DOCSLIB.ORG

Orthonormal Bases

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Orthonormal Bases 18.103 Fall 2013 Orthonormal Bases Consider an inner product space V with inner product hf; gi and norm kfk2 = hf; fi Proposition 1 (Continuity) If kun − uk ! 0 and kvn − vk ! 0 as n ! 1, then kunk ! kuk; hun; vni ! hu; vi: Proof. Note first that since kvn − vk ! 0, kvnk ≤ kvn − vk + kvk ≤ M < 1 for a constant M independent of n. Therefore, as n ! 1, jhun; vni − hu; vij = jhun − u; vni + hu; vn − vij ≤ Mkun − uk + kukkvn − vk ! 0 2 2 In particular, if un = vn, then kunk = hun; uni ! hu; ui = kuk . For u and v in V we say that u is perpendicular to v and write u ? v if hu; vi = 0. The Pythogorean theorem says that if u ? v, then ku + vk2 = kuk2 + kvk2 (1) Definition 1 'n is called an orthonormal sequence, n = 1; 2;::: , if h'n;'mi = 0 for n 6= m 2 and h'n;'ni = k'nk = 1. Suppose that 'n is an orthonormal sequence in an inner product space V . The following four consequences of the Pythagorean theorem (1) were proved in class (and are also in the text): N X If h = an'n, then n=1 N 2 X 2 khk = janj : (2) 1 1 N X If f 2 V and sN = hf; 'ni'n, then n=1 2 2 2 kfk = kf − sN k + ksN k (3) If VN = span f'1;'2;:::;'N g, then kf − sN k = min kf − gk (best approximation property) (4) g2VN If cn = hf; 'ni, then 1 2 X 2 kfk ≥ jcnj (Bessel's inequality): (5) n=1 Definition 2 A Hilbert space is defined as a complete inner product space (under the dis- tance d(u; v) = ku − vk). Theorem 1 Suppose that 'n is an orthonormal sequence in a Hilbert space H. Let 1 [ VN = span f'1;'2;:::;'N g;V = VN N=1 (V is the vector space of finite linear combinations of 'n.) The following are equivalent. a) V is dense in H (with respect to the distance d(f; g) = kf − gk), b) If f 2 H and hf; 'ni = 0 for all n, then f = 0. N X c) If f 2 H and sN = hf; 'ni'n, then ksN − fk ! 0 as N ! 1. n=1 d) If f 2 H, then 1 2 X 2 kfk = jhf; 'nij n=1 1 If the properties of the theorem hold, then f'ngn=1 is called an orthonormal basis or complete orthonormal system for H. (Note that the word \complete" used here does not mean the same thing as completeness of a metric space.) Proof. (a) =) (b). Let f satisfy hf; 'ni = 0, then by taking finite linear combinations, hf; vi = 0 for all v 2 V . Choose a sequence vj 2 V so that kvj − fk ! 0 as j ! 1. Then by Proposition 1 above 2 0 = hf; vji ! hf; fi =) kfk = 0 =) f = 0 2 N X (b) =) (c). Let f 2 H and denote cn = hf; 'ni, sN = cn'n. By Bessel's inequality 1 (5), 1 X 2 2 jcnj ≤ kfk < 1: 1 Hence, for M < N (using (2)) 2 N N 2 X X 2 ksN − sM k = cn'n = jcnj ! 0 as M; N ! 1: M+1 M+1 In other words, sN is a Cauchy sequence in H. By completeness of H, there is u 2 H such that ksN − uk ! 0 as N ! 1. Moreover, hf − sN ;'ni = 0 for all N ≥ n: Taking the limit as N ! 1 with n fixed yields hf − u; 'ni = 0 for all n: Therefore by (b), f − u = 0. (c) =) (d). Using (3) and (2), N 2 2 2 2 X 2 kfk = kf − sN k + ksN k = kf − sN k + jcnj ; (cn = hf; 'ni) 1 2 Take the limit as N ! 1. By (c), kf − sN k ! 0. Therefore, 1 2 X 2 kfk = jcnj 1 Finally, for (d) =) (a), N 2 2 X 2 kfk = kf − sN k + jcnj 1 2 2 Take the limit as N ! 