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The Rational and Jordan Forms Linear Algebra Notes
The Rational and Jordan Forms Linear Algebra Notes Satya Mandal November 5, 2005 1 Cyclic Subspaces In a given context, a "cyclic thing" is an one generated "thing". For example, a cyclic groups is a one generated group. Likewise, a module M over a ring R is said to be a cyclic module if M is one generated or M = Rm for some m 2 M: We do not use the expression "cyclic vector spaces" because one generated vector spaces are zero or one dimensional vector spaces. 1.1 (De¯nition and Facts) Suppose V is a vector space over a ¯eld F; with ¯nite dim V = n: Fix a linear operator T 2 L(V; V ): 1. Write R = F[T ] = ff(T ) : f(X) 2 F[X]g L(V; V )g: Then R = F[T ]g is a commutative ring. (We did considered this ring in last chapter in the proof of Caley-Hamilton Theorem.) 2. Now V acquires R¡module structure with scalar multiplication as fol- lows: Define f(T )v = f(T )(v) 2 V 8 f(T ) 2 F[T ]; v 2 V: 3. For an element v 2 V de¯ne Z(v; T ) = F[T ]v = ff(T )v : f(T ) 2 Rg: 1 Note that Z(v; T ) is the cyclic R¡submodule generated by v: (I like the notation F[T ]v, the textbook uses the notation Z(v; T ).) We say, Z(v; T ) is the T ¡cyclic subspace generated by v: 4. If V = Z(v; T ) = F[T ]v; we say that that V is a T ¡cyclic space, and v is called the T ¡cyclic generator of V: (Here, I di®er a little from the textbook.) 5. -
HILBERT SPACE GEOMETRY Definition: a Vector Space Over Is a Set V (Whose Elements Are Called Vectors) Together with a Binary Operation
HILBERT SPACE GEOMETRY Definition: A vector space over is a set V (whose elements are called vectors) together with a binary operation +:V×V→V, which is called vector addition, and an external binary operation ⋅: ×V→V, which is called scalar multiplication, such that (i) (V,+) is a commutative group (whose neutral element is called zero vector) and (ii) for all λ,µ∈ , x,y∈V: λ(µx)=(λµ)x, 1 x=x, λ(x+y)=(λx)+(λy), (λ+µ)x=(λx)+(µx), where the image of (x,y)∈V×V under + is written as x+y and the image of (λ,x)∈ ×V under ⋅ is written as λx or as λ⋅x. Exercise: Show that the set 2 together with vector addition and scalar multiplication defined by x y x + y 1 + 1 = 1 1 + x 2 y2 x 2 y2 x λx and λ 1 = 1 , λ x 2 x 2 respectively, is a vector space. 1 Remark: Usually we do not distinguish strictly between a vector space (V,+,⋅) and the set of its vectors V. For example, in the next definition V will first denote the vector space and then the set of its vectors. Definition: If V is a vector space and M⊆V, then the set of all linear combinations of elements of M is called linear hull or linear span of M. It is denoted by span(M). By convention, span(∅)={0}. Proposition: If V is a vector space, then the linear hull of any subset M of V (together with the restriction of the vector addition to M×M and the restriction of the scalar multiplication to ×M) is also a vector space. -
18.06 Linear Algebra, Problem Set 5 Solutions
18.06 Problem Set 5 Solution Total: points Section 4.1. Problem 7. Every system with no solution is like the one in problem 6. There are numbers y1; : : : ; ym that multiply the m equations so they add up to 0 = 1. This is called Fredholm’s Alternative: T Exactly one of these problems has a solution: Ax = b OR A y = 0 with T y b = 1. T If b is not in the column space of A it is not orthogonal to the nullspace of A . Multiply the equations x1 − x2 = 1 and x2 − x3 = 1 and x1 − x3 = 1 by numbers y1; y2; y3 chosen so that the equations add up to 0 = 1. Solution (4 points) Let y1 = 1, y2 = 1 and y3 = −1. Then the left-hand side of the sum of the equations is (x1 − x2) + (x2 − x3) − (x1 − x3) = x1 − x2 + x2 − x3 + x3 − x1 = 0 and the right-hand side is 1 + 1 − 1 = 1: Problem 9. If AT Ax = 0 then Ax = 0. Reason: Ax is inthe nullspace of AT and also in the of A and those spaces are . Conclusion: AT A has the same nullspace as A. This key fact is repeated in the next section. Solution (4 points) Ax is in the nullspace of AT and also in the column space of A and those spaces are orthogonal. Problem 31. The command N=null(A) will produce a basis for the nullspace of A. Then the command B=null(N') will produce a basis for the of A. -
Does Geometric Algebra Provide a Loophole to Bell's Theorem?
