Documents about Characteristic Polynomial
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 ...
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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. ...
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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 ...
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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 ...
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eigenvalues
Georgia Tech, MATH 2602
Excerpt: ... Lecture notes for Math 1502 - Georgia Institute of Technology April 7, 2000 Eigenvalues and eigenvectors definition examples ( 1 2) 1 0 1 25 = 2, x= (2) 1 x= = -1, x = 1 (-1) 66 ( -12 -12) 59 =2, (3) 4 1 (-1) = 3, x = (-4) 3 (2) 3 ( 1/2 1/3) 1/2 2/3 =1/6, x= = 1, x = diagonalization of an nn matrix If A is an nn matrix, and if there exists a basis of Rn consisting of eigenvectors for A, then A can be diagonalized Using diagonalization to compute powers of a matrix If A is an n n symmetric matrix, then there is an orthogonal basis of Rn consisting of eigenvectors computation of eigenvalues and eigenvectors the characteristic polynomial of a matrix is det(A-I) the eigenvalues are the roots of the characteristic polynomial how to find the eigenvectors which correspond to a given eigenvalue or each find the nonzero x which solve (A-lI)x=0 miscellaneous 0 is an eigenvalue if, and only if, the matrix is singular the sum of the eigenvalues is the trace, the product is t ...
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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 ...
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211lec20
ECCD, MATH 211
Excerpt: ... on that eigenvalues are the roots of the characteristic polynomial of A, which is, by denition pA () = det(A I). Note that this allows us to compute eigenvalues by factoring! Given an eigenvalue, , the corresponding eigenvectors are elements of the nullspace of A I. So these form a subspace which can be determined by a row reduction computation. 2. Planar systems We take up the specics of this approach for planar autonomous homogeneous linear systems with constant coecients. We shall look at three examples which show o the possible behaviors. In fact, these are specic examples, but they are completely representative. We shall state the general solution in each case and discuss the resulting phase plane pictures. 2 1 Example (distinct real eigenvalues) Use A = . 1 1 1 Example (complex eigenvalues) Use B = 2 1 . 1 2 1/3 0 . 0 1/3 5 1 . Example (a repeated eigenvalue, second type) Use D = 0 5 We discuss the solution of these test cases in great detail. . . Example (a r ...
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sol9
Berkeley, EE 221
Excerpt: ... EECS 221 A 1 Consider the system x = Ax + Bu, where: 3 6 13 7 B= 1 A = 2 4 2 1 2 4 3 (a) Lets compute the controllabitlity matrix: 3 2 5 C = B AB A2 B = 1 0 2 1 1 1 thus rank(C) = 3 and the pair (A, B) is controllable. Solutions # 9 (1) (2) (b) The characteristic polynomial of the system is A (s) = s3 + s2 + 2s + 2. Then, by denition, the controllable canonical form of this system is described by the matrices: 0 1 0 0 Ac = 0 Bc = 0 (3) 0 1 2 2 1 1 (c) Since we want to place all eigenvalues of the closed loop at 1, the desired characteristic polynomial is (s) = s3 + 3s2 + 3s + 1. Let F = [f1 f2 f3 ] and consider a state feedback given by u = F x, then: 0 1 0 (4) Ac + Bc F = 0 0 1 2 + f1 2 + f2 1 + f3 thus Ac +Bc F (s) = s3 + (1 f3 )s2 + (2 f2 )s + (2 f1 ). If we choose F = [1 1 2] we obt ...
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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 ...
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HW24
Concordia Moorhead, MATH 210
Excerpt: ... Lenarz - Math 210 1. Show that A = 4 1 3 1 HW # 24 4.4 Solutions and B = 1 0 0 1 Due: April 4, 2008 are not similar matrices. Solution: Note that det A = det B = 1, both A and B are invertible, and rank(A) = rank(B) = 2. But the characteristic polynomial s are dierent (and the eigenvalues are dierent): The characteristic polynomial of A is det(A I) = 4 1 3 1 = (4 )(1 ) (1)(3) = 2 5 + 1 and the characteristic polynomial of B is det(B I) = 1 0 0 1 = (1 )(1 ) (0)(0) = 2 2 + 1 2. Show that A = 3 1 5 7 and B = 2 1 4 6 are not similar matrices. Solution: Note that det A = det B = 16, both A and B are invertible, and rank(A) = rank(B) = 2. But the characteristic polynomial s are dierent (and the eigenvalues are dierent): The characteristic polynomial of A is det(A I) = 3 1 5 7 = (3 )(7 ) (1)(5) = 2 10 + 16 and the characteristic polynomial of B is de ...
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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 ...