1, then by (d) the rightmost term tends to kfk so that kf −sN k ! 0. Since sN 2 VN ⊂ V , V is dense in H. 3 Proposition 2 Let 'n be an orthonormal sequence in a Hilbert space H, and X 2 X 2 janj < 1; jbnj < 1 then 1 1 X X u = an'n; v = bn'n n=1 n=1 are convergent series in H norm and 1 X hu; vi = anbn (6) n=1 Proof. Let N N X X uN = an'n; vN = bn'n: 1 1 Then for M < N, N 2 X 2 kuN − uM k = janj ! 0 as M ! 1 M so that uN is a Cauchy sequence converging to some u 2 H. Similarly, vN ! v in H norm. Finally, N N N X X X huN ; vN i = haj'j; bk'ki = ajbkh'j;'ki = ajbj j;k=1 j;k=1 j=1 since h'j; fki = 0 for j 6= k and hfj; fji = 1. Taking the limit as N ! 1 and using the continuity property (1), huN ; vN i ! hu; vi, gives (6). 1 If H is a Hilbert space and f'ngn=1 is an orthonormal basis, then every element can be written 1 X f = an'n (series converges in norm) n=1 The mapping X fang 7! an'n n is a linear isometry from `2(N) to H that preserves the inner product. The inverse mapping is f 7! fang = fhf; 'nig 4 It is also useful to know that as soon as a linear mapping between Hilbert spaces is an isometry (preserves norms of vectors) it must also preserve the inner product. Indeed, the inner product function (of two variables u and v) can be written as a function of the norm function (of linear combinations of u and v). This is known as polarization: Polarization Formula. 2 2 2 2 hu; vi = a1ku + ivk + a2ku + vk + a3kuk + a4kvk (7) with a1 = i=2; a2 = 1=2; a3 = −(1 + i)=2; a4 = −(i + 1)=2 Proof. ku + ivk2 = hu + iv; u + ivi = kuk2 + hiv; ui + hu; ivi + kvk2 = kuk2 + i(hv; ui − hu; vi) + kvk2 Similarly, ku + vk2 = kuk2 + (hv; ui + hu; vi) + kvk2 Multiplying the first equation by i and adding to the second, we find that iku + ivk2 + ku + vk2 = (i + 1)kuk2 + 2hu; vi + (i + 1)kvk2 Solving for hu; vi yields (7). 5.
Load more

Recommended publications
  • 21. Orthonormal Bases
    21. Orthonormal Bases The canonical/standard basis 011 001 001 B C B C B C B0C B1C B0C e1 = B.C ; e2 = B.C ; : : : ; en = B.C B.C B.C B.C @.A @.A @.A 0 0 1 has many useful properties. • Each of the standard basis vectors has unit length: q p T jjeijj = ei ei = ei ei = 1: • The standard basis vectors are orthogonal (in other words, at right angles or perpendicular). T ei ej = ei ej = 0 when i 6= j This is summarized by ( 1 i = j eT e = δ = ; i j ij 0 i 6= j where δij is the Kronecker delta. Notice that the Kronecker delta gives the entries of the identity matrix. Given column vectors v and w, we have seen that the dot product v w is the same as the matrix multiplication vT w. This is the inner product on n T R . We can also form the outer product vw , which gives a square matrix. 1 The outer product on the standard basis vectors is interesting. Set T Π1 = e1e1 011 B C B0C = B.C 1 0 ::: 0 B.C @.A 0 01 0 ::: 01 B C B0 0 ::: 0C = B. .C B. .C @. .A 0 0 ::: 0 . T Πn = enen 001 B C B0C = B.C 0 0 ::: 1 B.C @.A 1 00 0 ::: 01 B C B0 0 ::: 0C = B. .C B. .C @. .A 0 0 ::: 1 In short, Πi is the diagonal square matrix with a 1 in the ith diagonal position and zeros everywhere else.