Discussion Does Geometric Algebra provide a loophole to Bell’s Theorem? Richard David Gill 1 1 Leiden University, Faculty of Science, Mathematical Institute; [email protected] Version October 30, 2019 submitted to Entropy Abstract: Geometric Algebra, championed by David Hestenes as a universal language for physics, was used as a framework for the quantum mechanics of interacting qubits by Chris Doran, Anthony Lasenby and others. Independently of this, Joy Christian in 2007 claimed to have refuted Bell’s theorem with a local realistic model of the singlet correlations by taking account of the geometry of space as expressed through Geometric Algebra. A series of papers culminated in a book Christian (2014). The present paper first explores Geometric Algebra as a tool for quantum information and explains why it did not live up to its early promise. In summary, whereas the mapping between 3D geometry and the mathematics of one qubit is already familiar, Doran and Lasenby’s ingenious extension to a system of entangled qubits does not yield new insight but just reproduces standard QI computations in a clumsy way. The tensor product of two Clifford algebras is not a Clifford algebra. The dimension is too large, an ad hoc fix is needed, several are possible. I further analyse two of Christian’s earliest, shortest, least technical, and most accessible works (Christian 2007, 2011), exposing conceptual and algebraic errors. Since 2015, when the first version of this paper was posted to arXiv, Christian has published ambitious extensions of his theory in RSOS (Royal Society - Open Source), arXiv:1806.02392, and in IEEE Access, arXiv:1405.2355. -
New Bell–Sheffer Polynomial Sets
axioms Article New Bell–Sheffer Polynomial Sets Pierpaolo Natalini 1,* and Paolo Emilio Ricci 2 1 Dipartimento di Matematica e Fisica, Università degli Studi Roma Tre, Largo San Leonardo Murialdo, 1, 00146 Roma, Italy 2 Sezione di Matematica, International Telematic University UniNettuno, Corso Vittorio Emanuele II, 39, 00186 Roma, Italy; [email protected] * Correspondence: [email protected] Received: 20 July 2018; Accepted: 2 October 2018; Published: 8 October 2018 Abstract: In recent papers, new sets of Sheffer and Brenke polynomials based on higher order Bell numbers, and several integer sequences related to them, have been studied. The method used in previous articles, and even in the present one, traces back to preceding results by Dattoli and Ben Cheikh on the monomiality principle, showing the possibility to derive explicitly the main properties of Sheffer polynomial families starting from the basic elements of their generating functions. The introduction of iterated exponential and logarithmic functions allows to construct new sets of Bell–Sheffer polynomials which exhibit an iterative character of the obtained shift operators and differential equations. In this context, it is possible, for every integer r, to define polynomials of higher type, which are linked to the higher order Bell-exponential and logarithmic numbers introduced in preceding papers. Connections with integer sequences appearing in Combinatorial analysis are also mentioned. Naturally, the considered technique can also be used in similar frameworks, where the iteration of exponential and logarithmic functions appear. Keywords: Sheffer polynomials; generating functions; monomiality principle; shift operators; combinatorial analysis 1. Introduction In recent articles [1,2], new sets of Sheffer [3] and Brenke [4] polynomials, based on higher order Bell numbers [2,5–7], have been studied. -
Spanning Sets the Only Algebraic Operations That Are Defined in a Vector Space V Are Those of Addition and Scalar Multiplication
i i “main” 2007/2/16 page 258 i i 258 CHAPTER 4 Vector Spaces 23. Show that the set of all solutions to the nonhomoge- and let neous differential equation S1 + S2 = {v ∈ V : y + a1y + a2y = F(x), v = x + y for some x ∈ S1 and y ∈ S2} . where F(x) is nonzero on an interval I, is not a sub- 2 space of C (I). (a) Show that, in general, S1 ∪ S2 is not a subspace of V . 24. Let S1 and S2 be subspaces of a vector space V . Let (b) Show that S1 ∩ S2 is a subspace of V . S ∪ S ={v ∈ V : v ∈ S or v ∈ S }, 1 2 1 2 (c) Show that S1 + S2 is a subspace of V . S1 ∩ S2 ={v ∈ V : v ∈ S1 and v ∈ S2}, 4.4 Spanning Sets The only algebraic operations that are defined in a vector space V are those of addition and scalar multiplication. Consequently, the most general way in which we can combine the vectors v1, v2,...,vk in V is c1v1 + c2v2 +···+ckvk, (4.4.1) where c1,c2,...,ck are scalars. An expression of the form (4.4.1) is called a linear combination of v1, v2,...,vk. Since V is closed under addition and scalar multiplica- tion, it follows that the foregoing linear combination is itself a vector in V . One of the questions we wish to answer is whether every vector in a vector space can be obtained by taking linear combinations of a finite set of vectors. The following terminology is used in the case when the answer to this question is affirmative: DEFINITION 4.4.1 If every vector in a vector space V can be written as a linear combination of v1, v2, ..., vk, we say that V is spanned or generated by v1, v2, ..., vk and call the set of vectors {v1, v2,...,vk} a spanning set for V . -