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Finalrev
UConn, MATH 2210
Excerpt: ... n 5.2: The characteristic polynomial and the characteristic equation; similar matrices; the characteristic polynomial and similar matrices; the multiplicity of a root of the characteristic polynomial and the dimension of the corresponding eigenspace. Section 5.3: Diagonalization; how diagonalization depends on the number of linearly independent eigenvectors and not on the number of eigenvalues. Section 5.4: Viewing the diagonalization of an n n matrix A as [T ]B where B is a basis of Rn consisting of eigenvectors of A and T : Rn - Rn ; x Ax; The matrix representation [T ]CB where T : V - W and B is a basis of V and C is a basis of W . The figures in this section are very helpful in getting a picture of what this is all about. Section 6.1: Dot product; properties of the dot product; length, norm, distance; orthogonality; Orthogonal complements of vector subspaces. Section 6.2: Orthogonal sets of vectors; orthogonal basis of a vector space; calculation of the coordinate vector of y in terms of an ort ...
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week8
Caltech, CDS 110
Excerpt: ... CDS110: Introduction to Control of Physical Systems Fall 2001 Q & A: Week 8 1 Example of pole placement x = Ax + Bu with eigenvalues i (A), we would like to find a control u = -Gx such that the eigenvalues of the system are in specified locations. If the system has the characteristic polynomial sn + a1 sn-1 + a2 sn-2 + . . . + an-1 s + an then the first companion form is -a1 1 0 . 0 -a2 0 1 0 . . . -an-1 . 0 . 0 . 1 -an 0 0 . 0 1 0 . . . Given the system Af = and the control canonical form is Ac = , Bf = 0 0 0 0 . 0 -an 1 0 0 -an-1 . . 0 0 . 0 . . . -a2 0 0 . 1 -a1 , Bc = 0 1 0 0 . . . To place the poles in either of these forms is straightforward and we simply use u = -Gf xf or u = -Gc xc If the desired characteristic polynomial is sn + a1 sn-1 + a2 sn-2 + . . . + an-1 s + an ^ ^ ^ ^ then the gains would simply be (this is the transpose of the book's definition and follows my lecture notes) ^ ^ ^ ^ Gf = a1 - a1 a2 - a2 . ...
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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 ...
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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 ...
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300_Exam2_StudyGuide
University of Hawaii - Hilo, MATH 300
Excerpt: ... Math 300 Fall 2008 Exam 2 Study Guide Dr. Brian Wissman The second exam will roughly cover sections 1.6 and 2.1 - 2.2. More specifically to do well on the exam you should know the following: Substitution for equations of the form, y (x) = F (ax + by + c) Substitution for Bernoulli Equations Reducible 2nd order Equations Superposition, show if y1 and y2 are solutions, so is y = c1 y1 + c2 y2 Solutions to 2nd order constant coefficient differential equations, real distinct roots, repeated roots. Derivation of the characteristic polynomial General solution to an Euler Equation, ax2 y + bxy + cy = 0 Structure of General Solutions to homogeneous problems Find solution to homogeneous IVP given a general solution What it means to be a particular solution to a non-homogeneous equation Structure of the solutions to a non-homogeneous problem Solve non-homogeneous IVP given a particular and general solution You are expected to know basic integration and differentiation techniques as they are needed in some way to solve ...
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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 ...
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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 , ...
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hw4
Cornell, WEB 224
Excerpt: ... Math 224 Homework Set 4 Due March 1, 2005 Reading: 4.8-4.9 Problems from the book: 4.8.4, 4.8.6, 4.8.8, 4.8.16, 4.9.3, 4.9.4. Problem 1 Prove that the matrix 1 1 x1 x2 x2 M = x2 1 2 . . xm-1 xm-1 2 1 . 1 . . . xm . . . x2 m . . . . . xm-1 m is invertible if and only if x1 , . . . , xm are distinct numbers. Hint: the "only if" part is easy - what is the det(M ) if xi = xj ? To complete the other half, prove that M T has a "zero kernel". To see that, a0 a1 assume that a = . . . Ker(M ) and consider the polynomial am-1 p(x) = am-1 xm-1 + am-2 xm-2 + . . . + a1 x + a0 . [Note that the above proves that the linear system of equations for weights w1 , . . . , wm always has a unique solution defining the quadrature based on a fixed set of distinct nodes x1 , . . . , xm .] Problem 2 Let A be an n n matrix. Recall the characteristic polynomial of A is det(xI - A). (a) Show that if A and B are n n matrices that satisfy A = P BP -1 then A and B have the same characteristic polynomi ...
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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 ...
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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 ...
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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 ...
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