DOCSLIB.ORG

Feyncalc Tutorial 2 (Dated: November 7, 2016)

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Feyncalc Tutorial 2 (Dated: November 7, 2016) FeynCalc Tutorial 2 (Dated: November 7, 2016) Last time we learned how to do Lorentz contractions with FeynCalc. We also did a simple calculation in scalar QED: two scalars annihilating into two photons at leading order (LO). In this tutorial we will first learn how to implement calculation with the spinors and gamma matrices. Then we are going to calculate the famous Compton scattering at LO. Let's first check whether you can load FeynCalc properly. Run the following command in your Mathematica notebook. If you get the similar output, you can start from section. I. If your FeynCalc cannot be loaded properly, we have a version for the class not requir- ing the installation, just ask. I. Review of the last tutorial Let's first test to see how much you have remembered from the last tutorial. Below are the three examples we had last time. Discuss with your teammates and make sure you understand the meaning of the FeynCalc functions and the results. µ ν 2 2 • Example 1: gµνp p with p = m . 1 µ 2 2 • Example 2: (p − k) (p − k)µ − (p − k) − p αβµν • Example 3: gαβ II. Spinors and Gamma matrices A. Notations The procedure is the same as last tutorial. We first need to convert the spinor expres- sions into something FeynCalc can understand, then use some FeynCalc functions to do the necessary simplifications. The conversion rules for spinor calculations are summarized in the following table. symbol in FeynCalc Spinor u with mass m u(p; m),u ¯(p; m) Spinor[p,m] Spinor v with mass m v(p; m),v ¯(p; m) Spinor[-p,m] µ Gamma Matrix γ , γµ GA[µ] 5 Gamma 5 γ , γ5 GA[5] or GA5 Dirac Slash p= = γ · p GS[p] Notice that there is an extra minus sign for spinor u and spinor v. It seems confusing that theu ¯ and u share the same notation in FeynCalc. Here FeynCalc uses a trick: since in reality every Fermion chain starts with a barred spinor and ends with a unbarred spinor, FeynCalc interprets the Spinor[p,m] at the beginning of a Fermion chain asu ¯, while interpreting the 2 Spinor[p,m] at the end of a Fermion chain as u. The case forv ¯ and v is the same. In summary, whether a Spinor is barred or unbarred depends on its position in a Fermion chain. Now let's do a simple example, µ ν • Example 4:u ¯(p2; m)γ (k= + m)γ v(p1; m). Since the γ-matrices are anti-commutative, the order of them is important. FeynCalc uses the dot multiplication \." to keep the order. As a test, you can remove all the \." in the input above, and see what the output looks like. Missing dots in Fermion chains is a very common mistake in real calculation, so please be very careful and double check the input every time you have a Fermion chain. B. DiracSimplify DiracSimplify[expr] simplifies products of γ-matrices in expr and expands non- commutative products. Repeated Lorentz indices are contracted. Here are some examples. µ • Example 5: simplify γ γµ. µ • Example 6: simplify γ (k= + m)γµ. • Example 7: simplify kγ= µk=. µ Check whether these results are the same as your expectation. Now try to simplify γ k==qγµ µ and γ k=p==qγµ and compare the results with your classmates. DiracSimplify[expr] also applies the Dirac equation. for example 3 • Example 8: simplifyu ¯(p2; m)(p=2 − p=1 + m) u(p1; m). The result is proportional to m! (Why?) C. Tr The trace of γ-matrices can be easily calculated with Tr. For example • Example 9: calculate Tr[γαγµγβγν]. • Example 10: calculate Tr[γαγµγβγνγ5]. Do these two outputs make sense to you? You can try the trace with more γ-matrices • Example 11: calculate Tr[γαγµγβγνγλ] and Tr[γαγµγβγνγλγ5]. Both of them equal to zero! (Why?) You are encouraged to calculate the trace with 6 and 8 γ-matrices. Count how many terms are in the outputs. III. Project: Compton scattering Compton scattering is the process e− + γ ! e− + γ. At LO, there are two Feynman diagrams, 4 p1 k1 p1 k1 k2 p2 k2 p2 ( a ) ( b ) The cross section can be expressed as 1 Z 1 X σ = dΠ jMj2; (1) 2s 2 4 e−,γ spin 