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ECE210a_lecture2_small - Vector spaces Vector spaces A...

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Unformatted text preview: Vector spaces Vector spaces A vector is a set of elements taken from a field, .77. Our most common choices for the field are 7?, (real numbers) or C (complex numbers). Addition: 3: + y (where as and y are vectors) Scalar multiplication: am, where oz 6 .7 and a: is a vector. Defining properties A vector space, V, satisfies the following properties: ForallaEJ-‘andallmandyEV, Al 3: —— y E V (closure) A2 (:1; + y) + z = x + (y + z) (distributive) A3 a: —— y = y —|— a: (commutative) A.4 There exists a unique 0 E V such that a: + 0 = x A.5 There exists —x e V such that a: + (—12) = 0 Roy Smith, ECE 210a, 2: 1 Vector spaces Vector spaces For all a and fl 6 fand all a: and y E V, M.1 as: E V M2 (amm = a(fix) M.3 a(x + y) = ax + ay M.4 (a + my: 2 am + fizz: M.5 There exists 1 E 7-" such that 1x = 1: Some examples 1. $1 Hill “More, v=1zn<orc> an ' aln, A: s s , m;eR(orC), V=RmX”(orCmX”) Roy Smith, ECE 210a, 2: 2 Vector spaces More examples 3. Functions, f : ’R,[0, 1] —> R, mapping [0, 1] on ’R, to R. 4. Real-valued polynomials of order n. Subspaces A subspace, S Q V (with the same field, addition and multiplication definitions), is also a vector space. Therefore, 1. x,yeS => w+ye$ 2.016.7-‘andz68 2) axes The key property of subspaces is closure. Other properties are inherited from the vector space. S = {0} is the trivial subspace. Roy Smith, ECE 210a, 2: 3 Vector spaces Span Given a set of vectors, 5: {111,U2,...,v,,}, 12,61}, the span of S is the set of all linear combinations. span(S) = {011111 + 02112 + - - - + anvn I 011‘ E «7} Properties Clearly a: E spanS => at E V. Furthermore, the spanS is also closed and is therefore a subspace. Proof: Roy Smith, ECE 210a, 2: 4 Vector spaces Question: What is the span of the following set of vectors? 1001 S: 2,0,0,2 3014 Sum of subspaces Given subspaces, X Q V and 32 Q V, X+y= {I+y|$€?€,y€y} If X = span(Sx) and 37 = span(Sy), then X + 3} = span(SX U 3y). Roy Smith, ECE 210a, 2: 5 Vector spaces Linear independence A set of vectors, S = {01, . . . ,v,, } is linearly independent if, Tl Emu-=0 2 a;=0, i=1,...,n. [:1 In other words, zero is the only solution. This is a property of a set of vectors7 not an individual vector. Example 1 n 1 0 0 01 Z (171}; = a1 0 + 052 1 + 053 0 = 012 = 0, {:1 0 0 1 (13 which clearly implies that a1 = a2 = 05;; = 0. Roy Smith, ECE 210a, 2: 6 Vector spaces Example 2. Consider a square matrix as a. collection of column vectors, 0*2 A 6 RM” = “*1 aw Column vector notation: an; := Now consider the set of column vectors in A, Show that S is a linearly independent set if and only if det(A) aé 0. Basic idea of the proof: If det(A) = 0 then there exists an eigenvector, a? 75 0, such that, As there exists at least one 2.,- 75 0, this equation has a non-trivial solution, implying linear dependence of the vectors. Roy Smith, ECE 210a, 2: 7 Vector spaces Example 3. Vandermonde matrices. 1 x1 xf - -~ 261171 1 x2 x? - -~ 267171 2 2 . . . l/mxn :2 2 : 2 : , Wlth 1;} 3A {IIJ' for all 2 7é j. 2 71 1 CB," 1' m I ~ . $le Show that if n = m, det(Van) 7é 0. (In the case Where n < m an analogous result can be stated and proven identically). Proof: Roy Smith, ECE 210a, 2: 8 Vector spaces Bases A basis of a vector space is a set of linearly independent vectors which span the vector space. Every nontrivial subspace has a basis. The number of vectors in a basis set is the dimension of the vector space. Example 1. The “standard” basis for R". 1 0 0 0 1 : . 7 : 7 7 0 == {517527---aen}- 0 0 1 Example 2. M2“: space of 2 X 2 real-valued matrices. {léfillfiélil‘ifillfifil} °r {Elllfiflllfléll‘ffil} Example 3. 