    [Show full text]
  • Orthonormal Bases in Hilbert Space APPM 5440 Fall 2017 Applied Analysis
    Orthonormal Bases in Hilbert Space APPM 5440 Fall 2017 Applied Analysis Stephen Becker November 18, 2017(v4); v1 Nov 17 2014 and v2 Jan 21 2016 Supplementary notes to our textbook (Hunter and Nachtergaele). These notes generally follow Kreyszig’s book. The reason for these notes is that this is a simpler treatment that is easier to follow; the simplicity is because we generally do not consider uncountable nets, but rather only sequences (which are countable). I have tried to identify the theorems from both books; theorems/lemmas with numbers like 3.3-7 are from Kreyszig. Note: there are two versions of these notes, with and without proofs. The version without proofs will be distributed to the class initially, and the version with proofs will be distributed later (after any potential homework assignments, since some of the proofs maybe assigned as homework). This version does not have proofs. 1 Basic Definitions NOTE: Kreyszig defines inner products to be linear in the first term, and conjugate linear in the second term, so hαx, yi = αhx, yi = hx, αy¯ i. In contrast, Hunter and Nachtergaele define the inner product to be linear in the second term, and conjugate linear in the first term, so hαx,¯ yi = αhx, yi = hx, αyi. We will follow the convention of Hunter and Nachtergaele, so I have re-written the theorems from Kreyszig accordingly whenever the order is important. Of course when working with the real field, order is completely unimportant. Let X be an inner product space. Then we can define a norm on X by kxk = phx, xi, x ∈ X.
    [Show full text]
  • Lecture 4: April 8, 2021 1 Orthogonality and Orthonormality
    Mathematical Toolkit Spring 2021 Lecture 4: April 8, 2021 Lecturer: Avrim Blum (notes based on notes from Madhur Tulsiani) 1 Orthogonality and orthonormality Definition 1.1 Two vectors u, v in an inner product space are said to be orthogonal if hu, vi = 0. A set of vectors S ⊆ V is said to consist of mutually orthogonal vectors if hu, vi = 0 for all u 6= v, u, v 2 S. A set of S ⊆ V is said to be orthonormal if hu, vi = 0 for all u 6= v, u, v 2 S and kuk = 1 for all u 2 S. Proposition 1.2 A set S ⊆ V n f0V g consisting of mutually orthogonal vectors is linearly inde- pendent. Proposition 1.3 (Gram-Schmidt orthogonalization) Given a finite set fv1,..., vng of linearly independent vectors, there exists a set of orthonormal vectors fw1,..., wng such that Span (fw1,..., wng) = Span (fv1,..., vng) . Proof: By induction. The case with one vector is trivial. Given the statement for k vectors and orthonormal fw1,..., wkg such that Span (fw1,..., wkg) = Span (fv1,..., vkg) , define k u + u = v − hw , v i · w and w = k 1 . k+1 k+1 ∑ i k+1 i k+1 k k i=1 uk+1 We can now check that the set fw1,..., wk+1g satisfies the required conditions. Unit length is clear, so let’s check orthogonality: k uk+1, wj = vk+1, wj − ∑ hwi, vk+1i · wi, wj = vk+1, wj − wj, vk+1 = 0. i=1 Corollary 1.4 Every finite dimensional inner product space has an orthonormal basis.
    [Show full text]
  • Ch 5: ORTHOGONALITY
    Ch 5: ORTHOGONALITY 5.5 Orthonormal Sets 1. a set fv1; v2; : : : ; vng is an orthogonal set if hvi; vji = 0, 81 ≤ i 6= j; ≤ n (note that orthogonality only makes sense in an inner product space since you need to inner product to check hvi; vji = 0. n For example fe1; e2; : : : ; eng is an orthogonal set in R . 2. fv1; v2; : : : ; vng is an orthogonal set =) v1; v2; : : : ; vn are linearly independent. 3. a set fv1; v2; : : : ; vng is an orthonormal set if hvi; vji = 0 and jjvijj = 1, 81 ≤ i 6= j; ≤ n (i.e. orthogonal and unit length). n For example fe1; e2; : : : ; eng is an orthonormal set in R . 4. given an orthogonal basis for a vector space V , we can always find an orthonormal basis for V by dividing each vector by its length (see Example 2 and 3 page 256) n 5. a space with an orthonormal basis behaves like the euclidean space R with the standard basis (it is easier to work with basis that are orthonormal, just like it is n easier to work with the standard basis versus other bases for R ) 6. if v is a vector in a inner product space V with fu1; u2; : : : ; ung an orthonormal basis, Pn Pn then we can write v = i=1 ciui = i=1hv; uiiui Pn 7. an easy way to find the inner product of two vectors: if u = i=1 aiui and v = Pn i=1 biui, where fu1; u2; : : : ; ung is an orthonormal basis for an inner product space Pn V , then hu; vi = i=1 aibi Pn 8.
