Elements of Linear Algebra: Q&A

Transcription

1 Elements of Linear Algebra: Q&A A matrix is a rectangular array of objects (elements that are numbers, functions, etc.) with its size indicated by the number of rows and columns, i.e., an m n matrix A with m rows and n columns. If A is an m n matrix, A T is an n m matrix. The determinant of a matrix is the absolute value of the sum of the diagonal elements. The determinant is only defined for a square matrix. The determinant of a matrix can be computed using the Laplace expansion where a row or column is expanded in terms of minors and cofactors. An orthogonal matrix is an invertible n n matrix Q with the property Q 1 = Q T. 75

2 Elements of Linear Algebra: Q&A Given a system of m linear equations in n variables x i (i = 1,, n), written as Ax = b, the system is either 1. Consistent, with a unique (one) solution x.. Consistent, with infinitely many possible solutions. 3. Inconsistent with no solutions. If n > m, the system has more unknowns than equations it is underdetermined. If the system is consistent, some of the variables can be chosen arbitrarily and the remaining variables defined in terms of the arbitrary ones. If n < m, the system has more equations than unknowns it is overdetermined. 76

3 Elements of Linear Algebra: Q&A P3.16. Invertible matrix properties Assume that A is an n n invertible matrix. Which statements are true? a. The system Ax = b has a unique solution for every vector b in R n. b. The rows (and columns) of A are linearly independent. c. det(a) = 0. d. A can be reduced (by elementary operations) to the identity matrix. e. The rank of A is n. f. The rows of A span R n. 77

4 Elements of Linear Algebra: Q&A P3.18. Linear Independence Consider the equations of combustion in which a mixture of CO, H, and CH 4 are burned with O to form CO, CO and H O. 1 CO + O = CO. 1 H + O = HO. CH 4 + O = CO + HO. 3 CH 4 + O = CO + HO. Treating the compounds as real variables, determine if the equations are independent. If not, write the dependent equation(s) in terms of the independent ones. 78

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7 Elements of Linear Algebra Eigenvalues & Eigenvectors As an engineer, you have undoubtedly been introduced to eigenvalues and possibly eigenvectors. We develop background here and will later make use of eigenvalues/vectors in the discussion of second and higherorder tensors. Given the linear equation A x= b the vector x is called the eigenvector (characteristic vector) and the scalar λ is the eigenvalue (characteristic value) of matrix A that characterizes the length (and sense) of the eigenvector x. The spectrum of A is the set of eigenvalues of A and the spectral radius of A is the absolute value of the largest eigenvalue. 81

8 Elements of Linear Algebra Example: Find eigenvalues and eigenvectors of Solution: 1. Compute roots of the characteristic polynomial 3 λ 0 0 D( λ) = 5 4 λ 0 = (3 λ)(4 λ)(1 λ) = λ roots: λ 1 = 3, λ = 4, λ 3 =1. 8

9 Elements of Linear Algebra These roots are the eigenvalues. They form the spectrum with a spectral radius of 4.. Compute the eigenvectors: λ 1 = 3 0= 0 1 5x1+ x = 0 set x1 1 5 or 10 = x = 3x1+ 6x x3 = 0 7 / 7 λ = 4 x1 = 0 0 5x1 = 0 set x 1 1 = x = 3x1+ 6x 3x3 = 0 83

10 Elements of Linear Algebra λ 3 = 1 x1 = 0 0 5x1+ 3x = 0 set x3 1 0 = x = 3x1+ 6x = 0 1 Properties of eigenvalues and eigenvectors of an n n square matrix A: 1. A has at least one eigenvalue and at most n numerically different eigenvalues, but it may have fewer than n.. If x is an eigenvector of a matrix A corresponding to an eigenvalue λ, so is kx, for any k 0, i.e., Ax = λx implies k(ax) = A(kx) = λ(kx). 84

11 Elements of Linear Algebra 3. M λ is the algebraic multiplicity, the number of times the root λ of the characteristic polynomial is repeated, and m λ is the geometric multiplicity, the number of independent eigenvectors corresponding to λ. According to property 1 above, the sum of algebraic multiplicities equals n and in general m λ M λ. 4. A real matrix may have complex eigenvalues that occur in conjugate pairs and complex eigenvectors. 5. The eigenvalues of a symmetric matrix (A T = A) are real. 6. The eigenvalues of a skew symmetric matrix (A T = A) are pure imaginary or zero. 85

