DOCSLIB.ORG

18.06 (Spring 14) Problem Set 9

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
18.06 (Spring 14) Problem Set 9 18.06 (Spring 14) Problem Set 9 This problem set is due Thursday, May 1, 2014 by 4pm in E17-131. The problems are out of the 4th edition of the textbook. This homework has 7 questions worth 70 points in total. Please WRITE NEATLY. You may discuss with others (and your TA), but you must turn in your own writing. 1. Problem 6.6.17 P.362. True or false, with a good reason: (a) A symmetric matrix can’t be similar to a nonsymmetric matrix. (b) An invertible matrix can’t be similar to a singular matrix. (c) A can’t be similar to −A unless A = 0. (d) A can’t be similar to A + I. solution (a) False. Let 0 0 1 −1 1 1 A = ;P = ;P −1 = 0 1 0 1 0 1 then we have 1 1 0 0 1 −1 0 1 P −1AP = = ; 0 1 0 1 0 1 0 1 which is not symmetric. (b) True. Let A; P be nonsingular matrices. Suppose −1P AP is singular, v 6= 0 is in the null space of P −1AP . We have P −1AP v = 0. Multiply by P on the left yields AP v = 0. Since P is nonsingular, v 6= 0, we have P v 6= 0 and is in the nullspace of A. We get a contradiction. (c) False. Let Let −1 0 1 0 0 1 A = ; −A = P = P −1 = 0 1 0 −1 1 0 then we have P −1AP = −A. The key is that A and −A could have the same set of eigenvalues. (d) True. Let v be a eigenvector of A with λ the corresponding eigenvalue(Both of them may be complex). Suppose there is a nonsingular P such that P −1AP = A + I. We have AP −1v + P −1v = (A + I)(P −1v) = P −1AP (P −1v) = P −1Av = λP −1v Therefore, we have AP −1v = λP −1Av − P −1v = (λ − 1)P −1v 1 Since P −1v 6= 0, λ − 1 is also an eigenvalue of A. Use the same reasoning, we get λ − k is an eigenvalue of A for any positive integer k. This contradict the fact that A can only have finite eigenvalues. Remark: The above argument works fine for matrices over R or C. If the field we are working on is finite, then the set fλ − kg may be finite. The argument above does not work and the statement may be false. However, it’s beyond the scope of 18.06. 2. Problem 6.6.18 P.362. If B is invertible, prove that AB is similar to BA. They have the same eigenvalues. solution Since B is invertible, B−1 exists and is invertible. (B−1)−1 = B. Therefore, (B−1)−1AB(B−1) = BA. AB and BA are similar. From P.355, we can conclude they have the same eigenvalues. T 3. Problem 6.7.1 P.371. Find the eigenvalues and unit eigenvectors v1, v2 of A A. Then find u1 = Av1/σ1: 1 2 10 20 5 15 A = ;AT A = ; AAT = : 3 6 20 40 15 45 T Verify that u1 is a unit eigenvector of AA . Complete the matrices U, Σ, V . 1 2 σ 0 T = u u 1 v v : 3 6 1 2 0 0 1 2 solution T 2 The characteristic polynomial for A A is λ − 50λ = 0. The eigenvalues are λ1 = 50, 1 T 1 T λ2 = 0. The corresponding unit eigenvectors are v1 = p (1; 2) , v2 = p (2; −1) . p 5 5 We have σ = 50. u = Av /σ = p1 (1; 3)T . Choose u from the null space of AT , 1 1 1 1 10 2 we have u = p1 (3; −1)T . Therefore, we have 2 10 p 1 1 −3 50 0 1 1 2 U = p ; Σ = ;V = p : 10 3 1 0 0 5 2 −1 4. Problem 6.7.14 P.372. Suppose A is invertible (with σ1 > σ2 > 0). Change A by as small a matrix as possible to produce a singular matrix A0. Hint: U and V do not change: σ1 0 T From A = u1 u2 v1 v2 , find the nearest A0. 0 σ2 2 solution The nearest matrix is obtained by letting σ2 to be 0, σ 0 T which is A = u u 1 v v . 0 1 2 0 0 1 2 0 0 T T The distance from A to A0 is u1 u2 v1 v2 = u2v2 . 