DOCSLIB.ORG

0 BBB@ 1 CCCA= X3 0 BBB@ 1 CCCA+ X4 0

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
0 BBB@ 1 CCCA= X3 0 BBB@ 1 CCCA+ X4 0 Math 252 Applied Linear Algebra Problem Notes on Take-Home Exams I to IX 1. 2 2 Linear System Find all solutions to x 5x =0, x +5x =0. Express the answer in one of 1 2 1 2 the three p ossible parametric forms: A point. This represents the intersection of two lines. A line. The parametric form is x = c + td , 1 1 y = c + td , 2 2 with 1 <t<1. A plane. In this case, all p oints x; y are answers. Solution to 1. To explain geometrically the exp ected kind of answer, observe that x +5x =0 is the 1 2 same as the rst equation x 5x = 0, therefore there are not two equations, but only one! The set of planar 1 2 points satisfying the two equations is exactly the set of points on the straight line x 5x =0 an in nite 1 2 number of points. The standard form of the solution is obtained by solving for x in terms of x , e.g., 1 2 x =5x , then write out the vector solution X as follows: 1 2 ! x 1 X = x 2 ! 5x 2 = x 2 ! 5 = x 2 1 Toeachvalue of x corresp onds a solution of the system of equations, i.e., there are in nitely many solutions, 2 represented go emetrically as a line. 1a. 2 4 Linear System Find all solutions to x x +7x x =0,2x +3x 8x + x =0. 1 2 3 4 1 2 3 4 Solution to 1a. Subtract two times the rst equation from the second to get 5x 22x +3x =0. Divide 2 3 4 3 22 x + x =0. Keep this as the replacement for the second equation. Add the new equation by5toget x 3 4 2 5 5 13 2 it to the rst equation to get its replacement x + x x =0. The replacement equations are therefore 1 3 4 5 5 2 13 x x =0; x + 3 4 1 5 5 22 3 x x + x =0; 2 3 4 5 5 which corresp ond exactly to the reduced row echelon form of the system. The general solution is 0 1 0 1 1 0 2 13 x 1 5 5 B C B C C B 22 3 x B C B C C B 2 5 5 + x : = x B C B C C B 4 3 @ A @ A A @ 1 0 x 3 0 1 x 4 2. 4 4 Linear System Find a fraction-free Gauss-elimination form and the reduced row-echelon form for the following equations: x 2x + x + x =2,3x +2x 2x = 8, 4x x x =1,5x +3x x =0. 1 2 3 4 1 3 4 2 3 4 1 3 4 1 Solution to 2. A fraction-free Gauss-elimination form can b e obtained from the reduced rowechelon form by multiplying each row by a suitable factor, to clear the fractions. In reality, there are in nitely many fraction-free forms, so there is no way to give an answer that everyone will arriveat. It turns out that the reduced rowechelon form is also fraction-free, so it can b e rep orted as the fraction-free answer! The reduced row-echelon form is obtained from the augmented matrix by row op erations, using the basic pivot algorithm. The answer for both questions: 1 0 0 1 0 0 4 C B 0 1 0 2 0 C B rref = C B A @ 0 0 0 1 7 0 0 0 0 1 3. 4 3 Linear System Find all solutions to the 4 3 system x + x x = 0, 4x x +5x = 0, 1 2 3 1 2 3 2x + x 2x =0,3x +2x 6x =0. 1 2 3 1 2 3 Solution to 3. The augmented matrix and its reduced rowechelon form are: 0 1 0 1 1 1 1 0 1 0 0 0 B C B C 4 1 5 0 0 1 0 0 B C B C aug = ; rref = B C B C @ A @ A 1 1 2 0 0 1 0 0 0 0 3 2 6 0 0 0 It follows that X = 0 is the only solution. 4. 3 4 Linear System Find all solutions to x x + x x = 2, 2x +3x x +2x = 5, 1 2 3 4 1 2 3 4 4x 2x +2x 3x =6. 