4 (4) where dΠ2 is the two-body phase space including (2π) δ (p1 + p2 − k1 − k2), the factor P (1=4) spin averages (sums) the spin of the initial (final) state electron and photon. The matrix element can be written as 1 X 2 1 X µν µν µ0ν0 µ0ν0 ∗ ∗ ∗ jMj = (M + M ) × (M + M ) × (p ) 0 (p ) (k ) 0 (k ) 4 4 a b a b µ 2 µ 2 ν 2 ν 2 e−,γ spin e−,γ spin 1 X µν µν µ0ν0 µ0ν0 ∗ = (M + M ) × (M + M ) × (−g 0 )(−g 0 ); 4 a b a b µµ νν e−spin (2) where Ma and Mb are for the two Feynman diagrams, respectively. In the second step of Eq. (2), we have summed over the polarization of the two photons. There are four terms in the sum in Eq. (2). Next I will show how to use FeynCalc to calculate one of them: 1 X µν µ0ν0 ∗ (M ) × (M ) × (−g 0 )(−g 0 ): 4 a a µµ νν (3) e−spin You need to add the contribution of the diagram (b) to get the full result in Eq. (2). From the Feynman rule of QED, we have µν ν i(p=1 + p=2 + m) µ M =u ¯ (k )(−ieγ ) (−ieγ )u 0 (p ) a s 1 (p + p )2 − m2 s 1 1 2 (4) 2 1 ν µ 0 = (−ie ) 2 2 × u¯s(k1)γ (p=1 + p=2 + m)γ us (p1); (p1 + p2) − m where s and s0 are the spins of the final and initial electrons, respectively. Now let's pass the calculation to FeynCalc. First we need to define the scalar products 5 where m is the electron mass. Next we type Eq. (4) in Mathematica and calculate its complex conjugate Then we can use FeynCalc function FermionSpinSum to sum over the spins of the initial and final electrons. Notice that the output of FermionSpinSum is a trace (why?). Although it appears as \tr" in the output, the internal function is \DiracTrace", as one can check by looking at its standard form. 6 We can do the trace by changing DiracTrace to Tr, which is explained in subsection II C. 2 where we have used momentum conservation k2 = p1 + p2 − k1 and identity s + t + u = 2m . The factor 1=4 is the factor in Eq. (3). So far, we have calculated the result for Eq. (3). It is your turn to write down the Feynman expression of diagram (b) and get the full result in Eq. (2). Compare your result with Eq. (13.119) on Matthew Schwartz's book to see whether the two results agree. (Hint: you need to replace p12 and p24 in Eq. (13.119) into SP[p1,p2] and SP[p1,k2] to compare the two results.) 7.
Load more

Recommended publications
  • Estimations of the Trace of Powers of Positive Self-Adjoint Operators by Extrapolation of the Moments∗
    Electronic Transactions on Numerical Analysis. ETNA Volume 39, pp. 144-155, 2012. Kent State University Copyright 2012, Kent State University. http://etna.math.kent.edu ISSN 1068-9613. ESTIMATIONS OF THE TRACE OF POWERS OF POSITIVE SELF-ADJOINT OPERATORS BY EXTRAPOLATION OF THE MOMENTS∗ CLAUDE BREZINSKI†, PARASKEVI FIKA‡, AND MARILENA MITROULI‡ Abstract. Let A be a positive self-adjoint linear operator on a real separable Hilbert space H. Our aim is to build estimates of the trace of Aq, for q ∈ R. These estimates are obtained by extrapolation of the moments of A. Applications of the matrix case are discussed, and numerical results are given. Key words. Trace, positive self-adjoint linear operator, symmetric matrix, matrix powers, matrix moments, extrapolation. AMS subject classifications. 65F15, 65F30, 65B05, 65C05, 65J10, 15A18, 15A45. 1. Introduction. Let A be a positive self-adjoint linear operator from H to H, where H is a real separable Hilbert space with inner product denoted by (·, ·). Our aim is to build estimates of the trace of Aq, for q ∈ R. These estimates are obtained by extrapolation of the integer moments (z, Anz) of A, for n ∈ N. A similar procedure was first introduced in [3] for estimating the Euclidean norm of the error when solving a system of linear equations, which corresponds to q = −2. The case q = −1, which leads to estimates of the trace of the inverse of a matrix, was studied in [4]; on this problem, see [10]. Let us mention that, when only positive powers of A are used, the Hilbert space H could be infinite dimensional, while, for negative powers of A, it is always assumed to be a finite dimensional one, and, obviously, A is also assumed to be invertible.
