Bilinear Forms Lecture Notes for MA1212
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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. -
Orthogonal Bases and the -So in Section 4.8 We Discussed the Problem of Finding the Orthogonal Projection P
Orthogonal Bases and the -So In Section 4.8 we discussed the problemR. of finding the orthogonal projection p the vector b into the V of , . , suhspace the vectors , v2 ho If v1 v,, form a for V, and the in x n matrix A has these basis vectors as its column vectors. ilt the orthogonal projection p is given by p = Ax where x is the (unique) solution of the normal system ATAx = A7b. The formula for p takes an especially simple and attractive Form when the ba vectors , . .. , v1 v are mutually orthogonal. DEFINITION Orthogonal Basis An orthogonal basis for the suhspacc V of R” is a basis consisting of vectors , ,v,, that are mutually orthogonal, so that v v = 0 if I j. If in additii these basis vectors are unit vectors, so that v1 . = I for i = 1. 2 n, thct the orthogonal basis is called an orthonormal basis. Example 1 The vectors = (1, 1,0), v2 = (1, —1,2), v3 = (—1,1,1) form an orthogonal basis for . We “normalize” ‘ R3 can this orthogonal basis 1w viding each basis vector by its length: If w1=—- (1=1,2,3), lvii 4.9 Orthogonal Bases and the Gram-Schmidt Algorithm 295 then the vectors /1 I 1 /1 1 ‘\ 1 2” / I w1 0) W2 = —— _z) W3 for . form an orthonormal basis R3 , . ..., v, of the m x ii matrix A Now suppose that the column vectors v v2 form an orthogonal basis for the suhspacc V of R’. Then V}.VI 0 0 v2.v .. -
Partitioned (Or Block) Matrices This Version: 29 Nov 2018
Partitioned (or Block) Matrices This version: 29 Nov 2018 Intermediate Econometrics / Forecasting Class Notes Instructor: Anthony Tay It is frequently convenient to partition matrices into smaller sub-matrices. e.g. 2 3 2 1 3 2 3 2 1 3 4 1 1 0 7 4 1 1 0 7 A B (2×2) (2×3) 3 1 1 0 0 = 3 1 1 0 0 = C I 1 3 0 1 0 1 3 0 1 0 (3×2) (3×3) 2 0 0 0 1 2 0 0 0 1 The same matrix can be partitioned in several different ways. For instance, we can write the previous matrix as 2 3 2 1 3 2 3 2 1 3 4 1 1 0 7 4 1 1 0 7 a b0 (1×1) (1×4) 3 1 1 0 0 = 3 1 1 0 0 = c D 1 3 0 1 0 1 3 0 1 0 (4×1) (4×4) 2 0 0 0 1 2 0 0 0 1 One reason partitioning is useful is that we can do matrix addition and multiplication with blocks, as though the blocks are elements, as long as the blocks are conformable for the operations. For instance: A B D E A + D B + E (2×2) (2×3) (2×2) (2×3) (2×2) (2×3) + = C I C F 2C I + F (3×2) (3×3) (3×2) (3×3) (3×2) (3×3) A B d E Ad + BF AE + BG (2×2) (2×3) (2×1) (2×3) (2×1) (2×3) = C I F G Cd + F CE + G (3×2) (3×3) (3×1) (3×3) (3×1) (3×3) | {z } | {z } | {z } (5×5) (5×4) (5×4) 1 Intermediate Econometrics / Forecasting 2 Examples (1) Let 1 2 1 1 2 1 c 1 4 2 3 4 2 3 h i A = = = a a a and c = c 1 2 3 2 3 0 1 3 0 1 c 0 1 3 0 1 3 3 c1 h i then Ac = a1 a2 a3 c2 = c1a1 + c2a2 + c3a3 c3 The product Ac produces a linear combination of the columns of A. -
Baire Category Theorem and Uniform Boundedness Principle)
Functional Analysis (WS 19/20), Problem Set 3 (Baire Category Theorem and Uniform Boundedness Principle) Baire Category Theorem B1. Let (X; k·kX ) be an innite dimensional Banach space. Prove that X has uncountable Hamel basis. Note: This is Problem A2 from Problem Set 1. B2. Consider subset of bounded sequences 1 A = fx 2 l : only nitely many xk are nonzerog : Can one dene a norm on A so that it becomes a Banach space? Consider the same question with the set of polynomials dened on interval [0; 1]. B3. Prove that the set L2(0; 1) has empty interior as the subset of Banach space L1(0; 1). B4. Let f : [0; 1) ! [0; 1) be a continuous function such that for every x 2 [0; 1), f(kx) ! 0 as k ! 1. Prove that f(x) ! 0 as x ! 1. B5.( Uniform Boundedness Principle) Let (X; k · kX ) be a Banach space and (Y; k · kY ) be a normed space. Let fTαgα2A be a family of bounded linear operators between X and Y . Suppose that for any x 2 X, sup kTαxkY < 1: α2A Prove that . supα2A kTαk < 1 Uniform Boundedness Principle U1. Let F be a normed space C[0; 1] with L2(0; 1) norm. Check that the formula 1 Z n 'n(f) = n f(t) dt 0 denes a bounded linear functional on . Verify that for every , F f 2 F supn2N j'n(f)j < 1 but . Why Uniform Boundedness Principle is not satised in this case? supn2N k'nk = 1 U2.( pointwise convergence of operators) Let (X; k · kX ) be a Banach space and (Y; k · kY ) be a normed space. -