1 2 3 4 Solution to 4. The augmented matrix and its reduced rowechelon form are: 0 1 0 1 1 0 0 1=2 5 1 1 1 1 2 B C B C aug = 2 3 1 2 5 ; rref = 0 1 0 1=4 4 @ A @ A 4 2 2 3 6 0 0 1 1=4 3 The standard form of the solution is obtained by identifying the lead variables and the arbitrary variables: Variables x , x and x are the lead variables, b ecause they corresp ond to a leading 1 in the RREF. 1 2 3 See the b oxed 1's ab ove. Variable x is the arbitrary variable, b ecause arbitrary variables are the variables left over after 4 removal of the lead variables. The standard form of the solution X is obtained by replacing each lead variable with its corresp onding equation, obtained from the RREF. The arbitrary variables are left untouched. Then: 1 0 1 0 1 0 0 1 1 1 x +5 5 x 4 1 2 2 C B C B C B B C 1 1 x +4 4 x C B C B C B B C 4 2 4 4 = x = + X = C B C B C B B C 4 1 A @ A @ A @ @ A x 3 3 3 x 4 3 4 x 1 0 x 4 4 2 4a. 3 4 Linear System Use Gauss-Jordan elimination to nd the general solution: x + 2x + 4x x = 3 1 2 3 4 3x + 4x + 5x x = 7 1 2 3 4 x + 3x + 4x + 5x = 4 1 2 3 4 Solution to 4a. The augmented matrix and its reduced rowechelon form are: 1 0 1 0 1 0 0 5 1 3 1 2 4 1 C B C B 3 4 5 1 7 ; rref = aug = 0 1 0 6 1 A @ A @ 1 3 4 5 4 0 0 1 2 0 Variables x , x and x are the lead variables, b ecause they corresp ond to a leading 1 in the RREF. 1 2 3 See the b oxed 1's ab ove. Variable x is the arbitrary variable, b ecause arbitrary variables are the variables left over after 4 removal of the lead variables. The standard form of the solution is obtained by replacing the lead variables x , x , x by their equations 1 2 3 x =5x +1, x = 6x +1, x =2x , but the arbitrary variable x is untouched. Then: 1 4 2 4 3 4 4 0 1 0 1 0 1 0 1 x 5x +1 5 1 1 4 B C B C B C B C x 6x +1 6 1 B C B C B C B C 2 4 X = = = x + B C B C B C B C 4 @ A @ A @ A @ A x 2x 2 0 3 4 x x 1 0 4 4 Imp ortant: This method for writing out X applies only in case the equations are in reduced echelon form. A matrix C is in reduced row echelon form provided each nonzero row starts with a leading 1, and ab ove and b elow that leading 1 app ear only zeros. 5. Linear Combinations Compute the result of the linear combination 2u + v 3w where 1 0 1 0 1 0 9 0 1 C B C B C B : ; w = 1 ; v = 0 u = 1 A @ A @ A @ 4 2 2 Solution to 5. The result of the linear combination is 1 0 1 0 1 0 1 0 25 9 0 1 C B C B C B C B : = 1 3 1 + 0 2u + v 3w =2 1 A @ A @ A @ A @ 10 4 2 2 5a. EqualityofVectors Let 0 1 0 1 0 1 1 x +1 9 B C B C B C u = x ; v = 0 ; w = 1 @ A @ A @ A 2 2 4x The linear combination p =2u + v 3w dep ends up on x. Is there a value of x such that 0 1 22:9 B C 0:8 p = ? @ A 11:2 3 Solution to 5a. The linear combination p is given by 0 1 0 1 0 1 0 1 1 x +1 9 x 24 B C B C B C B C p =2 x + 0 3 1 = 2x 3 : @ A @ A @ A @ A 2 2 4x 12x 2 There is a value of x such that 1 0 22:9 C B p = 0:8 A @ 11:2 exactly when the comp onents agree, i.e., x 24 = 22:9, 2x 3=0:8, 12x 3=11:2. This happ ens for x =1:1check all three equations!. 6. Largest Linearly Indep endent Set Extract from the list 1 1 0 1 0 1 0 0 0 1 2 1 C C B C B C B B ; 1 ; 0 ; 2 1 A A @ A @ A @ @ 1 1 0 0 a largest set of linearly indep endentvectors. Solution to 6. To extract a largest linearly indep endent set of vectors from a list v , v , v , v 4 requires an 1 2 3 algorithm b e followed.
