Chapter 1. Vector Algebra and Vector Space

Transcription

1 1. Vector Algebra 1.1. Scalars and vectors Chapter 1. Vector Algebra and Vector Space The simplest kind of physical quantity is one that can be completely specified by its magnitude, a single number, together with the units in which it is measured. Such a quantity is called a scalar and examples include temperature, time and density. they are described by numbers: Scalar = number (real or complex). A vector is a quantity that requires both a magnitude (> 0) and a direction in space to specify it completely; we may think of it as an arrow in space. A familiar example are force, velocity, displacement, momentum and electric field. Vectors are also used to describe quantities such as angular momentum and surface elements. Such quantities are represented geometrically by arrows. Vector = number (its size) and direction. Magnitude of vector: The length of the arrow representing a vector A is called the length or the magnitude of a (written a or a) or the norm of a ( a ). In three dimensions a = a = a x 2 + a y 2 + a z 2 where a i (i = x,y,z ) are the component of the vector. Unit Vector: A vector of length 1 is called a unit vector and is denoted by a or a Given a vector a, the unit vector in the direction of a is a = a a In the cartesian coordinate system (x, y, z), we may introduce the unit vectors i, j and k, which point along the positive x-, y- and z- axes respectively. A vector a may then be written as a sum of three vectors, each parallel to a different coordinate axis: a = a x i + a y j + a z k. The sum of two vectors a and b is found by simply adding their components, i.e. and their difference by subtracting them, Example: 1

2 and geometrically the addition and substraction can be shown as The Scalar Product: The scalar product (or dot product) of two vectors a and b is denoted by a. b and is given by a. b = a b cos (θ) where θ is the angle between the two vectors, placed 'tail to tail' or 'head to head'. It should be noted in particular that the Cartesian basis vectors i, j and k, being mutually orthogonal unit vectors, satisfy the i. i = j. j = k. k = 1 and i. j = j. k = i. k = 0. If the scalar product of two vectors is equal to zero, they are orthogonal to each other. It is represented geometrically as Example: Find the angle between the vectors a = i + 2j + 3k and b = 2i + 3j + 4k. The scalar product a and b has the value a.b = lx2+2x3+3x4=20, and the lengths of the vectors are a = a = = 14 and b = b = = 29. Thus, cos θ = a. b a b = and θ = 0.12 rad. 2

3 Example: Find the angle between i + j and i 3j The Vector Product: The vector product (or cross product) of two vectors a and b is denoted by axb and is given by axb = a b sin (θ)n where θ is the angle between the two vectors and n is defined as a unit vector in the positive normal direction to the plane of a and b. Example: Find the angle between the vectors A = 3 i - 2 j + 4 k and B = 2 i -3 j -2 k. Example: 3

4 Example: Example: Work done by force F over distance (direction) d W = F d Find work done by force F = i + 2j in moving from the origin to the point A = j. F d = (1 2) (0 1) = 2 Cross Products and Mechanics Angular motion: If a particle in a rigid body is rotating about an axis with angular velocity w (yes a vector!), then the linear velocity of a particle with position vector r, relative to the axis of motion is v = w r Angular Momentum: Suppose a planet of mass m is in orbit at position r with linear velocity v. Take the centre of rotation (e.g., the sun) as being the origin of coordinates. The angular momentum of the planet is then L = r (mv). Moment: The moment, or torque, m of a force F applied to a point with position vector r with respect to the origin is m = r F. The moment describes the turning effect of force about a fixed point, as shown in 4