3 Bilinear Forms & Euclidean/Hermitian Spaces
3 Bilinear Forms & Euclidean/Hermitian Spaces Bilinear forms are a natural generalisation of linear forms and appear in many areas of mathematics. Just as linear algebra can be considered as the study of `degree one' mathematics, bilinear forms arise when we are considering `degree two' (or quadratic) mathematics. For example, an inner product is an example of a bilinear form and it is through inner products that we define the notion of length in n p 2 2 analytic geometry - recall that the length of a vector x 2 R is defined to be x1 + ... + xn and that this formula holds as a consequence of Pythagoras' Theorem. In addition, the `Hessian' matrix that is introduced in multivariable calculus can be considered as defining a bilinear form on tangent spaces and allows us to give well-defined notions of length and angle in tangent spaces to geometric objects. Through considering the properties of this bilinear form we are able to deduce geometric information - for example, the local nature of critical points of a geometric surface. In this final chapter we will give an introduction to arbitrary bilinear forms on K-vector spaces and then specialise to the case K 2 fR, Cg. By restricting our attention to thse number fields we can deduce some particularly nice classification theorems. We will also give an introduction to Euclidean spaces: these are R-vector spaces that are equipped with an inner product and for which we can `do Euclidean geometry', that is, all of the geometric Theorems of Euclid will hold true in any arbitrary Euclidean space. -
Handout 9 More Matrix Properties; the Transpose
Handout 9 More matrix properties; the transpose Square matrix properties These properties only apply to a square matrix, i.e. n £ n. ² The leading diagonal is the diagonal line consisting of the entries a11, a22, a33, . ann. ² A diagonal matrix has zeros everywhere except the leading diagonal. ² The identity matrix I has zeros o® the leading diagonal, and 1 for each entry on the diagonal. It is a special case of a diagonal matrix, and A I = I A = A for any n £ n matrix A. ² An upper triangular matrix has all its non-zero entries on or above the leading diagonal. ² A lower triangular matrix has all its non-zero entries on or below the leading diagonal. ² A symmetric matrix has the same entries below and above the diagonal: aij = aji for any values of i and j between 1 and n. ² An antisymmetric or skew-symmetric matrix has the opposite entries below and above the diagonal: aij = ¡aji for any values of i and j between 1 and n. This automatically means the digaonal entries must all be zero. Transpose To transpose a matrix, we reect it across the line given by the leading diagonal a11, a22 etc. In general the result is a di®erent shape to the original matrix: a11 a21 a11 a12 a13 > > A = A = 0 a12 a22 1 [A ]ij = A : µ a21 a22 a23 ¶ ji a13 a23 @ A > ² If A is m £ n then A is n £ m. > ² The transpose of a symmetric matrix is itself: A = A (recalling that only square matrices can be symmetric). -
Representation of Bilinear Forms in Non-Archimedean Hilbert Space by Linear Operators