Load more

Recommended publications
  • Math 221: LINEAR ALGEBRA
    Math 221: LINEAR ALGEBRA Chapter 8. Orthogonality §8-3. Positive Definite Matrices Le Chen1 Emory University, 2020 Fall (last updated on 11/10/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary. Positive Definite Matrices Cholesky factorization – Square Root of a Matrix Positive Definite Matrices Definition An n × n matrix A is positive definite if it is symmetric and has positive eigenvalues, i.e., if λ is a eigenvalue of A, then λ > 0. Theorem If A is a positive definite matrix, then det(A) > 0 and A is invertible. Proof. Let λ1; λ2; : : : ; λn denote the (not necessarily distinct) eigenvalues of A. Since A is symmetric, A is orthogonally diagonalizable. In particular, A ∼ D, where D = diag(λ1; λ2; : : : ; λn). Similar matrices have the same determinant, so det(A) = det(D) = λ1λ2 ··· λn: Since A is positive definite, λi > 0 for all i, 1 ≤ i ≤ n; it follows that det(A) > 0, and therefore A is invertible. Theorem A symmetric matrix A is positive definite if and only if ~xTA~x > 0 for all n ~x 2 R , ~x 6= ~0. Proof. Since A is symmetric, there exists an orthogonal matrix P so that T P AP = diag(λ1; λ2; : : : ; λn) = D; where λ1; λ2; : : : ; λn are the (not necessarily distinct) eigenvalues of A. Let n T ~x 2 R , ~x 6= ~0, and define ~y = P ~x. Then ~xTA~x = ~xT(PDPT)~x = (~xTP)D(PT~x) = (PT~x)TD(PT~x) = ~yTD~y: T Writing ~y = y1 y2 ··· yn , 2 y1 3 6 y2 7 ~xTA~x = y y ··· y diag(λ ; λ ; : : : ; λ ) 6 7 1 2 n 1 2 n 6 .
    [Show full text]
  • 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).
    [Show full text]
  • On the Eigenvalues of Euclidean Distance Matrices
    “main” — 2008/10/13 — 23:12 — page 237 — #1 Volume 27, N. 3, pp. 237–250, 2008 Copyright © 2008 SBMAC ISSN 0101-8205 www.scielo.br/cam On the eigenvalues of Euclidean distance matrices A.Y. ALFAKIH∗ Department of Mathematics and Statistics University of Windsor, Windsor, Ontario N9B 3P4, Canada E-mail: [email protected] Abstract. In this paper, the notion of equitable partitions (EP) is used to study the eigenvalues of Euclidean distance matrices (EDMs). In particular, EP is used to obtain the characteristic poly- nomials of regular EDMs and non-spherical centrally symmetric EDMs. The paper also presents methods for constructing cospectral EDMs and EDMs with exactly three distinct eigenvalues. Mathematical subject classification: 51K05, 15A18, 05C50. Key words: Euclidean distance matrices, eigenvalues, equitable partitions, characteristic poly- nomial. 1 Introduction ( ) An n ×n nonzero matrix D = di j is called a Euclidean distance matrix (EDM) 1, 2,..., n r if there exist points p p p in some Euclidean space < such that i j 2 , ,..., , di j = ||p − p || for all i j = 1 n where || || denotes the Euclidean norm. i , ,..., Let p , i ∈ N = {1 2 n}, be the set of points that generate an EDM π π ( , ,..., ) D. An m-partition of D is an ordered sequence = N1 N2 Nm of ,..., nonempty disjoint subsets of N whose union is N. The subsets N1 Nm are called the cells of the partition. The n-partition of D where each cell consists #760/08. Received: 07/IV/08. Accepted: 17/VI/08. ∗Research supported by the Natural Sciences and Engineering Research Council of Canada and MITACS.
