Documents about Characteristic Polynomial

  • 2 Pages

    lec15

    Berkeley, MATH 110

    Excerpt: ... (t) splits over C. Theorem. The characteristic polynomial of any diagonalizable linear operator splits. Proof. Regard the operator as a matrix. Then the characteristic polynomial of that matrix is the same as that of a diagonal matrix, and the characteristic polynomial of a diagonal matrix can easily be written as a product of linear factors. Denition. Let be an eigenvalue of a linear operator with characteristic polynomial f (t). Then the multiplicity of is the largest integer m for which (t )m divides f (t). In particular, the multiplicity of any eigenvalue is at least 1. 3 Example. Find the multiplicities of the eigenvalues of A = 0 0 1 3 0 0 4 . 4 1 Denition. Let T be a linear operator with eigenvalue . Then the eigenspace of T corresponding to , E , is dened to be N (T I). Theorem. Let be an eigenvalue of an operator T on a nite dimensional vector space V with multiplicity m. Then 1 dim E m. Proof. Choose a basis for E , and expand thi ...

  • 2 Pages

    lec14

    Berkeley, MATH 110

    Excerpt: ... MATH 110 Lecture Notes 14 GSI Carter July 15, 2008 1 Eigenvalues and Eigenvectors Let T : V V be a linear transformation. Then a vector v V is called an eigenvector of T if v = 0 and there exists a scalar such that T (v) = v. The scalar is called the eigenvalue associated to v. To nd such vectors v, we are looking for scalars such that T (v) v = 0 has nonzero solutions, which is true if and only if the linear transformation T IV has a nonzero null space. Since this is a linear transformation from V to V , this is equivalent to det(T I) being zero. Thus we dene the characteristic polynomial of T , f (t) = det(T tI). The eigenvalues of T are exactly the roots of this polynomial. (You may have seen these denitions for matrices, but they work just as well for linear operators on arbitrary nite dimensional vector spaces.) Example. Find eigenvectors and eigenvalues for A = 1 4 1 1 . Example. Find eigenvectors and eigenvalues for A = 1 1 1 3 . Example. ...

  • 1 Pages

    rev3

    Kent State, MATH 41021

    Excerpt: ... Math 41021 Exam III Review Exam III will be given in class on Wednesday, April 29, 2009. It will cover 9.19.5, 9.79.8, 10.310.4, Homework #9#11, and material from class March 18 through April 20. (pages 67100 of the lecture notes). The exam will consist of statements of definitions and theorems (from the list below), computational problems, and proofs similar to those in the homework. The following is an outline of topics and types of problems that may be on the exam. Definitions/Theorems to State: Characteristic Polynomial Minimal Polynomial Eigenvector Eigenvalue Eigenstuff: characteristic polynomial ; definition, properties, and computation Cayley-Hamilton Theorem and applications (but not the proof) eigenvalues, eigenvectors, eigenspaces; definitions, computations, theory; relation to characteristic and minimal polynomials basis and dimension for eigenspaces independence of eigenvectors for distinct eigenvalues algebraic multiplicity and geometric multiplicity and the relationship b ...

  • 3 Pages

    Section 3.3

    Virginia Tech, MATH 2214

    Excerpt: ... on is important enough throughout the next few sections that we give it a name: De nition: If ay + by + c y = is a second-order homogeneous equation, then we call the polynomial ak + bk + c = the characteristic polynomial of the di erential equation. (Our textbook uses instead of k for the variable in the characteristic polynomial .) e Characteristic Polynomial e characteristic polynomial for a second-order homogeneous equation is simply the equation of a parabola. e number of roots of a parabola must be zero, one, or two, depending on how o en the parabola crosses the x-axis: If the quadratic formula, k= -b b - ac , a is used to nd the roots, the discriminant (b - ac) identi es the number of roots: a positive discriminant means that both roots are real numbers; a negative discriminant means that both roots are complex numbers. When the discriminant is zero, both roots are real, but they're the same. Two (di erent) real roots When the characteristic polynomial has two real roots k ...