0 σ2 5. Problem 10.2.3 P.506. Solve Az = 0 to find a vector in the nullspace of i 1 i A = : 1 i i Show that z is orthogonal to the columns of AH . Show that z is not orthogonal to the columns of AT . The good row space is no longer C(AT ). Now it is C(AH ). solution Use the elimination to get the upper triangular matrix i 1 i U = ; 0 2i i − 1 and the reduced row echelon form 1 1 0 2 (1 + i) R = 1 : 0 1 2 (1 + i) The free variable is the third variable. A basis of the homogeneous solution is given 1 1 T H T T by z = (− 2 (1 + i); − 2 (1 + i); 1) . The columns of A are (−i; 1; −i) , (1; −i; −i) . Since the inner product of any u; v is defined as uH v. We can conclude that the null space is orthogonal to C(AH ). Consider the inner product with the first row. 1 1 1 − i i − 1 (− (1 − i); − (1 − i); 1)(i; 1; i) = + + i = i 2 2 2 2 Therefore, z is not orthogonal to C(AT ). 6. Problem 10.2.6 P.507. True or false (give a reason if true or a counterexample if false): (a) If A is a real matrix then A + iI is invertible. (b) If A is a Hermitian matrix then A + iI is invertible. (c) If A is a unitary matrix then A + iI is invertible. 3 solution (a) False. Let 0 1 A = : −1 0 We have i 1 A + iI = : −1 i Let v = (i; 1)T . We have (A + iI)v = 0. Therefore, A + iI is not invertible. (b) True. If A is Hermitian, then A is unitarily diagonizable and all the eigenvalues are real. Therefore, all the eigenvalues of A+iI has imaginary part 1. Therefore, A + iI cannot have 0 as its eigenvalue and therefore invertible. (c) False. The counterexample in part (a) is also unitary. It can also serve as a counterexample for part (c). 7. Problem 10.2.19 P.508. The functions e−ix and eix are orthogonal on the interval R 2π 0 ≤ x ≤ 2π because their inner product is 0 = 0. solution Z 2π Z 2π Z 2π −ix ix −ix ¯ix −ix −ix −2ix i −2ix 2π he ; e i = e e dx = e e dx = e dx = e j0 = 0: 0 0 0 2 4.
Load more

Recommended publications
  • Math 4571 (Advanced Linear Algebra) Lecture #27
    Math 4571 (Advanced Linear Algebra) Lecture #27 Applications of Diagonalization and the Jordan Canonical Form (Part 1): Spectral Mapping and the Cayley-Hamilton Theorem Transition Matrices and Markov Chains The Spectral Theorem for Hermitian Operators This material represents x4.4.1 + x4.4.4 +x4.4.5 from the course notes. Overview In this lecture and the next, we discuss a variety of applications of diagonalization and the Jordan canonical form. This lecture will discuss three essentially unrelated topics: A proof of the Cayley-Hamilton theorem for general matrices Transition matrices and Markov chains, used for modeling iterated changes in systems over time The spectral theorem for Hermitian operators, in which we establish that Hermitian operators (i.e., operators with T ∗ = T ) are diagonalizable In the next lecture, we will discuss another fundamental application: solving systems of linear differential equations. Cayley-Hamilton, I First, we establish the Cayley-Hamilton theorem for arbitrary matrices: Theorem (Cayley-Hamilton) If p(x) is the characteristic polynomial of a matrix A, then p(A) is the zero matrix 0. The same result holds for the characteristic polynomial of a linear operator T : V ! V on a finite-dimensional vector space. Cayley-Hamilton, II Proof: Since the characteristic polynomial of a matrix does not depend on the underlying field of coefficients, we may assume that the characteristic polynomial factors completely over the field (i.e., that all of the eigenvalues of A lie in the field) by replacing the field with its algebraic closure. Then by our results, the Jordan canonical form of A exists.