5 the figure. Note that r denotes the position vector from origin of any point on the line of action of the force Differentiation of Vectors In many practical problems, we often deal with vectors that change with time, e.g. velocity, acceleration, etc. If a vector A depends on a scalar variable time t, then a can be represented as A(t) and A is then said to be a function of t. If A = a x i + a y j + a z k then a x, a y and a z will also dependent on the parameter t. A(t) = a x (t)i + a y (t)j + a z (t)k. Differentiation with respect to t gives Perhaps the simplest application of the above is to finding the velocity and acceleration of a particle in classical mechanics. If the time-dependent position vector of the particle with respect to the origin in Cartesian coordinates is given by r(t) = x(t) i + y(t) j + z(t) k then the velocity and the acceleration of the particle are given by the vector Example: The speed of the particle at t = 1 is simply v(1) = = 61 m/s. The acceleration of the particle is constant (i.e. independent of t), and its component in the direction s is given by Example: A particle moves in space so that at time t its position is stated as x = 2t+3, y = t 2 +3t, z = t 3 +2t 2. We are required to find the components of its velocity and acceleration in the direction of the vector 2i + 3j + 4 k when t=1. 5

6 First we can write the position as a vector r as r = (2t+3) i + (t 2 +3t) j + (t 3 +2t 2 ) k. Example: A particle moves in space at time t with position vector as r = t 3 i (4t 2-8t+1) j. Fint the corresponding quantities related with that particle. 6

7 1.3. Differentiation of composite vector expressions In composite vector expressions each of the vectors or scalars involved may be a function of some scalar variable u, and as we have seen. The derivatives of such expressions are easily found using the rules of ordinary differential calculus. Let consider assume that a and b are differentiable vector functions of a scalar u and that is a differentiable scalar function of u: the order of the factors in the terms on the RHS of final equation is, of course, just as important as it is in the original vector product. Example: A particle of mass m with position vector r relative to some origin 0 experiences a force F, which produces a torque T = r x F about O. The angular momentum of the particle about 0 is given by L = r x mv, where v is the particle's velocity. Show that the rate of change of angular momentum is equal to the applied torque. The rate of change of angular momentum is given by where in the last line we use Newton's second law, namely F = d(mv)/dt. Similiarly, let consider the following vector and apply tis properties. 7

8 1.4. Vector functions of several arguments The concept of the derivative of a vector is easily extended to cases where the vectors (or scalars) are functions of more than one independent scalar variable, u 1, u 2,...,u n. In this case, the partial derivatives of a in Cartesian coordinates is If If a = a (u 1, u 2,...,u n ) and each of the ui is also a function u i (v 1, v 2,...,v n ) of the variables v i then, the equation is generalized to As an example, the infinitesimal change in an electric field E in moving from a position r to a neighbouring one r + dr is given by Example: If a vector function F is given by F = 2uv i + (u 2-2v) j + (u+v 2 ) k. Find F u, F v, 2 F u 2 and 2 F u v. F F = 2vi + 2uj + k, u 1.5. Integration of vectors = 2ui 2j + 2vk, v 2 F = 2j and u2 2 F u v = 2i The integration of a vector (or of an expression involving vectors that may itself be either a vector or scalar) with respect to a scalar u can be regarded as the inverse of differentiation. We must remember, however, that (i) the integral has the same nature (vector or scalar) as the integrand, (ii) the constant of integration for indefinite integrals must be of the same nature as the integral. Example: If a vector F = (3t 2 +4t) i + (2t-5) j + 4t 3 k, then evaluate the integral of F over the range from 1 to 3. Example: If F = 3u i + u 2 j + (u+2) k and V = 2u i 3u j + (u-2) k, evaluate the integral of FxV over the range from 0 and 2 8

9 1.6. Scalar and Vector Fields If every point P(x,y,z) of a region R of space associated with it a scalar quantity φ (x, y, z), then φ (x, y, z) is a scalar function and a scalar field is said to exist in the region R. Examples of scalar fields are, temperature, potential, etc. Similiarly, if every point P(x, y, z) of region R has associated with it a vector quantity F(x,y,z), then F(x,y,z) is a vector function and a vector field is said to exist in the region R. Examples of vector fields are force, velocity, acceleration, etc Vector operators Certain differential operations may be performed on scalar and vector fields and have wide-ranging applications in the physical sciences. The most important operations are those of finding the gradient of a scalar field and the divergence and curl of a vector field. Central to all these differential operations is the vector operator, which is called del (or sometimes nabla) and in Cartesian coordinates is defined by Gradient of a scalar field The gradient of a scalar field φ(x, y, z) is defined by 9