    [Show full text]
  • Math 2331 – Linear Algebra 6.2 Orthogonal Sets
    6.2 Orthogonal Sets Math 2331 { Linear Algebra 6.2 Orthogonal Sets Jiwen He Department of Mathematics, University of Houston [email protected] math.uh.edu/∼jiwenhe/math2331 Jiwen He, University of Houston Math 2331, Linear Algebra 1 / 12 6.2 Orthogonal Sets Orthogonal Sets Basis Projection Orthonormal Matrix 6.2 Orthogonal Sets Orthogonal Sets: Examples Orthogonal Sets: Theorem Orthogonal Basis: Examples Orthogonal Basis: Theorem Orthogonal Projections Orthonormal Sets Orthonormal Matrix: Examples Orthonormal Matrix: Theorems Jiwen He, University of Houston Math 2331, Linear Algebra 2 / 12 6.2 Orthogonal Sets Orthogonal Sets Basis Projection Orthonormal Matrix Orthogonal Sets Orthogonal Sets n A set of vectors fu1; u2;:::; upg in R is called an orthogonal set if ui · uj = 0 whenever i 6= j. Example 82 3 2 3 2 39 < 1 1 0 = Is 4 −1 5 ; 4 1 5 ; 4 0 5 an orthogonal set? : 0 0 1 ; Solution: Label the vectors u1; u2; and u3 respectively. Then u1 · u2 = u1 · u3 = u2 · u3 = Therefore, fu1; u2; u3g is an orthogonal set. Jiwen He, University of Houston Math 2331, Linear Algebra 3 / 12 6.2 Orthogonal Sets Orthogonal Sets Basis Projection Orthonormal Matrix Orthogonal Sets: Theorem Theorem (4) Suppose S = fu1; u2;:::; upg is an orthogonal set of nonzero n vectors in R and W =spanfu1; u2;:::; upg. Then S is a linearly independent set and is therefore a basis for W . Partial Proof: Suppose c1u1 + c2u2 + ··· + cpup = 0 (c1u1 + c2u2 + ··· + cpup) · = 0· (c1u1) · u1 + (c2u2) · u1 + ··· + (cpup) · u1 = 0 c1 (u1 · u1) + c2 (u2 · u1) + ··· + cp (up · u1) = 0 c1 (u1 · u1) = 0 Since u1 6= 0, u1 · u1 > 0 which means c1 = : In a similar manner, c2,:::,cp can be shown to by all 0.
    [Show full text]
  • True False Questions from 4,5,6
    (c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 217: True False Practice Professor Karen Smith 1. For every 2 × 2 matrix A, there exists a 2 × 2 matrix B such that det(A + B) 6= det A + det B. Solution note: False! Take A to be the zero matrix for a counterexample. 2. If the change of basis matrix SA!B = ~e4 ~e3 ~e2 ~e1 , then the elements of A are the same as the element of B, but in a different order. Solution note: True. The matrix tells us that the first element of A is the fourth element of B, the second element of basis A is the third element of B, the third element of basis A is the second element of B, and the the fourth element of basis A is the first element of B. 6 6 3. Every linear transformation R ! R is an isomorphism. Solution note: False. The zero map is a a counter example. 4. If matrix B is obtained from A by swapping two rows and det A < det B, then A is invertible. Solution note: True: we only need to know det A 6= 0. But if det A = 0, then det B = − det A = 0, so det A < det B does not hold. 5. If two n × n matrices have the same determinant, they are similar. Solution note: False: The only matrix similar to the zero matrix is itself. But many 1 0 matrices besides the zero matrix have determinant zero, such as : 0 0 6.