  • 2 Pages

    hw9

    Kent State, MATH 41021

    Excerpt: ... MATH 4/51021 Homework #9 Due: Friday, April 10 Reading: For Fri., April 10: 9.7, 9.8; Problems 9.279.37 Problems to turn in: 1. Find the characteristic polynomial A (t) for the matrix 1 2 1 A = 2 1 -2 . -3 2 -1 2. Find the characteristic polynomial B (t) for 2 -3 5 1 B= 0 0 0 0 the matrix 7 4 6 3 . 5 2 4 -3 3. Find the characteristic polynomial C (t) for the matrix C= 1 2 3 1 and evaluate C (C) to verify that the Cayley-Hamilton Theorem holds for C. 4. Show that if A is an n n matrix, then An can be written as a linear combination of the matrices I, A, A2 , . . . , An-1 ; that is, there are scalars 0 , 1 , 2 , . . . , n-1 such that An = 0 I + 1 A + 2 A2 + + n-1 An-1 . 5. Show that an n n matrix A is invertible if and only if 0 is not an eigenvalue of A. 6. Let A be an invertible matrix. Show that if v is an eigenvector of A belonging to the eigenvalue , then v is also an eigenvector of A-1 belonging to the eigenvalue 1/. (In particular, 1/ is an eigenvalue of A-1 .) 7. Find all eigen ...

  • 3 Pages

    SVstability

    UT Arlington, EE 4343

    Excerpt: ... Copyright F.L. Lewis 1998 All rights reserved EE 4343/5329 - Control System Design Project LECTURE 9 Updated: Monday, June 28, 1999 STABILITY OF STATE VARIABLE SYSTEMS The linear time invariant (LTI) state-space form is & x = Ax + Bu, x (0) . y = Cx + Du The transfer function is given by H ( s ) = C ( sI A) 1 B + D . The denominator of this is the characteristic polynomial (s ) = sI A . The system poles are the roots of the characteristic equation (s ) = sI A = 0 . A state-space formulation allows one to get more information about the system than the input/output formulation, which is described only by a transfer function. Specifically, if A, B, C, D are known, then the internal states x(t) can be computed in addition to the input u(t) and output y(t). Thus, one is effectively able to look inside the black box in Fig. 1. The (internal) system poles (e.g. roots of (s) should be distinguished from the (input/output) poles of the transfer function. There may be some pole/zero cancel ...

  • 1 Pages

    hmwk78

    N. Illinois, MATH 423

    Excerpt: ... Math 423 Homework 7 due in class on Tuesday, 8/8/06 1. Curtis, page 201, #3 2. Curtis, page 201, #4 3. Curtis, page 201, #6 4. Curtis, page 215, #4 (Important: I changed to #4 instead of #5 because I wanted to do #5 in the class notes.) 5. Curtis, page 140, #3 6. Let f (x) = xn + an1 xn1 + . . . + a1 x + a0 F [x]. The companion matrix of f (x) is the matrix 0 0 0 0 a0 1 0 0 0 a1 0 1 0 0 a2 A= . . . . Show that the characteristic polynomial of A is f (x), and thus f (A) = 0. . . . . . . . . . . . . 0 0 0 0 an2 0 0 0 1 an1 Hint: Calculate the determinant by expanding by cofactors along the last column. Note that there is a bit of a discussion on the companion matrix on page 221 of Curtis. Math 423 Homework 8 due in class on Wednesday, 8/9/06 1. Over R, determine which of these matrices are similar by nding their Jordan canonical form. First show that for each matrix ...