    [Show full text]
  • MATH 2370, Practice Problems
    MATH 2370, Practice Problems Kiumars Kaveh Problem: Prove that an n × n complex matrix A is diagonalizable if and only if there is a basis consisting of eigenvectors of A. Problem: Let A : V ! W be a one-to-one linear map between two finite dimensional vector spaces V and W . Show that the dual map A0 : W 0 ! V 0 is surjective. Problem: Determine if the curve 2 2 2 f(x; y) 2 R j x + y + xy = 10g is an ellipse or hyperbola or union of two lines. Problem: Show that if a nilpotent matrix is diagonalizable then it is the zero matrix. Problem: Let P be a permutation matrix. Show that P is diagonalizable. Show that if λ is an eigenvalue of P then for some integer m > 0 we have λm = 1 (i.e. λ is an m-th root of unity). Hint: Note that P m = I for some integer m > 0. Problem: Show that if λ is an eigenvector of an orthogonal matrix A then jλj = 1. n Problem: Take a vector v 2 R and let H be the hyperplane orthogonal n n to v. Let R : R ! R be the reflection with respect to a hyperplane H. Prove that R is a diagonalizable linear map. Problem: Prove that if λ1; λ2 are distinct eigenvalues of a complex matrix A then the intersection of the generalized eigenspaces Eλ1 and Eλ2 is zero (this is part of the Spectral Theorem). 1 Problem: Let H = (hij) be a 2 × 2 Hermitian matrix. Use the Min- imax Principle to show that if λ1 ≤ λ2 are the eigenvalues of H then λ1 ≤ h11 ≤ λ2.
    [Show full text]
  • Math 223 Symmetric and Hermitian Matrices. Richard Anstee an N × N Matrix Q Is Orthogonal If QT = Q−1
    Math 223 Symmetric and Hermitian Matrices. Richard Anstee An n × n matrix Q is orthogonal if QT = Q−1. The columns of Q would form an orthonormal basis for Rn. The rows would also form an orthonormal basis for Rn. A matrix A is symmetric if AT = A. Theorem 0.1 Let A be a symmetric n × n matrix of real entries. Then there is an orthogonal matrix Q and a diagonal matrix D so that AQ = QD; i.e. QT AQ = D: Note that the entries of M and D are real. There are various consequences to this result: A symmetric matrix A is diagonalizable A symmetric matrix A has an othonormal basis of eigenvectors. A symmetric matrix A has real eigenvalues. Proof: The proof begins with an appeal to the fundamental theorem of algebra applied to det(A − λI) which asserts that the polynomial factors into linear factors and one of which yields an eigenvalue λ which may not be real. Our second step it to show λ is real. Let x be an eigenvector for λ so that Ax = λx. Again, if λ is not real we must allow for the possibility that x is not a real vector. Let xH = xT denote the conjugate transpose. It applies to matrices as AH = AT . Now xH x ≥ 0 with xH x = 0 if and only if x = 0. We compute xH Ax = xH (λx) = λxH x. Now taking complex conjugates and transpose (xH Ax)H = xH AH x using that (xH )H = x. Then (xH Ax)H = xH Ax = λxH x using AH = A.
    [Show full text]
  • Arxiv:1901.01378V2 [Math-Ph] 8 Apr 2020 Where A(P, Q) Is the Arithmetic Mean of the Vectors P and Q, G(P, Q) Is Their P Geometric Mean, and Tr X Stands for Xi
    MATRIX VERSIONS OF THE HELLINGER DISTANCE RAJENDRA BHATIA, STEPHANE GAUBERT, AND TANVI JAIN Abstract. On the space of positive definite matrices we consider dis- tance functions of the form d(A; B) = [trA(A; B) − trG(A; B)]1=2 ; where A(A; B) is the arithmetic mean and G(A; B) is one of the different versions of the geometric mean. When G(A; B) = A1=2B1=2 this distance is kA1=2− 1=2 1=2 1=2 1=2 B k2; and when G(A; B) = (A BA ) it is the Bures-Wasserstein metric. We study two other cases: G(A; B) = A1=2(A−1=2BA−1=2)1=2A1=2; log A+log B the Pusz-Woronowicz geometric mean, and G(A; B) = exp 2 ; the log Euclidean mean. With these choices d(A; B) is no longer a metric, but it turns out that d2(A; B) is a divergence. We establish some (strict) convexity properties of these divergences. We obtain characterisations of barycentres of m positive definite matrices with respect to these distance measures. 1. Introduction Let p and q be two discrete probability distributions; i.e. p = (p1; : : : ; pn) and q = (q1; : : : ; qn) are n -vectors with nonnegative coordinates such that P P pi = qi = 1: The Hellinger distance between p and q is the Euclidean norm of the difference between the square roots of p and q ; i.e. 1=2 1=2 p p hX p p 2i hX X p i d(p; q) = k p− qk2 = ( pi − qi) = (pi + qi) − 2 piqi : (1) This distance and its continuous version, are much used in statistics, where it is customary to take d (p; q) = p1 d(p; q) as the definition of the Hellinger H 2 distance.