Comment.Math.Univ.Carolin. 47,4 (2006)695–705 695 Representation of bilinear forms in non-Archimedean Hilbert space by linear operators Toka Diagana This paper is dedicated to the memory of Tosio Kato. Abstract. The paper considers representing symmetric, non-degenerate, bilinear forms on some non-Archimedean Hilbert spaces by linear operators. Namely, upon making some assumptions it will be shown that if φ is a symmetric, non-degenerate bilinear form on a non-Archimedean Hilbert space, then φ is representable by a unique self-adjoint (possibly unbounded) operator A. Keywords: non-Archimedean Hilbert space, non-Archimedean bilinear form, unbounded operator, unbounded bilinear form, bounded bilinear form, self-adjoint operator Classification: 47S10, 46S10 1. Introduction Representing bounded or unbounded, symmetric, bilinear forms by linear op- erators is among the most attractive topics in representation theory due to its significance and its possible applications. Applications include those arising in quantum mechanics through the study of the form sum associated with the Hamil- tonians, mathematical physics, symplectic geometry, variational methods through the study of weak solutions to some partial differential equations, and many oth- ers, see, e.g., [3], [7], [10], [11]. This paper considers representing symmetric, non- degenerate, bilinear forms defined over the so-called non-Archimedean Hilbert spaces Eω by linear operators as it had been done for closed, positive, symmetric, bilinear forms in the classical setting, see, e.g., Kato [11, Chapter VI, Theo- rem 2.23, p. 331]. Namely, upon making some assumptions it will be shown that if φ : D(φ) × D(φ) ⊂ Eω × Eω 7→ K (K being the ground field) is a symmetric, non-degenerate, bilinear form, then there exists a unique self-adjoint (possibly unbounded) operator A such that (1.1) φ(u, v)= hAu, vi, ∀ u ∈ D(A), v ∈ D(φ) where D(A) and D(φ) denote the domains of A and φ, respectively. -
Notes on Symmetric Bilinear Forms
Symmetric bilinear forms Joel Kamnitzer March 14, 2011 1 Symmetric bilinear forms We will now assume that the characteristic of our field is not 2 (so 1 + 1 6= 0). 1.1 Quadratic forms Let H be a symmetric bilinear form on a vector space V . Then H gives us a function Q : V → F defined by Q(v) = H(v,v). Q is called a quadratic form. We can recover H from Q via the equation 1 H(v, w) = (Q(v + w) − Q(v) − Q(w)) 2 Quadratic forms are actually quite familiar objects. Proposition 1.1. Let V = Fn. Let Q be a quadratic form on Fn. Then Q(x1,...,xn) is a polynomial in n variables where each term has degree 2. Con- versely, every such polynomial is a quadratic form. Proof. Let Q be a quadratic form. Then x1 . Q(x1,...,xn) = [x1 · · · xn]A . x n for some symmetric matrix A. Expanding this out, we see that Q(x1,...,xn) = Aij xixj 1 Xi,j n ≤ ≤ and so it is a polynomial with each term of degree 2. Conversely, any polynomial of degree 2 can be written in this form. Example 1.2. Consider the polynomial x2 + 4xy + 3y2. This the quadratic 1 2 form coming from the bilinear form HA defined by the matrix A = . 2 3 1 We can use this knowledge to understand the graph of solutions to x2 + 2 1 0 4xy + 3y = 1. Note that HA has a diagonal matrix with respect to 0 −1 the basis (1, 0), (−2, 1). -
Week 8-9. Inner Product Spaces. (Revised Version) Section 3.1 Dot Product As an Inner Product
Math 2051 W2008 Margo Kondratieva Week 8-9. Inner product spaces. (revised version) Section 3.1 Dot product as an inner product. Consider a linear (vector) space V . (Let us restrict ourselves to only real spaces that is we will not deal with complex numbers and vectors.) De¯nition 1. An inner product on V is a function which assigns a real number, denoted by < ~u;~v> to every pair of vectors ~u;~v 2 V such that (1) < ~u;~v>=< ~v; ~u> for all ~u;~v 2 V ; (2) < ~u + ~v; ~w>=< ~u;~w> + < ~v; ~w> for all ~u;~v; ~w 2 V ; (3) < k~u;~v>= k < ~u;~v> for any k 2 R and ~u;~v 2 V . (4) < ~v;~v>¸ 0 for all ~v 2 V , and < ~v;~v>= 0 only for ~v = ~0. De¯nition 2. Inner product space is a vector space equipped with an inner product. Pn It is straightforward to check that the dot product introduces by ~u ¢ ~v = j=1 ujvj is an inner product. You are advised to verify all the properties listed in the de¯nition, as an exercise. The dot product is also called Euclidian inner product. De¯nition 3. Euclidian vector space is Rn equipped with Euclidian inner product < ~u;~v>= ~u¢~v. De¯nition 4. A square matrix A is called positive de¯nite if ~vT A~v> 0 for any vector ~v 6= ~0. · ¸ 2 0 Problem 1. Show that is positive de¯nite. 0 3 Solution: Take ~v = (x; y)T . Then ~vT A~v = 2x2 + 3y2 > 0 for (x; y) 6= (0; 0). -