10 Clearly, when φ is a scalar function φ is a vector field whose x-, y- and z- components are the first partial derivatives of φ (x, y, z) with respect to x, y and z respectively. Example: Find the gradient of the scalar field φ = x y 2 z 3. The gradient of φ is given by φ = y 2 z 3 i + 2xyz 3 j + 3xy 2 z 2 k. Example: Find the gradient of the A.B where A = x 2 z i + xz j + y 2 z k and B = yz 2 i + xz j + x 2 z k. Directional Derivative: Starting from dφ = dr. φ, it can be reached to dφ ds = a gradφ where a isa unit vector in the direction of dr as a = dr ds. dφ ds is thus the projection of gradφ on the unit vector a is called the directional derivative of φ in the direction of a. It gives the rate of change of φ with distance measured in the direction of a. Example: Suppose the temperature T at the point (x,y,z) is given by T = x 2 - y 2 + xyz + 273; in which direction is the temperature increasing most rapidly at (-1,2,3), and what is the rate? Here, T = (2x+yz) i + (-2y+xz) j + xy k and at (-1,2,3) T = 4 i -7 j -2 k. The rate of increase is dt = T = = 69. We can also say that the temperature is ds decreasing most rapidly in the direction - T = 69. Heat flow in the direction - T (that is from hot to cold.) Example: Find the directional derivative of the function φ = x 2 z 2xy 2 + yz 2 at the point (1, 2,-1) in the direction of the vector given as A = 2 i + 3 j - 4 k. 10

11 Unit normal vector The equation of φ(x,y,z)=constant represents a surface in space. For example, 3x - 4y + 2z = 1 is the equation of the plane and x 2 + y 2 + z 2 = 4 represents a sphere centered on the origin and of radius 2. The unit vector N in the direction of grad φ is called the unit normal vector at P. Example: Find the unit normal vector to the surface x 3 y + 4xz 2 + xy 2 z + 2 = 0 at the point (1, 3, -1). Divergence of a vector field The divergence of a vector field a (x, y, z) is defined by Clearly,. a is a scalar field. Any vector field a for which. a = 0 is said to be solenoidal. Example: Find the divergence of the vector field a = x 2 y 2 i - y 2 z 2 j - x 2 z 2 k. The divergence of a is given by 11

12 The operator 2, the Laplacian Now if some vector field a is itself derived from a scalar field via a = φ then. a has the form. φ or, as it is usually written 2 φ, where 2 (del squared) is the scalar differential operator is a scalar differential operator which is called the Laplacian, after a French mathematician of the eighteenth century named Laplace. 2 φ is called the Laplacian of φ and appears in several important partial differential equations of mathematical physics. Example: Find the Laplacian of the scalar field φ = xy 2 z 3. The Laplacian of φ is given by Curl of a vector field The curl of a vector field a(x, y, z) is defined by The RHS can be written in a more memorable form as a determinant where it is understood that, on expanding the determinant, the partial derivatives in the second row act on the components of a in the third row. Clearly, xa is itself a vector field. Any vector field a for which xa = 0 is said to be irrotational. 12