    [Show full text]
  • 5 the Dirac Equation and Spinors
    5 The Dirac Equation and Spinors In this section we develop the appropriate wavefunctions for fundamental fermions and bosons. 5.1 Notation Review The three dimension differential operator is : ∂ ∂ ∂ = , , (5.1) ∂x ∂y ∂z We can generalise this to four dimensions ∂µ: 1 ∂ ∂ ∂ ∂ ∂ = , , , (5.2) µ c ∂t ∂x ∂y ∂z 5.2 The Schr¨odinger Equation First consider a classical non-relativistic particle of mass m in a potential U. The energy-momentum relationship is: p2 E = + U (5.3) 2m we can substitute the differential operators: ∂ Eˆ i pˆ i (5.4) → ∂t →− to obtain the non-relativistic Schr¨odinger Equation (with = 1): ∂ψ 1 i = 2 + U ψ (5.5) ∂t −2m For U = 0, the free particle solutions are: iEt ψ(x, t) e− ψ(x) (5.6) ∝ and the probability density ρ and current j are given by: 2 i ρ = ψ(x) j = ψ∗ ψ ψ ψ∗ (5.7) | | −2m − with conservation of probability giving the continuity equation: ∂ρ + j =0, (5.8) ∂t · Or in Covariant notation: µ µ ∂µj = 0 with j =(ρ,j) (5.9) The Schr¨odinger equation is 1st order in ∂/∂t but second order in ∂/∂x. However, as we are going to be dealing with relativistic particles, space and time should be treated equally. 25 5.3 The Klein-Gordon Equation For a relativistic particle the energy-momentum relationship is: p p = p pµ = E2 p 2 = m2 (5.10) · µ − | | Substituting the equation (5.4), leads to the relativistic Klein-Gordon equation: ∂2 + 2 ψ = m2ψ (5.11) −∂t2 The free particle solutions are plane waves: ip x i(Et p x) ψ e− · = e− − · (5.12) ∝ The Klein-Gordon equation successfully describes spin 0 particles in relativistic quan- tum field theory.
    [Show full text]
  • A Some Basic Rules of Tensor Calculus
    A Some Basic Rules of Tensor Calculus The tensor calculus is a powerful tool for the description of the fundamentals in con- tinuum mechanics and the derivation of the governing equations for applied prob- lems. In general, there are two possibilities for the representation of the tensors and the tensorial equations: – the direct (symbolic) notation and – the index (component) notation The direct notation operates with scalars, vectors and tensors as physical objects defined in the three dimensional space. A vector (first rank tensor) a is considered as a directed line segment rather than a triple of numbers (coordinates). A second rank tensor A is any finite sum of ordered vector pairs A = a b + ... +c d. The scalars, vectors and tensors are handled as invariant (independent⊗ from the choice⊗ of the coordinate system) objects. This is the reason for the use of the direct notation in the modern literature of mechanics and rheology, e.g. [29, 32, 49, 123, 131, 199, 246, 313, 334] among others. The index notation deals with components or coordinates of vectors and tensors. For a selected basis, e.g. gi, i = 1, 2, 3 one can write a = aig , A = aibj + ... + cidj g g i i ⊗ j Here the Einstein’s summation convention is used: in one expression the twice re- peated indices are summed up from 1 to 3, e.g. 3 3 k k ik ik a gk ∑ a gk, A bk ∑ A bk ≡ k=1 ≡ k=1 In the above examples k is a so-called dummy index. Within the index notation the basic operations with tensors are defined with respect to their coordinates, e.