Span, Linear Independence and Basis Rank and Nullity
Remarks for Exam 2 in Linear Algebra Span, linear independence and basis The span of a set of vectors is the set of all linear combinations of the vectors. A set of vectors is linearly independent if the only solution to c1v1 + ::: + ckvk = 0 is ci = 0 for all i. Given a set of vectors, you can determine if they are linearly independent by writing the vectors as the columns of the matrix A, and solving Ax = 0. If there are any non-zero solutions, then the vectors are linearly dependent. If the only solution is x = 0, then they are linearly independent. A basis for a subspace S of Rn is a set of vectors that spans S and is linearly independent. There are many bases, but every basis must have exactly k = dim(S) vectors. A spanning set in S must contain at least k vectors, and a linearly independent set in S can contain at most k vectors. A spanning set in S with exactly k vectors is a basis. A linearly independent set in S with exactly k vectors is a basis. Rank and nullity The span of the rows of matrix A is the row space of A. The span of the columns of A is the column space C(A). The row and column spaces always have the same dimension, called the rank of A. Let r = rank(A). Then r is the maximal number of linearly independent row vectors, and the maximal number of linearly independent column vectors. So if r < n then the columns are linearly dependent; if r < m then the rows are linearly dependent. -
Linear Algebra Midterm Exam, April 5, 2007 Write Clearly, with Complete
Mathematics 110 Name: GSI Name: Linear Algebra Midterm Exam, April 5, 2007 Write clearly, with complete sentences, explaining your work. You will be graded on clarity, style, and brevity. If you add false statements to a correct argument, you will lose points. Be sure to put your name and your GSI’s name on every page. 1. Let V and W be finite dimensional vector spaces over a field F . (a) What is the definition of the transpose of a linear transformation T : V → W ? What is the relationship between the rank of T and the rank of its transpose? (No proof is required here.) Answer: The transpose of T is the linear transformation T t: W ∗ → V ∗ sending a functional φ ∈ W ∗ to φ ◦ T ∈ V ∗. It has the same rank as T . (b) Let T be a linear operator on V . What is the definition of a T - invariant subspace of V ? What is the definition of the T -cyclic subspace generated by an element of V ? Answer: A T -invariant subspace of V is a linear subspace W such that T w ∈ W whenever w ∈ W . The T -cyclic subspace of V generated by v is the subspace of V spanned by the set of all T nv for n a natural number. (c) Let F := R, let V be the space of polynomials with coefficients in R, and let T : V → V be the operator sending f to xf 0, where f 0 is the derivative of f. Let W be the T -cyclic subspace of V generated by x2 + 1. -
Cyclic Vectors and Invariant Subspaces for the Backward Shift Operator Annales De L’Institut Fourier, Tome 20, No 1 (1970), P
ANNALES DE L’INSTITUT FOURIER R. G. DOUGLAS H. S. SHAPIRO A. L. SHIELDS Cyclic vectors and invariant subspaces for the backward shift operator Annales de l’institut Fourier, tome 20, no 1 (1970), p. 37-76 <http://www.numdam.org/item?id=AIF_1970__20_1_37_0> © Annales de l’institut Fourier, 1970, tous droits réservés. L’accès aux archives de la revue « Annales de l’institut Fourier » (http://annalif.ujf-grenoble.fr/) implique l’accord avec les conditions gé- nérales d’utilisation (http://www.numdam.org/conditions). Toute utilisa- tion commerciale ou impression systématique est constitutive d’une in- fraction pénale. Toute copie ou impression de ce fichier doit conte- nir la présente mention de copyright. Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques http://www.numdam.org/ Ann. Inst. Fourier, Grenoble 20,1 (1970), 37-76 CYCLIC VECTORS AND INVARIANT SUBSPACES FOR THE BACKWARD SHIFT OPERATOR (i) by R. G. DOUGLAS (2), H. S. SHAPIRO and A.L. SHIELDS 1. Introduction. Let T denote the unit circle and D the open unit disk in the complex plane. In [3] Beurling studied the closed invariant subspaces for the operator U which consists of multiplication by the coordinate function on the Hilbert space H2 = H^D). The operator U is called the forward (or right) shift, because the action of U is to transform a given function into one whose sequence of Taylor coefficients is shifted one unit to the right, that is, its action on sequences is U : (flo,^,^,...) ——>(0,flo,fli ,...). Strictly speaking, of course, the multiplication and the right shift operate on the distinct (isometric) Hilbert spaces H2 and /2. -