    [Show full text]
  • Rotation Matrix - Wikipedia, the Free Encyclopedia Page 1 of 22
    Rotation matrix - Wikipedia, the free encyclopedia Page 1 of 22 Rotation matrix From Wikipedia, the free encyclopedia In linear algebra, a rotation matrix is a matrix that is used to perform a rotation in Euclidean space. For example the matrix rotates points in the xy -Cartesian plane counterclockwise through an angle θ about the origin of the Cartesian coordinate system. To perform the rotation, the position of each point must be represented by a column vector v, containing the coordinates of the point. A rotated vector is obtained by using the matrix multiplication Rv (see below for details). In two and three dimensions, rotation matrices are among the simplest algebraic descriptions of rotations, and are used extensively for computations in geometry, physics, and computer graphics. Though most applications involve rotations in two or three dimensions, rotation matrices can be defined for n-dimensional space. Rotation matrices are always square, with real entries. Algebraically, a rotation matrix in n-dimensions is a n × n special orthogonal matrix, i.e. an orthogonal matrix whose determinant is 1: . The set of all rotation matrices forms a group, known as the rotation group or the special orthogonal group. It is a subset of the orthogonal group, which includes reflections and consists of all orthogonal matrices with determinant 1 or -1, and of the special linear group, which includes all volume-preserving transformations and consists of matrices with determinant 1. Contents 1 Rotations in two dimensions 1.1 Non-standard orientation
    [Show full text]
  • Matrices and Matrix Operations with TI-Nspire™ CAS
    Matrices and Matrix Operations with TI-Nspire™ CAS Forest W. Arnold June 2020 Typeset in LATEX. Copyright © 2020 Forest W. Arnold This work is licensed under the Creative Commons Attribution-Noncommercial-Share Alike 4.0 International License. To view a copy of this license, visit http://creativecommons. org/licenses/by-nc-sa/4.0/legalcode/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA. You can use, print, duplicate, share this work as much as you want. You can base your own work on it and reuse parts if you keep the license the same. Trademarks TI-Nspire is a registered trademark of Texas Instruments, Inc. Attribution Most of the examples in this article are from A First Course in Linear Algebra an Open Text by Lyrix Learning, base textbook version 2017 - revision A, by K. Kuttler. The text is licensed under the Creative Commons License (CC BY) and is available for download at the link https://lyryx.com/first-course-linear-algebra/. 1 Introduction The article Systems of Equations with TI-Nspire™ CAS: Matrices and Gaussian Elim- ination described how to represent and solve systems of linear equations with matrices and elementary row operations. By defining arithmetic and algebraic operations with matrices, applications of matrices can be expanded to include more than simply solving linear systems. This article describes and demonstrates how to use TI-Nspire’s builtin matrix functions to • add and subtract matrices, • multiply matrices by scalars, • multiply matrices, • transpose matrices, • find matrix inverses, and • use matrix equations. The TI-Nspire examples in this article require the CAS version of TI-Nspire.
    [Show full text]
  • Real Symmetric Matrices
    Chapter 2 Real Symmetric Matrices 2.1 Special properties of real symmetric matrices A matrixA M n(C) (orM n(R)) is diagonalizable if it is similar to a diagonal matrix. If this ∈ happens, it means that there is a basis ofC n with respect to which the linear transformation ofC n defined by left multiplication byA has a diagonal matrix. Every element of such a basis is simply multiplied by a scalar when it is multiplied byA, which means exactly that the basis consists of eigenvectors ofA. n Lemma 2.1.1. A matrixA M n(C) is diagonalizable if and only ifC possesses a basis consisting of ∈ eigenvectors ofA. 1 1 Not all matrices are diagonalizable. For exampleA= is not. To see this note that 1 0 1 � � (occurring twice) is the only eigenvalue ofA, but that all eigenvectors ofA are scalar multiples of 1 2 2 0 , soC (orR ) does not contain a basis consisting of eigenvectors ofA, andA is not similar to a diagonal matrix. � � We note that a matrix can fail to be diagonalizable only if it has repeated eigenvalues, as the following lemma shows. Lemma 2.1.2. LetA M n(R) and suppose thatA has distinct eigenvaluesλ 1,...,λ n inC. ThenA is ∈ diagonalizable. Proof. Letv i be an eigenvector ofA corresponding toλ i. We will show thatS={v 1,...,v n} is a linearly independent set. Sincev 1 is not the zero vector, we know that{v 1} is linearly independent. IfS is linearly dependent, letk be the least for which{v 1,...,v k} is linearly dependent.
    [Show full text]
  • Notes on Symmetric Matrices 1 Symmetric Matrices
    CPSC 536N: Randomized Algorithms 2011-12 Term 2 Notes on Symmetric Matrices Prof. Nick Harvey University of British Columbia 1 Symmetric Matrices We review some basic results concerning symmetric matrices. All matrices that we discuss are over the real numbers. Let A be a real, symmetric matrix of size d × d and let I denote the d × d identity matrix. Perhaps the most important and useful property of symmetric matrices is that their eigenvalues behave very nicely. Definition 1 Let U be a d × d matrix. The matrix U is called an orthogonal matrix if U TU = I. This implies that UU T = I, by uniqueness of inverses. Fact 2 (Spectral Theorem). Let A be any d × d symmetric matrix. There exists an orthogonal matrix U and a (real) diagonal matrix D such that A = UDU T: This is called a spectral decomposition of A. Let ui be the ith column of U and let λi denote the ith diagonal entry of D. Then fu1; : : : ; udg is an orthonormal basis consisting of eigenvectors of A, and λi is the eigenvalue corresponding to ui. We can also write d X T A = λiuiui : (1) i=1 The eigenvalues λ are uniquely determined by A, up to reordering. Caution. The product of two symmetric matrices is usually not symmetric. 1.1 Positive semi-definite matrices Definition 3 Let A be any d × d symmetric matrix. The matrix A is called positive semi-definite if all of its eigenvalues are non-negative. This is denoted A 0, where here 0 denotes the zero matrix.