12 Elements of Linear Algebra Eigenvectors & Diagonalization Similar matrices have the same spectrum (i.e., same eigenvalues), ˆ 1 A = T AT : n n matrix Â, is similar to A for some n n matrix T. This is an important property, particularly for numerical analysis, to diagonalize (or nearly diagonalize) matrices for computing approximations to eigenvalues and eigenvectors. The eigenvectors corresponding to a set of distinct eigenvalues form a linearly independent set. Thus, these eigenvectors form a basis. 1 If an n n matrix A has a basis of eigenvectors, then D= X AX is diagonal with eigenvalues of A as the entries on the main diagonal. 86

13 The vector algebra included operations involving sums and products of vectors. The definitions and operations defined in the linear algebra provide the basis for linear transformations and matrix operations useful in tensor analysis. The vector calculus allows us to apply the methods of differential and integral calculus in the general tensor analysis. We begin with the usual basic definitions and operations. Derivative of a Vector Function of a Scalar a(t) eˆ s a s a(t + t) da a( t+ t) a( t) = lim dt t 0 t s = a da a s ds = lim = eˆ dt t 0 s t dt s 87

14 Product Rules d da db ( ab ) = b+ a dt dt dt d da db ( a b) = b+ a (order preserved) dt dt dt Note that because a vector is composed of two distinct parts, magnitude and direction, a nonzero derivative could result from: a) a change in magnitude but not direction, b) a change in direction but not magnitude, or c) a change in both magnitude and direction as illustrated in the previous diagram. 88

15 For case b), a constant length vector, a = const da = 0 a dt = const d da da a ( aa = a ) a = a = 0 dt dt dt a = const da a a dt const a In general coordinates, the base vectors are not necessarily constant in magnitude or direction, i i da i dei a= aei ei + a dt dt By definition, the base vectors of Cartesian systems have constant magnitude and direction de / dt = 0. i 89

16 Example: Compute the acceleration of a body in a circular orbit. v(t) r(t) v(t + t) r(t + t) ω = ωeˆ r = reˆ r v = ω r = veˆ z t v(t) v v(t + t) 90

17 dv d dω dr a = = ( ω r) = r+ ω dt dt dt dt vdeˆ z = r + ω v = ω v r dt v v = eˆ ( ) [ ˆ ( ˆ ˆ z ω r = ez ez er)] r r v = [( eˆ ˆ ) ˆ ( ˆ ˆ ) ˆ z er ez ez ez er)] r v = eˆ r r 91

18 Example: Prove d d d d d a = dt dt dt dt dt 3 a a a a a 3 Solution: 0 0 d d d d d d d d d d a = + + dt dt dt dt dt dt dt dt dt dt 3 a a a a a a a a a a 3 3 da d a = a Q.E.D. 3 dt dt 9

19 Cartesian Coordinate Systems A general Cartesian coordinate system is oblique, i.e., the basis vectors are generally not all mutually orthogonal. As stated earlier, however, the basis vectors of a Cartesian system are constant in magnitude and direction. The usual convention is to refer to the familiar orthonormal Cartesian system as the Cartesian system, with basis vectors usually denoted as { ˆˆ i,, j kˆ }, { eˆ, eˆ, eˆ }, { ˆi } x y z i 93

20 z ˆk r xyz = x x x = x x x 1 3 (,, ) (,, ) ( 1,, 3) r = ˆi x j j î ĵ y x 94

21 In any coordinate system, the differential distance between two points is given by the differential arclength, computed from dr dr. In particular, for the Cartesian system, z i i dr dr = ( ds) = dx dx = ( dx) + ( dy) + ( dz) ds dy dz dx y x 95

22 Curvilinear Coordinates Define a coordinate system 1 3 ( q, q, q ) with the coordinate transformation from the Cartesian system, q = q ( x, x, x ) 1 3 q = q ( x, x, x ) q = q ( x, x, x ) 96

23 3 x r q 3 3 q 1 x x q = const r q 1 = const r q 1 q r 1 q q 3 = const q 97

24 If the transformation is linear, it defines a Cartesian system. If the transformation is nonlinear, it defines a curvilinear system. The Jacobian of the transformation is defined by the following determinant, J x x x x x x q q q q q q = = = j x x x x 1 x x 3 x i q 1 q q 3 q q q q x x x x x x q q q q q q

25 If J 0, then J -1 (inverse Jacobian) is defined and the inverse transformation is also defined, x = x ( q, q, q ) 1 3 x = x ( q, q, q ) x = x ( q, q, q ) The position arrow is r i = r( q ), i = 1,,3 and a differential displacement is then r 1 r r 3 r dr = dq + dq + dq = dq 1 3 i q q q q i 99

26 i The vectors r / q are tangent to the coordinate curves defined by the intersection of the coordinate surfaces (q i = const). Using these vectors, we define a unitary basis, r ei =, i = 1,,3 i q Note, in general, the orientation and magnitude of the basis vectors are not constant, e.g., 100