Worksheet 1, for the MATLAB Course by Hans G. Feichtinger, Edinburgh, Jan
Worksheet 1, for the MATLAB course by Hans G. Feichtinger, Edinburgh, Jan. 9th Start MATLAB via: Start > Programs > School applications > Sci.+Eng. > Eng.+Electronics > MATLAB > R2007 (preferably). Generally speaking I suggest that you start your session by opening up a diary with the command: diary mydiary1.m and concluding the session with the command diary off. If you want to save your workspace you my want to call save today1.m in order to save all the current variable (names + values). Moreover, using the HELP command from MATLAB you can get help on more or less every MATLAB command. Try simply help inv or help plot. 1. Define an arbitrary (random) 3 × 3 matrix A, and check whether it is invertible. Calculate the inverse matrix. Then define an arbitrary right hand side vector b and determine the (unique) solution to the equation A ∗ x = b. Find in two different ways the solution to this problem, by either using the inverse matrix, or alternatively by applying Gauss elimination (i.e. the RREF command) to the extended system matrix [A; b]. In addition look at the output of the command rref([A,eye(3)]). What does it tell you? 2. Produce an \arbitrary" 7 × 7 matrix of rank 5. There are at east two simple ways to do this. Either by factorization, i.e. by obtaining it as a product of some 7 × 5 matrix with another random 5 × 7 matrix, or by purposely making two of the rows or columns linear dependent from the remaining ones. Maybe the second method is interesting (if you have time also try the first one): 3. -
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proceedings OF the AMERICAN MATHEMATICAL SOCIETY Volume 78, Number 1, January 1980 THE INACCESSIBLE INVARIANT SUBSPACES OF CERTAIN C0 OPERATORS JOHN DAUGHTRY Abstract. We extend the Douglas-Pearcy characterization of the inaccessi- ble invariant subspaces of an operator on a finite-dimensional Hubert space to the cases of algebraic operators and certain C0 operators on any Hubert space. This characterization shows that the inaccessible invariant subspaces for such an operator form a lattice. In contrast to D. Herrero's recent result on hyperinvariant subspaces, we show that quasisimilar operators in the classes under consideration have isomorphic lattices of inaccessible in- variant subspaces. Let H be a complex Hubert space. For T in B(H) (the space of bounded linear operators on H) the set of invariant subspaces for T is given the metric dist(M, N) = ||PM — PN|| where PM (PN) is the orthogonal projection on M (N) and "|| ||" denotes the norm in B(H). An invariant subspace M for T is "cyclic" if there exists x in M such that { T"x) spans M. R. G. Douglas and Carl Pearcy [3] have characterized the isolated invariant subspaces for T in the case of finite-dimensional H (see [9, Chapters 6 and 7], for the linear algebra used in this article): An invariant subspace M for T is isolated if and only if M n M, = {0} or M, for every noncyclic summand M, in the primary decomposition for T. In [1] we showed how to view this result as a sharpening of the previously known conditions for the isolation of a solution to a quadratic matrix equation. -
Mathematics of Information Hilbert Spaces and Linear Operators
Chair for Mathematical Information Science Verner Vlačić Sternwartstrasse 7 CH-8092 Zürich Mathematics of Information Hilbert spaces and linear operators These notes are based on [1, Chap. 6, Chap. 8], [2, Chap. 1], [3, Chap. 4], [4, App. A] and [5]. 1 Vector spaces Let us consider a field F, which can be, for example, the field of real numbers R or the field of complex numbers C.A vector space over F is a set whose elements are called vectors and in which two operations, addition and multiplication by any of the elements of the field F (referred to as scalars), are defined with some algebraic properties. More precisely, we have Definition 1 (Vector space). A set X together with two operations (+; ·) is a vector space over F if the following properties are satisfied: (i) the first operation, called vector addition: X × X ! X denoted by + satisfies • (x + y) + z = x + (y + z) for all x; y; z 2 X (associativity of addition) • x + y = y + x for all x; y 2 X (commutativity of addition) (ii) there exists an element 0 2 X , called the zero vector, such that x + 0 = x for all x 2 X (iii) for every x 2 X , there exists an element in X , denoted by −x, such that x + (−x) = 0. (iv) the second operation, called scalar multiplication: F × X ! X denoted by · satisfies • 1 · x = x for all x 2 X • α · (β · x) = (αβ) · x for all α; β 2 F and x 2 X (associativity for scalar multiplication) • (α+β)·x = α·x+β ·x for all α; β 2 F and x 2 X (distributivity of scalar multiplication with respect to field addition) • α·(x+y) = α·x+α·y for all α 2 F and x; y 2 X (distributivity of scalar multiplication with respect to vector addition).