    [Show full text]
  • On the Uniqueness of Euclidean Distance Matrix Completions Abdo Y
    View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Linear Algebra and its Applications 370 (2003) 1–14 www.elsevier.com/locate/laa On the uniqueness of Euclidean distance matrix completions Abdo Y. Alfakih Bell Laboratories, Lucent Technologies, Room 3J-310, 101 Crawfords Corner Road, Holmdel, NJ 07733-3030, USA Received 28 December 2001; accepted 18 December 2002 Submitted by R.A. Brualdi Abstract The Euclidean distance matrix completion problem (EDMCP) is the problem of determin- ing whether or not a given partial matrix can be completed into a Euclidean distance matrix. In this paper, we present necessary and sufficient conditions for a given solution of the EDMCP to be unique. © 2003 Elsevier Inc. All rights reserved. AMS classification: 51K05; 15A48; 52A05; 52B35 Keywords: Matrix completion problems; Euclidean distance matrices; Semidefinite matrices; Convex sets; Singleton sets; Gale transform 1. Introduction All matrices considered in this paper are real. An n × n matrix D = (dij ) is said to be a Euclidean distance matrix (EDM) iff there exist points p1,p2,...,pn in some i j 2 Euclidean space such that dij =p − p for all i, j = 1,...,n. It immediately follows that if D is EDM then D is symmetric with positive entries and with zero diagonal. We say matrix A = (aij ) is symmetric partial if only some of its entries are specified and aji is specified and equal to aij whenever aij is specified. The unspecified entries of A are said to be free. Given an n × n symmetric partial matrix A,ann × n matrix D is said to be a EDM completion of A iff D is EDM and Email address: [email protected] (A.Y.
    [Show full text]
  • Rotations in 3-Space
    Rotations in 3-space A. Eremenko March 13, 2020 These facts of the combinations of rotations, and what they produce, are hard to grasp intuitively. It is rather strange, because we live in three dimensions, but it is hard for us to appreciate what happens if we turn this way and then that way. Perhaps, if we were fish or birds and had a real appreciation of what happens when we turn somersaults in space, we could more easily appreciate these things. Feynman’s lectures on physics, vol. 3, 6.5. O(3) pervades all the essential properties of the physical world. But we remain intellectually blind to this symmetry, even if we encounter it frequently and use it in everyday life, for instance when we experience or engender mechanical movements, such as walking. This is due in part to the non commutativity of O(3), which is difficult to grasp. M. Gromov, Notices AMS, 45, 846-847. Rotations in 3-space are more complicated than rotations in a plane. One reason for this is that rotations in 3-space in general do not commute. (Give an example!) We consider rotations with fixed center which we place at the origin. Then rotations are linear transformations of the space represented by orthogonal matrices with determinant 1. Orthogonality means that all distances and angles are preserved. The determinant of an orthogonal matrix is always 1 or -1. (Think why.) The 1 condition that determinant equals 1 means that orientation is preserved. An orthogonal transformation with determinant -1 will map a right shoe onto a left shoe; such orthogonal transformations are called sometimes improper rotations, we cannot actually perform them on real objects.