    [Show full text]
  • MATRICES WHOSE HERMITIAN PART IS POSITIVE DEFINITE Thesis by Charles Royal Johnson in Partial Fulfillment of the Requirements Fo
    MATRICES WHOSE HERMITIAN PART IS POSITIVE DEFINITE Thesis by Charles Royal Johnson In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1972 (Submitted March 31, 1972) ii ACKNOWLEDGMENTS I am most thankful to my adviser Professor Olga Taus sky Todd for the inspiration she gave me during my graduate career as well as the painstaking time and effort she lent to this thesis. I am also particularly grateful to Professor Charles De Prima and Professor Ky Fan for the helpful discussions of my work which I had with them at various times. For their financial support of my graduate tenure I wish to thank the National Science Foundation and Ford Foundation as well as the California Institute of Technology. It has been important to me that Caltech has been a most pleasant place to work. I have enjoyed living with the men of Fleming House for two years, and in the Department of Mathematics the faculty members have always been generous with their time and the secretaries pleasant to work around. iii ABSTRACT We are concerned with the class Iln of n><n complex matrices A for which the Hermitian part H(A) = A2A * is positive definite. Various connections are established with other classes such as the stable, D-stable and dominant diagonal matrices. For instance it is proved that if there exist positive diagonal matrices D, E such that DAE is either row dominant or column dominant and has positive diag­ onal entries, then there is a positive diagonal F such that FA E Iln.
    [Show full text]
  • 216 Section 6.1 Chapter 6 Hermitian, Orthogonal, And
    216 SECTION 6.1 CHAPTER 6 HERMITIAN, ORTHOGONAL, AND UNITARY OPERATORS In Chapter 4, we saw advantages in using bases consisting of eigenvectors of linear opera- tors in a number of applications. Chapter 5 illustrated the benefit of orthonormal bases. Unfortunately, eigenvectors of linear operators are not usually orthogonal, and vectors in an orthonormal basis are not likely to be eigenvectors of any pertinent linear operator. There are operators, however, for which eigenvectors are orthogonal, and hence it is possible to have a basis that is simultaneously orthonormal and consists of eigenvectors. This chapter introduces some of these operators. 6.1 Hermitian Operators § When the basis for an n-dimensional real, inner product space is orthonormal, the inner product of two vectors u and v can be calculated with formula 5.48. If v not only represents a vector, but also denotes its representation as a column matrix, we can write the inner product as the product of two matrices, one a row matrix and the other a column matrix, (u, v) = uT v. If A is an n n real matrix, the inner product of u and the vector Av is × (u, Av) = uT (Av) = (uT A)v = (AT u)T v = (AT u, v). (6.1) This result, (u, Av) = (AT u, v), (6.2) allows us to move the matrix A from the second term to the first term in the inner product, but it must be replaced by its transpose AT . A similar result can be derived for complex, inner product spaces. When A is a complex matrix, we can use equation 5.50 to write T T T (u, Av) = uT (Av) = (uT A)v = (AT u)T v = A u v = (A u, v).
    [Show full text]
  • Unitary-And-Hermitian-Matrices.Pdf
    11-24-2014 Unitary Matrices and Hermitian Matrices Recall that the conjugate of a complex number a + bi is a bi. The conjugate of a + bi is denoted a + bi or (a + bi)∗. − In this section, I’ll use ( ) for complex conjugation of numbers of matrices. I want to use ( )∗ to denote an operation on matrices, the conjugate transpose. Thus, 3 + 4i = 3 4i, 5 6i =5+6i, 7i = 7i, 10 = 10. − − − Complex conjugation satisfies the following properties: (a) If z C, then z = z if and only if z is a real number. ∈ (b) If z1, z2 C, then ∈ z1 + z2 = z1 + z2. (c) If z1, z2 C, then ∈ z1 z2 = z1 z2. · · The proofs are easy; just write out the complex numbers (e.g. z1 = a+bi and z2 = c+di) and compute. The conjugate of a matrix A is the matrix A obtained by conjugating each element: That is, (A)ij = Aij. You can check that if A and B are matrices and k C, then ∈ kA + B = k A + B and AB = A B. · · You can prove these results by looking at individual elements of the matrices and using the properties of conjugation of numbers given above. Definition. If A is a complex matrix, A∗ is the conjugate transpose of A: ∗ A = AT . Note that the conjugation and transposition can be done in either order: That is, AT = (A)T . To see this, consider the (i, j)th element of the matrices: T T T [(A )]ij = (A )ij = Aji =(A)ji = [(A) ]ij. Example. If 1 2i 4 1 + 2i 2 i 3i ∗ − A = − , then A = 2 + i 2 7i .