13 Example: If A = (y 4 -x 2 z 2 ) i + (x 2 +y 2 ) j - yzx 2 k, determine the curl of A at the point (1,3,-2) Line Integrals: Suppose an object moves along same path (say from A to B as in figure) with the force F acting on it varying as it moves. For example F might be the force on a charged particle in electric field; then F would be vary from point to point, that is, it would be a function of x, y, z. However, on a curve, x, y, z are related by the equation of curve. Thus along a curve, there is one independent variable; then we write F and dr = dx i+ dy j+ dz k as a functions of a single variable. The integral of dw = F.dr along the given curve gives the total work done by F in moving an abject. Such an integral is called a line integral. A line integral means an integral along a curve (or line), that is, a single integral as contrasted to a double integral over a surface or area, or a triple integral over a volume. In general, we may encounter line integrals of the forms C dr, a. dr C C a x dr where φ is a scalar field and a is a vector field. The three integrals themselves are respectively vector, scalar and vector in nature. As we will see below, in physical applications line integrals of the second type are by far the most common. Each of the line integrals is evaluated over some curve C that may be either open (A and B being distinct points) or closed (the curve C forms a loop, so that A and B are coincident). In the case where C is closed, the line integral is written C to indicate this. The curve may be given either parametrically by r(u) = x(u) i + y(u) j + z(u) k or by means of simultaneous equations relating x, y, z for the given path (in Cartesian coordinates). In practice, x, y, and z are often expressed in terms of parametric equations of a fourth 13

14 variable (say u), i.e. x = x(u); y = y(u); and z = z(u). From these, dx, dy and dz can be written in terms of u and the integral evaluated in terms of this parameter. Evaluating line integrals The method of evaluating a line integral is to reduce it to a set of scalar integrals. It is usual to work in Cartesian coordinates, in which case dr = dx i + dy j + dz k. The first type of line integral then becomes simply C dr = i C (x, y, z)dy + j (x, y, z)dy C + j (x, y, z)dy C The second and third line integrals in (11.1) can also be reduced to a set of scalar integrals by writing the vector field a in terms of its Cartesian components as a = a x i + a y j + a z k, where a x, a y, and a z are each (in general) functions of x, y, z. The second line integral can then be written as C a. dr = C a x i + a y j + a z k. dx i + dy j + dz k = a x dx + a y dy + a z dz C = a x dx C + a y dy + a z dz C C Example: Given the force F = xy i y 2 j, find the work done by F along the paths indicated in figure from (0, 0) to (2, 1). If we write the integrand in terms of one variable, we can obtain y = ½ x, dy = ½ dx. Substituting these values, we get We could just as well use y as the independent variable and put x = 2y, dy = 2 dy, and integrate from 0 to 1. Along path 2 (a parabola), y = 1/4 x 2, dy = ½ xdx, then we get 14

15 Along path 3, we integrate first from (0,0) to (0,1) and then from (0,1) to (2,1) and add the results. Along (0,0) to (0,1), x = 0 and dx = 0 so we must use y as the variable. Then we have Along (0,1) to (2,1), y = 1, dy = 0, so we use x as the variable. We have Then the total W 3 = -1/3 + 2 = 5/3. Path 4 illustrates still another technique. Instead of using either x or y as the integration variable, we can use a parameter t. For x = 2t 3, y = t 2, we have dx = 6t 2 dt, dy = 2tdt. At the origin, t = 0 and at (2, 1), t = 1. Substituting these values, we get Example: If V = xy 2 z, evaluate the integral of V dr for the curve having parametric equations x = 3u; y = 2u 2 ; z = u 3 between A(0,0,0) and B(3,2,1). x = 3u, dx = 3du; y = 2u 2, dy = 4udu; z = u 3, dz =3 u 2 du, thus the position vector becomes dr(u) = 3du i + 4udu j +3 u 2 du k and the points A(0,0,0) corresponds u = 0, and B(3,2,1) corresponds u = Conservative fields and potentials We saw that, for paths in the xy-plane, line integrals whose integrands have certain properties are independent of the path taken. For line integrals of the form a dr, there exists a class of vector fields for which the line integral between two points is independent of the path taken. Such vector fields are called conservative. A vector field a that has continuous partial derivatives in a simply connected region R is conservative if, and only if, any of the following is true. (i) The integral a. dr from A to B lie in the region R, is independent of the path from A to B. Hence the integral of a. dr around any closed loop in R is zero. (ii) There exists a single-valued function φ of position such that a = φ. (iii) a = 0 or curla = 0 (iv) a dr is an exact differential. Let us assume that (i) above is true. If the line integral from A to B is independent of the path taken between the points then its value must be a function only of the positions of A and B. We may therefore write 15