5'2”: space of symmetric 2 )< 2 real-valued matrices. What is the dimension of this space? Roy Smith, ECE 210a, 2: 9 Vector spaces Example 4 The space of solutions, x(t) E C [07 1] (continuous functions on [0,1]), that satisfy, 2 (1 E1501) +x(t) = 0. What is the dimension of this space? Find a basis. Unique representations Givenabasis, B = {b1,...,bn} E V. For every x E V, there exists ai, i = 1, . . . ,n, such that7 I? x = E 01in and the a. are unique. [71 This choice of basis for any given space is not unique. Roy Smith, ECE 210a, 2: 10 Vector spaces Representation in terms of a basis Our standard vector notation is actually a representation in terms of a particular basis. For example, 1 1 0 0 5 = 1 0 + 5 1 + (—2) 0 2 161+ 562 + (—2)e3. —2 0 O 1 The “vector” is really just a list of the (1,- factors. The standard notation may confuse this. More precise notation: If we have a basis, B = {b1,...,b,,} G V, then the v E V, expressed with respect to the basis, B, is denoted by [v] 5‘ ’01 72 [’U]b, = = Zia-bi. i:l vIL This is important when we express the same vector in different bases. Roy Smith, ECE 210a, 2: 11 Vector spaces Example: Consider, B = {sint, cost }, defined on t E ’R[0, 1]. This is actually a basis for all continuous solutions of on the interval, R[O, 1]. Any solution can be expressed as, x(t) = a1 sint + a2 cost. Therefore, [x0013 = [3:]- All solutions are therefore associated with a vector in R2 and it is often easier to manipulate them in R2 than in their original form. Roy Smith, ECE 210a, 2: 12 Linear transformations Vector spaces Given two vector spaces, U and V, over a field, .73, a linear trwusfownatim is a mapping, T, from u to V (written T : u —> V) satisfying, 1. T(ar+ y) = T(x) + T(y) for all :1: and y G Ll. 2. T(am) = 047(x) for all as E L! and a E .7-'. In the case Where T : u —> U (i.e. mapping back into the same space), T is called a linear operator. The zero transformation is: 0(x) = 0, or more precisely 0([zr]u) = My. The idcntity operator is: I(a:) = 1:. Example 1. Matrix multiplication: T : R” —> Rm. T(:c) = Ax, Where A E Rm“. This is obviously linear; A(om: + y) = ozAm + Ag. Example 2. Consider L1 as the space of continuous functions, f (t) : ’R,[0, 00) —> R[0, oo). Define the operator: T( f) := /\ f (t)dt. (more precisely written as: T( f) (11)) 0 This linear because, Adriaf(t)+g(t)]dt = 0‘ lJilif(1t)dt+ / 1:9(t)dt, f0ra11f(t)and9(t) eu. 0 Example 3. Define Q6 : R2 —> R2 as a rotation of a vector in R2 by a fixed angle of 0. mcosO—ysiné] _ [cos0 —sin0] [ac] ac, then 99%) .— [xsin0+y0050 sin6 0050 y . Ifu=[ y This is linear: Q9(ozu + v) = 019901) + 96(7))- Note the correspondence with matrix multiplication. Roy Smith, ECE 210a, 2: 13 Vector spaces Roy Smith, ECE 210a, 2: 14 Vector spaces Space of linear transformations Given vector spaces U and V over .7, the set of (1,1 l linear transformations, [.(Ll,V) :2 {T:Z/l—)V | Tis linear} is itself a vector space over .7. Basis foru: Bu = {u1,...,un} (dim(Ll) = n) Basis for V: BV = {111, . . .,vm} (dim(V) = m) We can define a linear transformation T : U —> V by its action on each of the basis vectors in U. T will map u; (the jth basis vector in U) to some vector in V, Now any a: E U is expressed in terms of the basis Bu, 931 TI MB“ 22%” = j:l 3777 Roy Smith, ECE 210a, 2: 15 Vector spaces Coordinate matrix representations So now, [mama = [7 (qufl m = Z[T($juj)le = .2ng [7%)le = 2961' 206007: Bv We now express a vector y E V, the result of a linear transformation, y = ’1'(rc)7 x E u, in terms of the basis By, m [lev = 2.712% = {:1 From above7 we can identify each of the y values, 71 7'! yr = E 95,7060 = E 017:ij- j:1 1:1 This has an obvious matrix interpretation: 3/] yin 3/1 (111 ‘ ' ' aln $1 $1 =: 3V [T] Bu 5 . (Meyer’s notation: [’1'] n] 4 13V) ym aml ' ’ ' arm! {1}” x71 Roy Smith, ECE 210a, 2: 16 Vector spaces Key facts about linear transformations 1. Every linear transformation can be represented by a matrix. 2. The action of a linear transformation T : u —> V, on x E u, can be represented by matrix multiplication between the coordinates of the bases. l7($)le = Bi; [7] Ba lxlBu' 3. dim([,(u, 12)) = (dim(u)) (dim(V)). Example Differentiation for polynomials of degree 3. we» == did—f) Select the basis:Bu = B]; = { 1,t,t2,t3 } We can show that, 13v [T] Bu 2 COCO OOOH COMO OOOOO Roy Smith, ECE 210a, 2: 17 Vector spaces More about notation Consider operators, ’P : X —> 3} and Q : W —> X. y=’Px and x=Qw. Note that this is taken to mean that 7’ maps a: E X to y E 37. .7163] :vEX wEW ‘ In matrix form these equations are: [y] M = By [’P] EX [:5] By and [it] b’l- = Bx [Q] 3W [wlifw This is equivalent to, [ley = By [73le Bar [Q] 13W [w] W. And we usually write the associated matrix/vector mapping as y = PQw. Roy Smith, ECE 210a, 2: 18 ...
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