A Symplectic Banach Space with No Lagrangian Subspaces
transactions of the american mathematical society Volume 273, Number 1, September 1982 A SYMPLECTIC BANACHSPACE WITH NO LAGRANGIANSUBSPACES BY N. J. KALTON1 AND R. C. SWANSON Abstract. In this paper we construct a symplectic Banach space (X, Ü) which does not split as a direct sum of closed isotropic subspaces. Thus, the question of whether every symplectic Banach space is isomorphic to one of the canonical form Y X Y* is settled in the negative. The proof also shows that £(A") admits a nontrivial continuous homomorphism into £(//) where H is a Hilbert space. 1. Introduction. Given a Banach space E, a linear symplectic form on F is a continuous bilinear map ß: E X E -> R which is alternating and nondegenerate in the (strong) sense that the induced map ß: E — E* given by Û(e)(f) = ü(e, f) is an isomorphism of E onto E*. A Banach space with such a form is called a symplectic Banach space. It can be shown, by essentially the argument of Lemma 2 below, that any symplectic Banach space can be renormed so that ß is an isometry. Any symplectic Banach space is reflexive. Standard examples of symplectic Banach spaces all arise in the following way. Let F be a reflexive Banach space and set E — Y © Y*. Define the linear symplectic form fiyby Qy[(^. y% (z>z*)] = z*(y) ~y*(z)- We define two symplectic spaces (£,, ß,) and (E2, ß2) to be equivalent if there is an isomorphism A : Ex -» E2 such that Q2(Ax, Ay) = ß,(x, y). A. Weinstein [10] has asked the question whether every symplectic Banach space is equivalent to one of the form (Y © Y*, üy). -
Properties of Bilinear Forms on Hilbert Spaces Related to Stability Properties of Certain Partial Differential Operators
JOURNAL OF hlATHEhlATICAL ANALYSIS AND APPLICATIONS 20, 124-144 (1967) Properties of Bilinear Forms on Hilbert Spaces Related to Stability Properties of Certain Partial Differential Operators N. SAUER National Research Institute for Mathematical Sciences, Pretoria, South Africa Submitted by P. Lax INTRODUCTION The continuous dependence of solutions of positive definite, self-adjoint partial differential equations on the domain was investigated by Babugka [I], [2]. In a recent paper, [3], Babugka and Vjibornjr also studied the continuous dependence of the eigenvalues of strongly-elliptic, self-adjoint partial dif- ferential operators on the domain. It is the aim of this paper to study formalisms in abstract Hilbert space by means of which results on various types of continuous dependences can be obtained. In the case of the positive definite operator it is found that the condition of self-adjointness can be dropped. The properties of operators of this type are studied in part II. In part III the behavior of the eigenvalues of a self- adjoint elliptic operator under various “small changes” is studied. In part IV some applications to partial differential operators and variational theory are sketched. I. BASIC: CONCEPTS AND RESULTS Let H be a complex Hilbert space. We use the symbols X, y, z,... for vectors in H and the symbols a, b, c,... for scalars. The inner product in H is denoted by (~,y) and the norm by I( .v 11 = (x, x)llz. Weak and strong convergence of a sequence (x~} to an element x0 E H are denoted by x’, - x0 and x’, + x,, , respectively. A functional f on H is called: (i) continuous if it is continuous in the strong topology on H, (ii) weakly continuous (w.-continuous) if it is continuous in the weak topology on H. -
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.