27 Oblique-Cartesian system: Basis vectors have constant magnitude and orientation Curvilinear system: Basis vectors generally have non-constant magnitude and orientation 101

28 The coordinate transformation was written for a general system in terms of the original Cartesian system. We almost always write the transformations in this manner. In terms of the original Cartesian system, the unitary basis is given by, e i j r x = = ˆ i i i j, i = 1,,3 q q This is a linear system that is easily written in matrix format. The coefficient matrix is the Jacobian matrix, ˆ e e = e x / q x / q x / q ˆ x / q x / q x / q i i ˆ 3 x / q x / q x / q i 3 10

29 Fundamental Metric Tensor In a unitary system, the square of the differential distance separating two infinitesimally spaced points is, i j dr dr = ( ds) = ( ei ej) dq dq Now define the components of the fundamental metric tensor as, g ij i j Then, e e i j dr dr = ( ds) = gijdq dq 103

30 In matrix format, the fundamental metric tensor is, g g g G g g g = 1 3 g g g Properties of the fundamental metric tensor: 1. Symmetric, i.e., e e = e e g = g i j j i ij ji. The norm (magnitude) of the unitary base vectors is, 1/ 1/ ei = ( ei ei) = ( gii) (no summation) 104

31 3. Describes the curvature of the space, a) A flat space has no curvature and is called Euclidean. In this case, all the g ij components are constant. b) A curved space is called Riemannian In this case, the g ij components are not constant. An example is Lobachevskian space. This space has hyperbolic curvature. We can compare these two spaces by Euclidean looking at the geometry of a triangle in each. For Euclidean geometry, we know the sum of the interior angles of a triangle is γ always 180. α β α + β + γ =

32 In Lobachevskian geometry, that sum is always less than 180, the difference being proportional to the area of the triangle. (Penrose, Roger, The Emperor s New Mind, p. 156). Lobachevskian α + β + γ 180 = const area α γ β 106

33 Example: Find the unitary basis vectors and components of the fundamental metric tensor for elliptic-cylindrical, coordinates defined by the following inverse transformation (a = constant): x = acosh q cos q, x = asinh q sin q, x = q In terms of the Cartesian basis, the unitary basis is, x e i i i q i = ˆ sinh cos ˆ ˆ 1 i = a q q 1+ acosh q sin q x e i i i q i 1 1 = ˆ cosh sin ˆ ˆ i = a q q 1+ asinh q cos q i x e = ˆi = ˆi q 3 i 3 107

34 Components of the fundamental metric tensor are: 1 1 g = e e = a [sinh ( q )cos ( q ) + cosh ( q )sin ( q ) = g g = 1, g = g = g = g = Components and Bases Recall, i a= ( a e ) e = ( a e ) e j Now set a= e, e = ( e e ) e i j j i i i i With the definition for the components of G, we have, 108

35 g e e ij i j g e e ij i j contravariant component of the fundamental metric. covariant component of the fundamental metric. Then according to the cogredient and contragredient transformation laws raising and lowering of the indices is accomplished with the following, e j = g ij e i and a j = g ij a i e j = g ij e i and a j = g ij a i. Note that when dealing with a unitary basis, cogredient components and vectors are referred to as covariant components and contragredient components and vectors are referred to as contravariant components. 109

36 Also, if we dot both sides of the e j transformation equation, (e j = g ij e i ) e k, then we get the neat result k ik δ (4) j = gg ij For this relation, note the sum over i, e.g., δ δ = g g + g g + g g = 1, g11g g1g g31g = + + = 0. Now with a given unitary basis e i, both sets of fundamental metric components can be generated via, 110

37 Now with a given unitary basis e i, both sets of fundamental metric components can be generated via, g ij = e i e j e i ej e i k ij i j e = g = e e (ijk cyclic) [ eee ]

38 The cross product step is avoided by using the linear transformation (e j = g ij e i ) e k or in matrix notation, e g g g e e = g1 g g 3 e 3 e3 g31 g3 g33 e (5) (6) and a g g g a a = g1 g g 3 a 3 a3 g31 g3 g33 a (7) 11

39 To determine the e j in terms of the e i the matrix equation (6) must be inverted. Let, g g ij det G, M ij = minor of g ij, C ij ( 1) i+j M ij = cofactor of g ij Employing Cramer s rule, e e g g (no summation) e1m 11 em1+ e3m31 eici1 = e g g3 = = e g g g g We obtain similar expressions for e and e 3. In general then C j ij e = ei. g 113