    [Show full text]
  • RANDOM DOUBLY STOCHASTIC MATRICES: the CIRCULAR LAW 3 Easily Be Bounded by a Polynomial in N
    The Annals of Probability 2014, Vol. 42, No. 3, 1161–1196 DOI: 10.1214/13-AOP877 c Institute of Mathematical Statistics, 2014 RANDOM DOUBLY STOCHASTIC MATRICES: THE CIRCULAR LAW By Hoi H. Nguyen1 Ohio State University Let X be a matrix sampled uniformly from the set of doubly stochastic matrices of size n n. We show that the empirical spectral × distribution of the normalized matrix √n(X EX) converges almost − surely to the circular law. This confirms a conjecture of Chatterjee, Diaconis and Sly. 1. Introduction. Let M be a matrix of size n n, and let λ , . , λ be × 1 n the eigenvalues of M. The empirical spectral distribution (ESD) µM of M is defined as 1 µ := δ . M n λi i n X≤ We also define µcir as the uniform distribution over the unit disk 1 µ (s,t) := mes( z 1; (z) s, (z) t). cir π | |≤ ℜ ≤ ℑ ≤ Given a random n n matrix M, an important problem in random matrix theory is to study the× limiting distribution of the empirical spectral distri- bution as n tends to infinity. We consider one of the simplest random matrix ensembles, when the entries of M are i.i.d. copies of the random variable ξ. When ξ is a standard complex Gaussian random variable, M can be viewed as a random matrix drawn from the probability distribution P(dM)= 1 tr(MM ∗) 2 e− dM on the set of complex n n matrices. This is known as arXiv:1205.0843v3 [math.CO] 27 Mar 2014 πn × the complex Ginibre ensemble.
    [Show full text]
  • PHYSICS 116A Homework 8 Solutions 1. Boas, Problem 3.9–4
    PHYSICS 116A Homework 8 Solutions 1. Boas, problem 3.9–4. Given the matrix, 0 2i 1 − A = i 2 0 , − 3 0 0 find the transpose, the inverse, the complex conjugate and the transpose conjugate of A. Verify that AA−1 = A−1A = I, where I is the identity matrix. We shall evaluate A−1 by employing Eq. (6.13) in Ch. 3 of Boas. First we compute the determinant by expanding in cofactors about the third column: 0 2i 1 − i 2 det A i 2 0 =( 1) − =( 1)( 6) = 6 . ≡ − − 3 0 − − 3 0 0 Next, we evaluate the matrix of cofactors and take the transpose: 2 0 2i 1 2i 1 − − 0 0 − 0 0 2 0 0 0 2 i 0 0 1 0 1 adj A = − − − = 0 3 i . (1) − 3 0 3 0 − i 0 − 6 6i 2 i 2 0 2i 0 2i − − − 3 0 − 3 0 i 2 − According to Eq. (6.13) in Ch. 3 of Boas the inverse of A is given by Eq. (1) divided by det(A), so 1 0 0 3 −1 1 i A = 0 2 6 . (2) 1 i 1 − − 3 It is straightforward to check by matrix multiplication that 0 2i 1 0 0 1 0 0 1 0 2i 1 1 0 0 − 3 3 − i 2 0 0 1 i = 0 1 i i 2 0 = 0 1 0 . − 2 6 2 6 − 3 0 0 1 i 1 1 i 1 3 0 0 0 0 1 − − 3 − − 3 The transpose, complex conjugate and the transpose conjugate∗ can be written down by inspec- tion: 0 i 3 0 2i 1 0 i 3 − − − ∗ † AT = 2i 2 0 , A = i 2 0 , A = 2i 2 0 .
