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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. $1
Hill “More, v=1zn<orc> an ' aln,
A: s s , m;eR(orC), V=RmX”(orCmX”) Roy Smith, ECE 210a, 2: 2 Vector spaces More examples 3. Functions, f : ’R,[0, 1] —> R, mapping [0, 1] on ’R, to R.
4. Realvalued polynomials of order n. Subspaces A subspace, S Q V (with the same ﬁeld, addition and multiplication deﬁnitions), is also a
vector space. Therefore, 1. x,yeS => w+ye$
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 nontrivial 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 == {517527aen}
0 0 1 Example 2. M2“: space of 2 X 2 realvalued matrices. {léﬁllﬁélil‘iﬁllﬁﬁl} °r {Elllﬁﬂllﬂéll‘ffil} Example 3. 5'2”: space of symmetric 2 )< 2 realvalued 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, = = Ziabi.
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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 Spring '08
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