40 Continuing in matrix format, you will probably recognize where this is leading, from the previous section on linear algebra. Since, g ij = g ji, the fundamental metric tensor is symmetric and C ij = C ji, then, j 1 T e = C ij e i e = e i so j 1 [ ] G [ ] g 1 T 1 Cij G. g = We designate the elements of G 1 with superscripts, i.e., G = g. 1 ij 114

41 So what have we accomplished with all this? If G = [g ij ] is known, we can use linear transformations and the rules of linear algebra to determine the dual basis and covariant components without formulae that involve cross products. In fact, knowing what we now know about systems of linear equations we could have anticipated this result from the matrix representation of Eq. (6), i.e., j j 1 [ e ] = G e e = G [ e ] i Another thing to note is the result in Eq. (4) is also anticipated since, in matrix notation, the Kronecker delta is the unit matrix, i. 115

42 1 0 0 i δ j = Note that the product in Eq. (4) is just, δ k j = = ik gg 1 ij I GG. 116

43 The General Permutation Symbol In the Cartesian system, the cross product is well defined analytically and geometrically. What about general coordinates? We define the general permutation symbol by the operation k ei ej = Eijke (for a right-handed system) where Eijk = ei ej ek [ ee 1 e3]. Using det( g ) = g = [ eee ], we then write E ij 1 3 ijk 1 = gε and E ε. g ijk ijk ijk 117

44 Physical Components of a Vector Recall a physical component of a vector is defined by i i aˆ eˆ i = aei (no summation). Then, i i aˆ eˆ eˆ = ae eˆ i i i i i i e i i g ˆ i ii a = ae ˆ i a = a (no summation). ei gii Therefore, the physical component, in terms of the contravariant and covariant components is, i i ii aˆ = a g and similarily aˆ = a g (no summation). ii i i 118

45 Orthogonal Curvilinear Coordinate Systems Because of the many conveniences of orthogonal systems, most space-coordinate systems used in engineering analysis are orthogonal. Many of these systems are also curvilinear systems, in particular, the spherical and cylindrical systems with which you are familiar. In this section we will look at orthogonal curvilinear systems and how they relate to our original Cartesian system. Scale Factors Define the scale factors h = e = g, h = e = g, h = e = g

46 Orthogonal Curvilinear Coordinate Systems With these definitions, then, e1 = h1eˆ1, e = heˆ, e3 = h3eˆ3 For a general curvilinear system, we earlier showed that a differential displacement is written as, 1 3 dr = dq e1+ dq e + dq e3. Now using the scale factors, 1 3 dr = ( h1dq ) eˆ1+ ( h ˆ ˆ dq ) e + ( h3dq ) e3. So, for the arclength, the differential distances are dr dr = ( ds) = ( h dq ) + ( h dq ) + ( h dq ) ,, 3 3 ds = h dq ds = h dq ds = h dq 10

47 r + dr 3 q r dr hdq = ( ds) q q hdq = ( ds) 1 hdq = ( ds)

48 The scale factors scale the q j to the appropriate magnitude and dimension for an orthogonal curvilinear system. In terms of the original Cartesian system, h h h r x x x = + + q q q q r x x x = + + q q q q 1 3 r x x x = + + q q q q / 1/ 1/,., 1

49 Differential Volume Element In many applications, especially finite-volume and finiteelement methods, you often must determine the volume of a differential element. For instance, a finite-volume form of the mass conservation equation in fluid mechanics requires a computation of the flux of mass through the boundaries, which must balance the creation of mass inside the volume. In most applications, the differential cell (volume) is of some variable shape determined by a curvilinear coordinate system. Here we introduce a general expression for determining a differential volume. Recall how the scalar triple product is related to the volume of a parallelepiped (with appropriate sign): 13

50 [e 1 e e 3 ] = volume of parallelepiped (with appropriate sign. In general, dv = ds ds ds 1 3 = dq e dq e dq e = dq dq dq 1 3 [ eee ] 1 3 = = dq dq dq g dq dq dq J For an orthogonal curvilinear system, dv = dq e dq e dq e = = hhhdqdq dq hhhdqdq dq [ eee ˆˆˆ] 14

51 Finally, for the Cartesian system, the familiar result 1 3 dv = dx dx dx = dx dy dz Note we can gain a bit of insight into the physical meaning of the Jacobian J. Combining the general expression for the differential volume element with that for the Cartesian system, we find, J dv = = dq dq dq dx dy dz dq dq dq This shows that the Jacobian of the transformation is the ratio of a differential volume in the Cartesian system to that of the general system. You can also see (if you haven t already discovered this) how the Jacobian is related to the fundamental metric,.i.e., J = g. 15