    [Show full text]
  • Supplement: Symmetric and Hermitian Matrices
    Supplement: Symmetric and Hermitian Matrices A Bunch of Definitions Definition: A real n × n matrix A is called symmetric if AT = A. Definition: A complex n × n matrix A is called Hermitian if A∗ = A, where A∗ = AT , the conjugate transpose. Definition: A complex n × n matrix A is called normal if A∗A = AA∗, i.e. commutes with its conjugate transpose. It is quite a surprising result that these three kinds of matrices are always diagonalizable; and moreover, one can construct an orthonormal basis (in standard inner product) for Rn/Cn, consisting of eigenvectors of A. Hence the matrix P that gives diagonalization A = P DP −1 will be orthogonal/unitary, namely: T −1 T Definition: An n × n real matrix P is called orthogonal if P P = In, i.e. P = P . ∗ −1 ∗ Definition: An n × n complex matrix P is called unitary if P P = In, i.e. P = P . Diagonalization using these special kinds of P will have special names: Definition: A matrix A is called orthogonally diagonalizable if A is similar to a diagonal matrix D with an orthogonal matrix P , i.e. A = P DP T . A matrix A is called unitarily diagonalizable if A is similar to a diagonal matrix D with a unitary matrix P , i.e. A = P DP ∗. Then we have the following big theorems: Theorem: Every real n × n symmetric matrix A is orthogonally diagonalizable Theorem: Every complex n × n Hermitian matrix A is unitarily diagonalizable. Theorem: Every complex n × n normal matrix A is unitarily diagonalizable. To prove the above results, it is convenient to introduce the concept of adjoint operator, which allows us to discuss effectively the \transpose" operation in a general inner product space.
    [Show full text]
  • Symmetric and Hermitian Matrices
    “notes2” i i 2013/4/9 page 73 i i Chapter 5 Symmetric and Hermitian Matrices In this chapter, we discuss the special classes of symmetric and Hermitian matrices. We will conclude the chapter with a few words about so-called Normal matrices. Before we begin, we mention one consequence of the last chapter that will be useful in a proof of the unitary diagonalization of Hermitian matrices. Let A be an m n matrix with m n, and assume (for the moment) that A has linearly independent× columns. Then≥ if the Gram-Schmidt process is applied to the columns of A, the result can be expressed in terms of a matrix factorization A = Q˜ R˜, where the orthogonal vectors are the columns in Q˜, and R˜ is unit upper triangular n n with the inner products as the entries in R˜. H × i 1 qk ai For example, consider that qi := ai k−=1 H qk. Rearranging the equa- − qk qk tion, P i 1 H n − qk ai ai = q + q + 0q i qH q k k kX=1 k k k=Xi+1 where the right hand side is a linear combination of the columns of the first i (only!) columns of the matrix Q˜. So this equation can be rewritten H q1 ai H q1 q1 H q2 ai H q2 q2 . H qi−1ai ˜ H ai = Q qi−1qi−1 . 1 0 . . 0 Putting such equations together for i = 1 ,...,n , we arrive at the desired result, A = Q˜ R˜. 73 “notes2” i i 2013/4/9 page 74 i i 74 Chapter5.
    [Show full text]
  • Matrix Nearness Problems and Applications∗
    Matrix Nearness Problems and Applications∗ Nicholas J. Higham† Abstract A matrix nearness problem consists of finding, for an arbitrary matrix A, a nearest member of some given class of matrices, where distance is measured in a matrix norm. A survey of nearness problems is given, with particular emphasis on the fundamental properties of symmetry, positive definiteness, orthogonality, normality, rank-deficiency and instability. Theoretical results and computational methods are described. Applications of nearness problems in areas including control theory, numerical analysis and statistics are outlined. Key words. matrix nearness problem, matrix approximation, symmetry, positive definiteness, orthogonality, normality, rank-deficiency, instability. AMS subject classifications. Primary 65F30, 15A57. 1 Introduction Consider the distance function (1.1) d(A) = min E : A + E S has property P ,A S, k k ∈ ∈ where S denotes Cm×n or Rm×n, is a matrix norm on S, and P is a matrix property k · k which defines a subspace or compact subset of S (so that d(A) is well-defined). Associated with (1.1) are the following tasks, which we describe collectively as a matrix nearness problem: Determine an explicit formula for d(A), or a useful characterisation. • ∗This is a reprint of the paper: N. J. Higham. Matrix nearness problems and applications. In M. J. C. Gover and S. Barnett, editors, Applications of Matrix Theory, pages 1–27. Oxford University Press, 1989. †Department of Mathematics, University of Manchester, Manchester, M13 9PL, England ([email protected]). 1 Determine X = A + E , where E is a matrix for which the minimum in (1.1) • min min is attained.
    [Show full text]