, v = Au Av defines an inner product; it is called the inner product on R n generated by A. Recalling that the Euclidean inner product u v can be written as the matrix product v T u, the above formula can be written in the alternative form u, v = (Av) T Au, or equivalently, u, v = v T A T Au 14
Inner Product Generated by the Identity Matrix The inner product on R n generated by the n n identity matrix is the Euclidean inner product: Let A = I, we have u, v = Iu Iv = u v The weighted Euclidean inner product u, v = 3u 1 v 1 + u v is the inner product on R generated by 15 the inner product on R generated by since 3 0 0 A = [ ] [ ] 1 1 1 1 1 1 3 0 0 3 0 0 3 0 0 3, v u v u u u v v u u v v v u + = = =
Inner Product Generated by the Identity Matrix In general, the weighted Euclidean inner product u, v = w 1 u 1 v 1 + w u v + + w n u n v n is the inner product on R n generated by 16
An Inner Product on M If 1 1 U = u u and V v v u u = v v 3 4 3 4 are any two matrices, then U, V = tr(u T V) = tr(v T U) = u 1 v 1 + u v + u 3 v 3 + u 4 v 4 defines an inner product on M For example, if 1 1 0 U = and V 3 4 = 3 then U, V = 1(-1)+(0) + 3(3) + 4() = 16 The norm of a matrix U relative to this inner product is U =< U, U > = u + u + u + u 1/ 1 3 4 and the unit sphere in this space consists of all matrices U whose entries satisfy the equation U = 1, which on squaring yields u 1 + u + u 3 + u 4 = 1 17
An Inner Product on P If p = a 0 + a 1 x + a x and q = b 0 + b 1 x + b x are any two vectors in P, then the following formula defines an inner product on P : p, q = a 0 b 0 + a 1 b 1 + a b The norm of the polynomial p relative to this inner product is 1/ p =< p, p > = a + a + a 0 1 and the unit sphere in this space consists of all polynomials p in P whose coefficients satisfy the equation p = 1, which on squaring yields a 0 + a 1 + a =1 18
Theorem 6.1. (Properties of Inner Products) If u, v, and w are vectors in a real inner product space, and k is any scalar, then: 0, v = v, 0 = 0 u, v + w = u, v + u, w u, kv =k u, v u v, w = u, w v, w u, v w = u, v u, w 19
Example u v, 3u + 4v = u, 3u + 4v v, 3u+4v = u, 3u + u, 4v v, 3u v, 4v = 3 u, u + 4 u, v 6 v, u 8 v, v = 3 u + 4 u, v 6 u, v 8 v = 3 u u, v 8 v 0
Example We are guaranteed without any further proof that the five properties given in Theorem 6.1. are true for the inner product on R n generated by any matrix A. u, v + w = (v+w) T A T Au =(v T +w T )A T Au =(v T A T Au) + (w T A T Au) = u, v + u, w [Property of transpose] [Property of matrix multiplication] 1
Lecture 0: 6. Angle and Orthogonality Wei-Ta Chu 011/1/5
Theorem 6..1 Theorem 6..1 (Cauchy-Schwarz Inequality) If u and v are vectors in a real inner product space, then u, v u v The inequality can be written in the following two forms The Cauchy-Schwarz inequality for R n follows as a special case of this theorem by taking u, v to be the Euclidean inner product u v. 3
Angle Between Vectors Cauchy-Schwarz inequality can be used to define angles in general inner product cases. We define to be the angle between u and v. 4
Example Let R 4 have the Euclidean inner product. Find the cosine of the angle θ between the vectors u = (4, 3, 1, -) and v = (-, 1,, 3). 5
Theorem 6.. Theorem 6.. (Properties of Length) If u, v and w are vectors in an inner product space V, and if k is any scalar, then : u + v u + v (Triangle inequality for vectors) d(u,v) d(u,w) + d(w,v) (Triangle inequality for distances) Proof of (a) (Property of absolute value) (Theorem 6..1) 6
Orthogonality Definition Two vectors u and v in an inner product space are called orthogonal if u, v = 0. Example ( U, V = tr(u T V) = tr(v T U) = u 1 v 1 + u v + u 3 v 3 + u 4 v 4 ) If M has the inner project defined previously, then the matrices U 1 = 1 0 1 and V 0 = 0 are orthogonal, since U, V = 1(0) + 0() + 1(0) + 1(0) = 0. 0 7
Orthogonal Vectors in P Let P have the inner product < p, q >= p( x) q( x) dx and let p = x and q = x. 1 Then p q 1 1/ 1 1/ 1/ =< p, p > = xxdx = x dx = 1 1 1 1/ 1 1/ 1/ 4 =< q, q > = x x dx = x dx = 1 1 3 < p, q >= xx dx = x dx = 0 1 1 1 1 because p, q = 0, the vectors p = x and q = x are orthogonal relative to the given inner product. 1 3 5 8
Theorem 6..3 (Generalized Theorem of Pythagoras) If u and v are orthogonal vectors in an inner product space, then Proof u + v = u + v 9
Orthogonality Definition 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 of all vectors in V that are orthogonal to W is called the orthogonal complement ( 正交 補餘 ) of W. If V is a plane through the origin of R 3 with Euclidean inner product, then the set of all vectors that are orthogonal to every vector in V forms the line L through the origin that is perpendicular to V. L V 30
Theorem 6..4 Theorem 6..4 (Properties of Orthogonal Complements) If W is a subspace of a finite-dimensional inner product space V, then: W is a subspace of V. (read W perp ) The only vector common to W and W is 0; that is,w W = 0. 31
Proof of Theorem 6..4(a) Note first that 0, w =0 for every vector w in W, so W contains at least the zero vector. We want to show that the sum of two vectors in W is orthogonal to every vector in W (closed under addition) and that any scalar multiple of a vector in W is orthogonal to every vector in W (closed under scalar multiplication). 3
Proof of Theorem 6..4(a) Let u and v be any vector in W, let k be any scalar, and let w be any vector in W. Then from the definition of W, we have u, w =0 and v, w =0. Using the basic properties of the inner product, we have u+v, w = u, w + v, w = 0 + 0 = 0 ku, w = k u, w = k(0) = 0 Which proves that u+v and ku are in W. 33
Proof of Theorem 6..4(b) The only vector common to W and W is 0; that is,w W = 0. If v is common to W and W, then v, v =0, which implies that v=0 by Axiom 4 for inner products. 34
Theorem 6..5 Theorem 6..5 If W is a subspace of a finite-dimensional inner product space V, then the orthogonal complement of W is W; that is (W ) = W 35
Example (Basis for an Orthogonal Complement) Let W be the subspace of R 6 spanned by the vectors w 1 =(1, 3, -, 0,, 0), w =(, 6, -5, -, 4, -3), w 3 =(0, 0, 5, 10, 0, 15), w 4 =(, 6, 0, 8, 4, 18). Find a basis for the orthogonal complement of W. Solution The space W spanned by w 1, w, w 3, and w 4 is the same as the row space of the matrix Since the row space and null space of A are orthogonal complements, this problem reduces to finding a basis for the null space of A. form a basis for this null space. 36