  • 5 Pages

    4027s07ex2a

    LSU, M 4027

    Excerpt: ... Name: Exam 2 Instructions. Answer each of the questions on your own paper, except for problem 2, where you may record your answers in the box provided. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. Put your name on each page of your paper. You are allowed a single sheet of paper with notes on it. 1. [20 Points] Find the general solution, expressed in terms of real valued functions, of each of the following dierential equations. (a) y 4y + 13y = 0 Solution. The characteristic polynomial is p() = 2 4 + 13 = ( 2)2 + 9, so the roots are 1, 2 = 2 3i. Hence the general solution is y = c1 e2t cos 3t + c2 e2t sin 3t. (b) y 4y + 4y = 0 Solution. The characteristic polynomial is p() = 2 4 + 4 = ( 2)2 , so the only root is = 2, with multiplicity 2. Hence the general solution is y = c1 e2t + c2 te2t . (c) y 4y 5y = 0 Solution. The characteristic po ...

  • 1 Pages

    32l

    University of Hawaii - Hilo, MATH 311

    Excerpt: ... Math 311 Lecture 32 Recall: O is an eigenvalue of L or A and v is an eigenvector for O iff v 0 and L(v) = Ov or A(v) = Ov. LEMMA. For any eigenvalue O of a matrix A, the set of its eigenvectors along with 0 is a subspace, called the eigenspace of O. PROOF. Given O: v is an eigenvector of O iff Av = Ov iff Ov Av = 0 iff (OIA)v = 0 iff v is in the nullspace of (OIA). Hence the eigenspace for O is the nullspace for OIA and nullspaces are always subspaces. PROCEDURE. To find a basis for the eigenvectors for O, find a basis for the nullspace of OIA. E DEFINITION. For any matrix A, det(OIA) is the characteristic polynomial of A. THEOREM. The eigenvalues of a matrix A are exactly the real roots of its characteristic polynomial . PROOF. O is an eigenvalue of A iff Av = Ov for some v 0 iff Ov Av = 0 for some v 0 iff (OIA)v = 0 for some v 0 iff (OIA)v = 0 has a nontrivial solution iff det(OIA) = 0 iff O is a root of the characteristic polynomial . E PROCEDURE. To find all eigenvalues: find and factor the characterist ...

  • 2 Pages

    lec18

    Berkeley, MATH 110

    Excerpt: ... MATH 110 Lecture Notes 18 GSI Carter July 22, 2008 1 1.1 The Jordan Canonical Form Nilpotent Matrices Here we assume that T is a nilpotent operator on an n-dimensional vector space V with characteristic polynomial f (t) = (t)n . (This was the case reduced to last time, up to addition of a matrix I for some scalar .) We have a chain of subspaces V R(T ) R(T 2 ) R(T n1 ) R(T n ) = {0}. This gives us another chain N (T ) N (T ) R(T ) N (T ) R(T 2 ) N (T ) R(T n ) = {0}. We can construct an ordered basis v1 , . . . , vk of N (T ) by starting with a basis for N (T )R(T n1 ), expanding it to a basis for N (T ) R(T n2 ), expanding that to a basis for N (T ) R(T n3 ), and so on. Then for each i there exist cycles vi = T ri xi , T ri 1 xi , . . . , T xi , xi whose lengths decrease (not strictly) as i increases. Let Wi = span{T ri xi , . . . , T xi , xi }. By a previous prelim problem, this cycle is a basis for Wi . Th ...

  • 1 Pages

    10-29-03

    Washington, MATH 252

    Excerpt: ... Homework Assignment 7 (Math 252: Modular Abelian Varieties) William A. Stein Oct. 29 (Due: Nov. 5) 1. Let L be a lattice in C and let n be a positive integer. Prove that the number of sublattices of L of index n is equal to the sum of the positive divisors of n. (Hint: Use the characterization of sublattices of index n from the notes, and reduce to the prime power case.) 2. In this problem you probably want to use a computer, though it isn't strictly necessary. (a) Show that k = 24 is the smallest weight such that dim Sk (1) > 1. (b) Find a basis for S24 (1). (c) Compute a matrix for the Hecke operator T2 with respect to your basis. (d) Compute the characteristic polynomial of your matrix. (e) Prove that the characteristic polynomial of your matrix is irreducible, and hence verify Maeda's conjecture in this case. (f) Find the smallest k such that Sk (1) has dimension 3. (g) Compute a matrix for T2 on Sk (1). (h) Verify that the characteristic polynomial of T2 has Galois group the full symmetric group S3 , ...