    [Show full text]
  • 18.700 JORDAN NORMAL FORM NOTES These Are Some Supplementary Notes on How to Find the Jordan Normal Form of a Small Matrix. Firs
    18.700 JORDAN NORMAL FORM NOTES These are some supplementary notes on how to find the Jordan normal form of a small matrix. First we recall some of the facts from lecture, next we give the general algorithm for finding the Jordan normal form of a linear operator, and then we will see how this works for small matrices. 1. Facts Throughout we will work over the field C of complex numbers, but if you like you may replace this with any other algebraically closed field. Suppose that V is a C-vector space of dimension n and suppose that T : V → V is a C-linear operator. Then the characteristic polynomial of T factors into a product of linear terms, and the irreducible factorization has the form m1 m2 mr cT (X) = (X − λ1) (X − λ2) ... (X − λr) , (1) for some distinct numbers λ1, . , λr ∈ C and with each mi an integer m1 ≥ 1 such that m1 + ··· + mr = n. Recall that for each eigenvalue λi, the eigenspace Eλi is the kernel of T − λiIV . We generalized this by defining for each integer k = 1, 2,... the vector subspace k k E(X−λi) = ker(T − λiIV ) . (2) It is clear that we have inclusions 2 e Eλi = EX−λi ⊂ E(X−λi) ⊂ · · · ⊂ E(X−λi) ⊂ .... (3) k k+1 Since dim(V ) = n, it cannot happen that each dim(E(X−λi) ) < dim(E(X−λi) ), for each e e +1 k = 1, . , n. Therefore there is some least integer ei ≤ n such that E(X−λi) i = E(X−λi) i .
    [Show full text]
  • Trace Inequalities for Matrices
    Bull. Aust. Math. Soc. 87 (2013), 139–148 doi:10.1017/S0004972712000627 TRACE INEQUALITIES FOR MATRICES KHALID SHEBRAWI ˛ and HUSSIEN ALBADAWI (Received 23 February 2012; accepted 20 June 2012) Abstract Trace inequalities for sums and products of matrices are presented. Relations between the given inequalities and earlier results are discussed. Among other inequalities, it is shown that if A and B are positive semidefinite matrices then tr(AB)k ≤ minfkAkk tr Bk; kBkk tr Akg for any positive integer k. 2010 Mathematics subject classification: primary 15A18; secondary 15A42, 15A45. Keywords and phrases: trace inequalities, eigenvalues, singular values. 1. Introduction Let Mn(C) be the algebra of all n × n matrices over the complex number field. The singular values of A 2 Mn(C), denoted by s1(A); s2(A);:::; sn(A), are the eigenvalues ∗ 1=2 of the matrix jAj = (A A) arranged in such a way that s1(A) ≥ s2(A) ≥ · · · ≥ sn(A). 2 ∗ ∗ Note that si (A) = λi(A A) = λi(AA ), so for a positive semidefinite matrix A, we have si(A) = λi(A)(i = 1; 2;:::; n). The trace functional of A 2 Mn(C), denoted by tr A or tr(A), is defined to be the sum of the entries on the main diagonal of A and it is well known that the trace of a Pn matrix A is equal to the sum of its eigenvalues, that is, tr A = j=1 λ j(A). Two principal properties of the trace are that it is a linear functional and, for A; B 2 Mn(C), we have tr(AB) = tr(BA).
    [Show full text]
  • Lecture 28: Eigenvalues Allowing Complex Eigenvalues Is Really a Blessing
    Math 19b: Linear Algebra with Probability Oliver Knill, Spring 2011 cos(t) sin(t) 2 2 For a rotation A = the characteristic polynomial is λ − 2 cos(α)+1 − sin(t) cos(t) which has the roots cos(α) ± i sin(α)= eiα. Lecture 28: Eigenvalues Allowing complex eigenvalues is really a blessing. The structure is very simple: Fundamental theorem of algebra: For a n × n matrix A, the characteristic We have seen that det(A) = 0 if and only if A is invertible. polynomial has exactly n roots. There are therefore exactly n eigenvalues of A if we count them with multiplicity. The polynomial fA(λ) = det(A − λIn) is called the characteristic polynomial of 1 n n−1 A. Proof One only has to show a polynomial p(z)= z + an−1z + ··· + a1z + a0 always has a root z0 We can then factor out p(z)=(z − z0)g(z) where g(z) is a polynomial of degree (n − 1) and The eigenvalues of A are the roots of the characteristic polynomial. use induction in n. Assume now that in contrary the polynomial p has no root. Cauchy’s integral theorem then tells dz 2πi Proof. If Av = λv,then v is in the kernel of A − λIn. Consequently, A − λIn is not invertible and = =0 . (1) z =r | | | zp(z) p(0) det(A − λIn)=0 . On the other hand, for all r, 2 1 dz 1 2π 1 For the matrix A = , the characteristic polynomial is | | ≤ 2πrmax|z|=r = . (2) z =r 4 −1 | | | zp(z) |zp(z)| min|z|=rp(z) The right hand side goes to 0 for r →∞ because 2 − λ 1 2 det(A − λI2) = det( )= λ − λ − 6 .