    [Show full text]
  • C Self-Adjoint Operators and Complete Orthonormal Bases
    C Self-adjoint operators and complete orthonormal bases The concept of the adjoint of an operator plays a very important role in many aspects of linear algebra and functional analysis. Here we will primar­ ily focus on s e ll~adjoint operators and how they can be used to obtain and characterize complete orthonormal sets of vectors in a Hilbert space. The main fact we will present here is that self-adjoint operators typically have orthogonal eigenvectors, and under some conditions, these eigenvectors can form a complete orthonormal basis for the Hilbert space. In particular, we will use this technique to find a variety of orthonormal bases for L2 [a, b] that satisfy different types of boundary conditions. We begin with the algebraic aspects of self-adjoint operators and their eigenvectors. Those algebraic properties are identical in the finite and infinite dimensional cases. The completeness arguments in infinite dimensions are more delicate, we will only state those results without complete proofs. We begin first by defining the adjoint. D efinition C.l. (Finite dimensional or bounded operator case) Let A H I -; H2 be a bounded operator between the Hilbert spaces HI and H 2. The adjomt A' : H2 ---; HI of Xis defined by the req1Lirement J (V2, AVI ) ~ = (A''V2, v])], (C.l) fOT all VI E HI--- and'V2 E H2· Note that for emphasis, we have written (. , .)] and (. , ')2 for the inner products in HI and H2 respectively. It remains to be shown that the relation (C.l) defines an operator A' from A uniquely. We will not do this here in general but will illustrate it with several examples.
    [Show full text]
  • Worksheet and Solutions
    Tuesday, December 6 ∗∗ Orthogonal bases & Orthogonal projections ∗∗ Solutions 1. A basis B is called an orthonormal basis if it is orthogonal and each basis vector has norm equal to 1. (a) Convert the orthogonal basis 80− 1 0 1 0 19 < 1 1 1 = B = @ 1 A , @ 1 A , @1A : 0 −2 1 ; into an orthonormal basis C . Solution. We just scale each basis vector by its length. The new, orthonormal, basis is 8 0−11 0 1 1 0119 < 1 1 1 = C = p @ 1 A , p @ 1 A , p @1A 2 6 3 : 0 −2 1 ; 011 0−41 (b) Find the coordinates of the vectors v1 = @3A and v2 = @ 2 A in the basis C . 5 7 Solution. The formula is the same as for a general orthogonal basis: writing u1, u2, and u3 for the basis vectors, the formula for each coordinate of v1 in the basis C is u · v i 1 = · 2 ui v1. kuik We compute to find 2 p −6 p 9 p u1 · v1 = p = 2, u2 · v1 = p = − 6, u3 · v1 = p = 3 3, 2 6 3 0 p 1 p2 B C so (v1)C = @−p 6A. Similarly, we get 3 3 r 6 p −16 2 5 u1 · v2 = p = 3 2, u2 · v2 = p = −8 , u3 · v2 = p , 2 6 3 3 0 p 1 3p 2 B C so (v2)C = @−8 p2/3A. 5/ 3 Orthonormal bases are very convenient for calculations. 2. (The Gram-Schmidt process) There is a standard process for converting a basis into an 3 orthogonal basis.
    [Show full text]
  • Inner Product Spaces
    CHAPTER 6 Woman teaching geometry, from a fourteenth-century edition of Euclid’s geometry book. Inner Product Spaces In making the definition of a vector space, we generalized the linear structure (addition and scalar multiplication) of R2 and R3. We ignored other important features, such as the notions of length and angle. These ideas are embedded in the concept we now investigate, inner products. Our standing assumptions are as follows: 6.1 Notation F, V F denotes R or C. V denotes a vector space over F. LEARNING OBJECTIVES FOR THIS CHAPTER Cauchy–Schwarz Inequality Gram–Schmidt Procedure linear functionals on inner product spaces calculating minimum distance to a subspace Linear Algebra Done Right, third edition, by Sheldon Axler 164 CHAPTER 6 Inner Product Spaces 6.A Inner Products and Norms Inner Products To motivate the concept of inner prod- 2 3 x1 , x 2 uct, think of vectors in R and R as x arrows with initial point at the origin. x R2 R3 H L The length of a vector in or is called the norm of x, denoted x . 2 k k Thus for x .x1; x2/ R , we have The length of this vector x is p D2 2 2 x x1 x2 . p 2 2 x1 x2 . k k D C 3 C Similarly, if x .x1; x2; x3/ R , p 2D 2 2 2 then x x1 x2 x3 . k k D C C Even though we cannot draw pictures in higher dimensions, the gener- n n alization to R is obvious: we define the norm of x .x1; : : : ; xn/ R D 2 by p 2 2 x x1 xn : k k D C C The norm is not linear on Rn.