  • 8 Pages

    Lecture 4

    UT Arlington, EE 4314

    Excerpt: ... . when the number of poles of the transfer function is equal to the number of poles of the state-space representation of the system, or, in other words, when the state variable system is a minimal representation of the transfer function). Problem 1 0 1 0 & x+ u, Let x = Ax + Bu = 1 0 1 y = Cx = [- 1 1]x . 1 a. The characteristic polynomial is s -1 D( s) = = s 2 - 1 = ( s + 1)( s - 1) . The poles are at s=-1, s=1, so the system is -1 s not AS. It is unstable. The natural modes are e - t , e t . b. The transfer function is 1 s -1 = H ( s ) = C ( sI - A) -1 B = , ( s - 1)( s + 1) s + 1 which has poles at s=-1. Therefore, the system is BIBO stable. Note that the unstable pole at s=1 has cancelled with a zero at s=1. c. d. e. f. g. Does this system have transmission zeros? Is this system controllable? Is this system observable? Does this system have decoupling zeros? If YES, what are their values and what is their nature (input, output or input/output decoupling zeros)? Problem 2 What is the ...

  • 1 Pages

    charAB

    Los Angeles Southwest College, MATH 700

    Excerpt: ... The Characteristic Polynomial of a Product Ralph Howard This is a note to prove the most notorious of all qualifying exam questions: Theorem Let A and B be n n matrices over a eld F. Then the characteristic polynomial s of AB and BA are the same. Remark 1 If one of the two matrices, say A, is invertible then A1 (AB)A = BA. Thus AB and BA are similar which certainly implies AB and BA have the same characteristic polynomial . However if A and B are both nonsingular then AB and BA do not have to be similar. For example if 1 0 0 1 A= , B= 0 0 0 0 then AB = B = 0, but BA = 0 so that AB and BA are not similar. Remark 2 I leaned of the following proof from Tom Markham who said it is originally due to Paul Halmos. Proof: For any square matrix let C let charC () := det(I C) be the characteristic polynomial of C. Let r be the rank of A. Then by doing row and column reduction there are invertible n n matrices P and Q so that I 0 A=P Q 0 0 where I is the r r identity matrix. Now express B as B = Q ...

  • 4 Pages

    helper_11-16-05

    BU, MATH 242

    Excerpt: ... MA 242 The characteristic polynomial and characteristic equation November 16, 2005 Last class we saw that the number is an eigenvalue for the matrix A if and only if det(A - I) = 0. Now I want to use the computer to examine the characteristic polynomial for various matrices. Example. Let A= 3 1 1 3 . Example. Let A= 3 -2 -1 0 . 1 MA 242 Example. Let A= November 16, 2005 2 1 1 1 -1 2 -1 1 0 . Example. Let A= 0 -2 -2 3 2 1 -2 2 -2 0 5 4 -4 6 -6 -6 . 2 MA 242 November 16, 2005 Theorem. If a matrix is upper/lower triangular, then its eigenvalues are the entries along the main diagonal. In theory, any polynomial p() can be factored into irreducible linear and quadratic factors using real numbers. For example, consider the polynomial p() = 9 + 88 + 367 + 946 + 1435 + 984 - 483 - 1602 - 132 - 40. This polynomial factors into p() = (2 + 2 + 2)2 (2 + 3 + 10)2 ( + 1)2 ( - 1). The algebraic multiplicity of an eigenvalue 0 is the number of times that the factor ( - 0 ) appears in the fac ...