    [Show full text]
  • Approximating Spectral Sums of Large-Scale Matrices Using Stochastic Chebyshev Approximations∗
    Approximating Spectral Sums of Large-scale Matrices using Stochastic Chebyshev Approximations∗ Insu Han y Dmitry Malioutov z Haim Avron x Jinwoo Shin { March 10, 2017 Abstract Computation of the trace of a matrix function plays an important role in many scientific com- puting applications, including applications in machine learning, computational physics (e.g., lat- tice quantum chromodynamics), network analysis and computational biology (e.g., protein fold- ing), just to name a few application areas. We propose a linear-time randomized algorithm for approximating the trace of matrix functions of large symmetric matrices. Our algorithm is based on coupling function approximation using Chebyshev interpolation with stochastic trace estima- tors (Hutchinson’s method), and as such requires only implicit access to the matrix, in the form of a function that maps a vector to the product of the matrix and the vector. We provide rigorous approximation error in terms of the extremal eigenvalue of the input matrix, and the Bernstein ellipse that corresponds to the function at hand. Based on our general scheme, we provide algo- rithms with provable guarantees for important matrix computations, including log-determinant, trace of matrix inverse, Estrada index, Schatten p-norm, and testing positive definiteness. We experimentally evaluate our algorithm and demonstrate its effectiveness on matrices with tens of millions dimensions. 1 Introduction Given a symmetric matrix A 2 Rd×d and function f : R ! R, we study how to efficiently compute d X Σf (A) = tr(f(A)) = f(λi); (1) i=1 arXiv:1606.00942v2 [cs.DS] 9 Mar 2017 where λ1; : : : ; λd are eigenvalues of A.
    [Show full text]
  • Math 217: True False Practice Professor Karen Smith 1. a Square Matrix Is Invertible If and Only If Zero Is Not an Eigenvalue. Solution Note: True
    (c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 217: True False Practice Professor Karen Smith 1. A square matrix is invertible if and only if zero is not an eigenvalue. Solution note: True. Zero is an eigenvalue means that there is a non-zero element in the kernel. For a square matrix, being invertible is the same as having kernel zero. 2. If A and B are 2 × 2 matrices, both with eigenvalue 5, then AB also has eigenvalue 5. Solution note: False. This is silly. Let A = B = 5I2. Then the eigenvalues of AB are 25. 3. If A and B are 2 × 2 matrices, both with eigenvalue 5, then A + B also has eigenvalue 5. Solution note: False. This is silly. Let A = B = 5I2. Then the eigenvalues of A + B are 10. 4. A square matrix has determinant zero if and only if zero is an eigenvalue. Solution note: True. Both conditions are the same as the kernel being non-zero. 5. If B is the B-matrix of some linear transformation V !T V . Then for all ~v 2 V , we have B[~v]B = [T (~v)]B. Solution note: True. This is the definition of B-matrix. 21 2 33 T 6. Suppose 40 2 05 is the matrix of a transformation V ! V with respect to some basis 0 0 1 B = (f1; f2; f3). Then f1 is an eigenvector. Solution note: True. It has eigenvalue 1. The first column of the B-matrix is telling us that T (f1) = f1.