    [Show full text]
  • Orthonormal Bases
    Math 108B Professor: Padraic Bartlett Lecture 3: Orthonormal Bases Week 3 UCSB 2014 In our last class, we introduced the concept of \changing bases," and talked about writ- ing vectors and linear transformations in other bases. In the homework and in class, we saw that in several situations this idea of \changing basis" could make a linear transformation much easier to work with; in several cases, we saw that linear transformations under a certain basis would become diagonal, which made tasks like raising them to large powers far easier than these problems would be in the standard basis. But how do we find these \nice" bases? What does it mean for a basis to be\nice?" In this set of lectures, we will study one potential answer to this question: the concept of an orthonormal basis. 1 Orthogonality To start, we should define the notion of orthogonality. First, recall/remember the defini- tion of the dot product: n Definition. Take two vectors (x1; : : : xn); (y1; : : : yn) 2 R . Their dot product is simply the sum x1y1 + x2y2 + : : : xnyn: Many of you have seen an alternate, geometric definition of the dot product: n Definition. Take two vectors (x1; : : : xn); (y1; : : : yn) 2 R . Their dot product is the product jj~xjj · jj~yjj cos(θ); where θ is the angle between ~x and ~y, and jj~xjj denotes the length of the vector ~x, i.e. the distance from (x1; : : : xn) to (0;::: 0). These two definitions are equivalent: 3 Theorem. Let ~x = (x1; x2; x3); ~y = (y1; y2; y3) be a pair of vectors in R .
    [Show full text]
  • 3. Hilbert Spaces
    3. Hilbert spaces In this section we examine a special type of Banach spaces. We start with some algebraic preliminaries. Definition. Let K be either R or C, and let Let X and Y be vector spaces over K. A map φ : X × Y → K is said to be K-sesquilinear, if • for every x ∈ X, then map φx : Y 3 y 7−→ φ(x, y) ∈ K is linear; • for every y ∈ Y, then map φy : X 3 y 7−→ φ(x, y) ∈ K is conjugate linear, i.e. the map X 3 x 7−→ φy(x) ∈ K is linear. In the case K = R the above properties are equivalent to the fact that φ is bilinear. Remark 3.1. Let X be a vector space over C, and let φ : X × X → C be a C-sesquilinear map. Then φ is completely determined by the map Qφ : X 3 x 7−→ φ(x, x) ∈ C. This can be see by computing, for k ∈ {0, 1, 2, 3} the quantity k k k k k k k Qφ(x + i y) = φ(x + i y, x + i y) = φ(x, x) + φ(x, i y) + φ(i y, x) + φ(i y, i y) = = φ(x, x) + ikφ(x, y) + i−kφ(y, x) + φ(y, y), which then gives 3 1 X (1) φ(x, y) = i−kQ (x + iky), ∀ x, y ∈ X. 4 φ k=0 The map Qφ : X → C is called the quadratic form determined by φ. The identity (1) is referred to as the Polarization Identity.
    [Show full text]
  • 17. Inner Product Spaces Definition 17.1. Let V Be a Real Vector Space
    17. Inner product spaces Definition 17.1. Let V be a real vector space. An inner product on V is a function h ; i: V × V −! R; which is • symmetric, that is hu; vi = hv; ui: • bilinear, that is linear (in both factors): hλu, vi = λhu; vi; for all scalars λ and hu1 + u2; vi = hu1; vi + hu2; vi; for all vectors u1, u2 and v. • positive that is hv; vi ≥ 0: • non-degenerate that is if hu; vi = 0 for every v 2 V then u = 0. We say that V is a real inner product space. The associated quadratic form is the function Q: V −! R; defined by Q(v) = hv; vi: Example 17.2. Let A 2 Mn;n(R) be a real matrix. We can define a function n n h ; i: R × R −! R; by the rule hu; vi = utAv: The basic rules of matrix multiplication imply that this function is bi- linear. Note that the entries of A are given by t aij = eiAej = hei; eji: In particular, it is symmetric if and only if A is symmetric that is At = A. It is non-degenerate if and only if A is invertible, that is A has rank n. Positivity is a little harder to characterise. Perhaps the canonical example is to take A = In. In this case if t P u = (r1; r2; : : : ; rn) and v = (s1; s2; : : : ; sn) then u Inv = risi. Note 1 P 2 that if we take u = v then we get ri . The square root of this is the Euclidean distance.
    [Show full text]