16 a. dr = A (B) which defines a single-valued scalar function of position φ. One example of conservative filed is the work done by an applying force F. The integration of F.dr must be independent of the path, that is, curlf = 0 for conservative fields and curlf is different from zero for nonconservative fields. Suppose that for a given function F there is a function W(x,y,z) such that Then F = W x, y, z = i W x F x = W x, F y = W y, + j W x F z = W z, + k W x Example: Show that F = (2xy - z 3 ) i + x 2 j (3xz 2 +1) k is conservative, and find a scalar potential such that F = So, F is conservative. Then or 16

17 Physical examples of line integrals There are many physical examples of line integrals, but perhaps the most common is the expression for the total work done by a force F when it moves its point of application from a point A to a point B along a given curve C. Naturally, other physical quantities can be expressed in such a way. For example, the electrostatic potential energy gained by moving a charge q along a path C in an electric field E is q E dr. We may also note that Ampere s law concerning the magnetic field B associated with a currentcarrying wire can be written as B dr = μ 0 I where I is the current enclosed by a closed path C traversed in a right-handed sense with respect to the current direction. Magnetostatics also provides a physical example of the third type of line integral. If a loop of wire C carrying a current I is placed in a magnetic field B then the force df on a small length dr of the wire is given by df = I dr B, and so the total (vector) force on the loop is F = I dr B Surface integrals As with line integrals, integrals over surfaces can involve vector and scalar fields and, equally, can result in either a vector or a scalar. The form of surface integrals have forms S ds, a. ds S S a x ds (a) A closed surface and (b) an open surface. In each case a normal to the surface is shown: ds = n ds. All the above integrals are taken over some surface S, which may be either open or closed, and are therefore, in general, double integrals. Following the notation for line integrals, for surface integrals over a closed surface S is replaced by S. 17

18 Example: A scalar field V = xyz exists over the curved surface S defined by x 2 + y 2 = 4 between the planes z = 0 and z = 3 in the first octant. Evaluate S V ds over this surface. We have to evaluate this integral over the prescribed surface. Changing to cylindrical coordinates with ρ = 2. 18

19 Example: A vector field F = y i + 2j + k exist over a surface S defined by x 2 + y 2 + z 2 = 9 bounded by x = 0, y = 0, z = 0, in the first octant. Evaluate S F.dS over the surface indicated. Before integrating over the surface, we convert to spherical coordinates as Limits of θ = 0 to π/2; and φ = 0 to π/2; Physical examples of surface integrals 19

20 There are many examples of surface integrals in the physical sciences. Surface integrals of the form occur in computing the total electric charge on a surface or the mass of a shell, ρ(r) ds, given the charge or mass density ρ(r). For a vector field a, the surface integral a ds is called the flux of a through S. Examples of physically important flux integrals are numerous. For example, let us consider a surface S in a fluid with density ρ(r) that has a velocity field v(r). The mass of fluid crossing an element of surface area ds in time dt is dm = ρv ds dt. Therefore the net total mass flux of fluid crossing S M = ρ(r)v(r) ds. As another example, the electromagnetic flux of energy out of a given volume V bounded by a surface S is (E H) ds. The solid angle, to be defined below, subtended at a point O by a surface (closed or otherwise) can also be represented by an integral of this form, although it is not strictly a flux integral (unless we imagine isotropic rays radiating from O). The integral = r. ds r 3 = r. ds r 2 gives the solid angle Ω subtended at O by a surface S if r is the position vector measured from O of an element of the surface Volume integrals: Volume integrals are defined in an obvious way and are generally simpler than line or surface integrals since the element of volume dv is a scalar quantity. We may encounter volume integrals of the forms V dv, a dv V Clearly, the first form results in a scalar, whereas the second form yields a vector. Two closely related physical examples, one of each kind, are provided by the total mass of a fluid contained in a volume V, ρ(r) dv, and the total linear momentum of that same fluid, given by ρ(r) v(r)dv where v(r) is the velocity field in the fluid. 20