    [Show full text]
  • Exam 3 Review
    Exam 3 review The exam will cover the material through the singular value decomposition. Linear transformations and change of basis will be covered on the final. The main topics on this exam are: • Eigenvalues and eigenvectors du At • Differential equations dt = Au and exponentials e • Symmetric matrices A = AT: These always have real eigenvalues, and they always have “enough” eigenvectors. The eigenvector matrix Q can be an orthogonal matrix, with A = QLQT . • Positive definite matrices • Similar matrices B = M−1 AM. Matrices A and B have the same eigen­ values; powers of A will “look like” powers of B. • Singular value decomposition Sample problems 1. This is a question about a differential equation with a skew symmetric matrix. Suppose 2 0 −1 0 3 du = Au = 4 1 0 −1 5 u. dt 0 1 0 The general solution to this equation will look like l1t l2t l3t u(t) = c1e x1 + c2e x2 + cee x3. a) What are the eigenvalues of A? The matrix A is singular; the first and third rows are dependent, so one eigenvalue is l1 = 0. We might also notice that A is antisymmet­ ric (AT = −A) and realize that its eigenvalues will be imaginary. To find the other two eigenvalues, we’ll solve the equation jA − lIj = 0. � � � −l −1 0 � � � � 1 −l −1 � = −l3 − 2l = 0. � � � 0 1 −l � p p We conclude l2 = 2 i and l3 = − 2 i. 1 At this point we know that our solution will look like: p p 2 it − 2 it u(t) = c1x1 + c2e x2 + cee x3.
    [Show full text]
  • Spectral Properties of Sign Symmetric Matrices∗
    Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 13, pp. 90-110, March 2005 ELA www.math.technion.ac.il/iic/ela SPECTRAL PROPERTIES OF SIGN SYMMETRIC MATRICES∗ DANIEL HERSHKOWITZ† AND NATHAN KELLER‡ Abstract. Spectral properties of sign symmetric matrices are studied.A criterion for sign symmetry of shifted basic circulant permutation matrices is proven, and is then used to answer the question which complex numbers can serve as eigenvalues of sign symmetric 3 × 3matrices.The results are applied in the discussion of the eigenvalues of QM-matrices.In particular, it is shown that for every positive integer n there exists a QM-matrix A such that Ak is a sign symmetric P -matrix for all k ≤ n, but not all the eigenvalues of A are positive real numbers. Key words. Spectrum, Sign symmetric matrices, Circulant matrices, Basic circulant permuta- tion matrices, P -matrices, PM-matrices, Q-matrices, QM-matrices. AMS subject classifications. 15A18, 15A29. 1. Introduction. A square complex matrix A is called a Q-matrix [Q0-matrix] if the sums of principal minors of A of the same order are positive [nonnegative]. Equivalently, Q-matrices can be defined as matrices whose characteristic polynomials have coefficients with alternating signs. A square complex matrix is called a P -matrix [P0-matrix] if all its principal minors are positive [nonnegative]. A square complex matrix is said to be stable if its spectrum lies in the open right half plane. Let A be an n×n matrix. For subsets α and β of {1, ..., n} we denote by A(α|β)the submatrix of A with rows indexed by α and columns indexed by β.If|α| = |β| then we denote by A[α|β] the corresponding minor.
    [Show full text]
  • Definite Matrices
    John Nachbar September 2, 2014 Definite Matrices1 1 Basic Definitions. An N × N symmetric matrix A is positive definite iff for any v 6= 0, v0Av > 0. For example, if a b A = b c then the statement is that for any v = (v1; v2) 6= 0, 0 a b v1 v Av = [ v1 v2 ] b c v2 av1 + bv2 = [ v1 v2 ] bv1 + cv2 2 2 = av1 + 2bv1v2 + cv2 > 0 2 2 (The expression av1 +2bv1v2 +cv2 is called a quadratic form.) An N ×N symmetric matrix A is negative definite iff −A is positive definite. The definition of a positive semidefinite matrix relaxes > to ≥, and similarly for negative semi-definiteness. If N = 1 then A is just a number and a number is positive definite iff it is positive. For N > 1 the condition of being positive definite is somewhat subtle. For example, the matrix 1 3 A = 3 1 is not positive definite. If v = (1; −1) then v0Av = −4. Loosely, a matrix is positive definite iff (a) it has a diagonal that is positive and (b) off diagonal terms are not too large in absolute value relative to the terms on the diagonal. I won't formalize this assertion but it should be plausible given the example. The canonical positive definite matrix is the identity matrix, where all the off diagonal terms are zero. A useful fact is the following. If S is any M ×N matrix then A = S0S is positive semi-definite. To see this, note that S0S is symmetric N × N. To see that it is N positive semi-definite, note that for any v 2 R , v0Av = v0[S0S]v = (v0S0)(Sv) = (Sv)0(Sv) ≥ 0: 1cbna.
    [Show full text]