    [Show full text]
  • Sam Roweis' Notes on Matrix Identities
    matrix identities sam roweis (revised June 1999) note that a,b,c and A,B,C do not depend on X,Y,x,y or z 0.1 basic formulae A(B + C) = AB + AC (1a) (A + B)T = AT + BT (1b) (AB)T = BT AT (1c) 1 1 1 if individual inverses exist (AB)− = B− A− (1d) 1 T T 1 (A− ) = (A )− (1e) 0.2 trace, determinant and rank AB = A B (2a) j j j jj j 1 1 A− = (2b) j j A j j A = evals (2c) j j Y Tr [A] = evals (2d) X if the cyclic products are well defined; Tr [ABC :::] = Tr [BC ::: A] = Tr [C ::: AB] = ::: (2e) rank [A] = rank AT A = rank AAT (2f) biggest eval condition number = γ = r (2g) smallest eval derivatives of scalar forms with respect to scalars, vectors, or matricies are indexed in the obvious way. similarly, the indexing for derivatives of vectors and matrices with respect to scalars is straightforward. 1 0.3 derivatives of traces @Tr [X] = I (3a) @X @Tr [XA] @Tr [AX] = = AT (3b) @X @X @Tr XT A @Tr AXT = = A (3c) @X @X @Tr XT AX = (A + AT )X (3d) @X 1 @Tr X− A 1 T 1 = X− A X− (3e) @X − 0.4 derivatives of determinants @ AXB 1 T T 1 j j = AXB (X− ) = AXB (X )− (4a) @X j j j j @ ln X j j = (X 1)T = (XT ) 1 (4b) @X − − @ ln X(z) 1 @X j j = Tr X− (4c) @z @z T @ X AX T T 1 T T T 1 j j = X AX (AX(X AX)− + A X(X A X)− ) (4d) @X j j 0.5 derivatives of scalar forms @(aT x) @(xT a) = = a (5a) @x @x @(xT Ax) = (A + AT )x (5b) @x @(aT Xb) = abT (5c) @X @(aT XT b) = baT (5d) @X @(aT Xa) @(aT XT a) = = aaT (5e) @X @X @(aT XT CXb) = CT XabT + CXbaT (5f) @X @ (Xa + b)T C(Xa + b) = (C + CT )(Xa + b)aT (5g) @X 2 the derivative of one vector y with respect to another vector x is a matrix whose (i; j)th element is @y(j)=@x(i).
    [Show full text]
  • Combining Systems: the Tensor Product and Partial Trace
    Appendix A Combining systems: the tensor product and partial trace A.1 Combining two systems The state of a quantum system is a vector in a complex vector space. (Technically, if the dimension of the vector space is infinite, then it is a separable Hilbert space). Here we will always assume that our systems are finite dimensional. We do this because everything we will discuss transfers without change to infinite dimensional systems. Further, when one actually simulates a system on a computer, one must always truncate an infinite dimensional space so that it is finite. Consider two separate quantum systems. Now consider the fact that if we take them to- gether, then we should be able to describe them as one big quantum system. That is, instead of describing them by two separate state-vectors, one for each system, we should be able to represent the states of both of them together using one single state-vector. (This state vector will naturally need to have more elements in it than the separate state-vectors for each of the systems, and we will work out exactly how many below.) In fact, in order to describe a situation in which the two systems have affected each other — by interacting in some way — and have become correlated with each other as a result of this interaction, we will need to go beyond using a separate state vector for each. If each system is in a pure state, described by a state-vector for it alone, then there are no correlations between the two systems.
    [Show full text]
  • Similar Matrices and Jordan Form
    Similar matrices and Jordan form We’ve nearly covered the entire heart of linear algebra – once we’ve finished singular value decompositions we’ll have seen all the most central topics. AT A is positive definite A matrix is positive definite if xT Ax > 0 for all x 6= 0. This is a very important class of matrices; positive definite matrices appear in the form of AT A when computing least squares solutions. In many situations, a rectangular matrix is multiplied by its transpose to get a square matrix. Given a symmetric positive definite matrix A, is its inverse also symmet­ ric and positive definite? Yes, because if the (positive) eigenvalues of A are −1 l1, l2, · · · ld then the eigenvalues 1/l1, 1/l2, · · · 1/ld of A are also positive. If A and B are positive definite, is A + B positive definite? We don’t know much about the eigenvalues of A + B, but we can use the property xT Ax > 0 and xT Bx > 0 to show that xT(A + B)x > 0 for x 6= 0 and so A + B is also positive definite. Now suppose A is a rectangular (m by n) matrix. A is almost certainly not symmetric, but AT A is square and symmetric. Is AT A positive definite? We’d rather not try to find the eigenvalues or the pivots of this matrix, so we ask when xT AT Ax is positive. Simplifying xT AT Ax is just a matter of moving parentheses: xT (AT A)x = (Ax)T (Ax) = jAxj2 ≥ 0.
    [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]