21 Example: Evaluate V F dv where V is the region bounded by the planes x = 0, x = 2, y = 0,, y = 3, z = 0, z = 4 and the vector field F = xy i + zj x 2 k Appendix Curvilinear Coordinates In order to study solutions of the wave equation, the heat equation, or even Schrödinger s equation in different geometries, we need to see how differential operators, such as the Laplacian, appear in these geometries. The most common coordinate systems arising in physics are polar coordinates, cylindrical coordinates, and spherical coordinates. These reflect the common geometrical symmetries encountered in our studies. In this section we will consider general coordinate systems and how the differential operators are written in the new coordinate system. Cylindrical Polar Coordinates (,, z): Let P be a point with cylindrical coordinates (,, z) as shown. The position of the P is a function of the three variables (,, z). 21

22 Spherical Polar Coordinates (r,, ): The position of the P is a function of the three variables (r,, ). Where Scale Factors Collecting the recent results together, we have: a) For cylindrical polar coordinates, the unit base vectors are b) For spherical polar coordinates, the unit base vectors are where in each case h is called scale factor. 22

23 23

24 General Curvilinear Coordinate system ( u, v, w) 24

25 Transformation Equations Elements of arc ds and element of volume dv in orthogonal curvilinear coordinates 25

26 Example: Grad, div and Curl in orthogonal curvilinear coordinates Exercises: A particle moves along the curve x = 2 t 2, y = t 2-4t, and z = 3t - 5, where t is the time. Find the components of the particle s velocity and acceleration at time t = 1 in the direction i 3 j + 3 k. 4. Find the unit vector normal to the surface x 2 + y 2 z = 1 at the point P(1,1,1). 26

27 5. Find the directional derivative of = x 2 yz + 4xz 3 at (1,-2,-1) in the direction 2 i j - 2 k. 6. Consider an electric dipole moment p at the origin. It produces an electric field of = p.r 4πε r 3 outside the dipole. Nothing that E = -, find the electric field at r. 7. Find constants a, b, c such that A = ( x + 2y + a z) i +( b x - 3y - z) j +(4 x + 2cy + 2 z) k is irrotational The Vector Space R n The Vector Space R n The world of everyday experience has three space dimensions. But often we encounter setting s in which more dimensions occur. If we want to specify not only the location of a particle but the time in which it occupies a particular point, we need four coordinates (x, y, z, t). And specifying the location of each particle in a system of particles may require any number of coordinates. The natural setting for such problems is R n, the space of points having n coordinates. Theorem: If n is a positive integer, an n-vector is an n-tuple (x 1, x 2,.., x n ), with each coordinate x j a real number. The set of all n-vectors is denoted R n. R 1 is the real line, consisting of all real numbers. We can think of real numbers as 1-vectors, but there is is no advantage to doing this. R 2 consists of ordered pairs (x, y) of real numbers, and each such ordered pair (or 2-vector) can be identified with a point in the plane. R 3 consists of all 3-vectors, or points in 3-space. If n > 4, we can no longer draw a set of mutuall y independent coordinate axes, one for each coordinate, but we can still work with vectors in R n according to rules we will now describe. Algebra of R n 1. Two n-vectors are added by adding their respective components: (x 1, x 2,.., x n )+ (y 1, y 2,.., y n ) = (x 1 + y 1, x 2 + y 2,.., x n + y n ) 2. An n-vector is multiplied by a scalar by multiplying each component by the scalar: (x 1, x 2,.., x n )= ( x 1, x 2,.., x n ) The zero vector in R n is the n-vector 0 = (0, 0,..., 0) having each coordinate equal to zero. The algebraic rules in R n mirror those we saw for R 3. Properties Let F, G, and H be in R n, and let and be real numbers. Then F+G = G+F. F+(G+H) = (F+G) + H. F+O=F. ( + ) F = F+ F. ( ) F = ( F). (F+G) = F+ G. O=O. 27

28 Because of these properties of the operations of addition of n-vectors, and multiplicatio n of an n-vector by a scalar, we call R n a vector space. In the next section we will clarify the sense in which R n can be said to have dimension n. The length (norm, magnitude) of X = (x 1, x 2,.., x n ) is defined by a direct generalization from the plane and 3-space: There is no analogue of the cross product for vectors in R n when n > 3. However, the dot product readily extends to n-vectors. Dot Product of n-vectors The dot product of (x 1, x 2,.., x n ) and (y 1, y 2,.., y n ) is defined as (x 1, x 2,.., x n ).(y 1, y 2,.., y n ) = x 1 y 1 + x 2 y x n y n Subspace: A set of n-vectors containing the zero vector, as well as sums of vectors in the set an d scalar multiples of vectors in the set, is called a subspace of R n. A set S of n-vectors is a subspace of R n if: 0 is in S. The sum of any vectors in S is in S. The product of any vector in S with any real number is also in S. Example: Let K consist of all scalar multiples of (-1, 4, 2, 0) in R 4. We want to know if K is a subspace of R 4. First, 0 is in K, because 0 = 0(-1, 4, 2, 0) = (0, 0, 0, 0). Next, if F and G are in K, then F = (-1, 4, 2, 0) for some and G = (-1, 4, 2, 0) for some, so F + G = ( + ) (-l, 4, 2, 0) is a scalar multiple of (-1, 4, 2, 0) and therefore is in K. Finally, if F = (-1, 4, 2, 0) is any vector in K, and is any scalar, then F = ( ) (-1, 4, 2, 0) is a scalar multiple of (-1, 4, 2, 0), and hence is in K. Thus K is a subspace of R 4. 28

29 Cauchy Inequality: X Y X Y if X, Y R n Orthogonal Vectors Orthogonal and Orthonormal Sets Example: 29

30 Linear Independence In dependence Let S = {X 1,X 2,...,X k } be a set of k vectors in R n, let a 1, a 2,..., a k be scalars, and suppose a 1 X 1 + a 2 X a k X k = O. If the only solution for the scalars are a 1 = 0, a 2 = 0,..., a k = 0, then the set S is called a linearly independent set, and we say that the vectors X 1, X 2,..., X k are linearly independent. Purely symbolically: means the vectors X 1, X 2,..., X k are linearly independent. Example: 30

31 Example: Linear Dependence If S = {X 1,X 2,...,X k } is not linearly independent, it is called linearly dependent, and we say that the vectors X 1,X 2,...,X k are linearly dependent. That is, there are scalars a 1, a 2,..., a k, not all zero, such that a 1 X 1 + a 2 X a k X k = O. Equivalently: if a set of vectors is linearly dependent, then one of the vectors is a linear combination of the others. For example, if the relation between vectors is 2X 1 X 2 + X 3 = 0 X 2 = 2X 1 + X 3. On the other hand, if a set of vectors is linearly independent, then none of the vectors is a linear combination of the others. 31

32 Example: Example: n Vectors in R n 32

33 Example: Basis: 33

34 Example: Example: References: 1. Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson and S. J. Bence, 3th edition Advanced Engineering Mathematics, K.A. Stroud, 4th edition. 4. Mathematical Methods in the Physical Sciences, Mary L. Boas, 2 nd edition. 34