MATH 304 Linear Algebra Lecture 8: Vector spaces. Subspaces.

Transcription

1 MATH 304 Linear Algebra Lecture 8: Vector spaces. Subspaces.

2 Linear operations on vectors Let x = (x 1, x 2,...,x n ) and y = (y 1, y 2,...,y n ) be n-dimensional vectors, and r R be a scalar. Vector sum: x + y = (x 1 + y 1, x 2 + y 2,...,x n + y n ) Scalar multiple: rx = (rx 1, rx 2,...,rx n ) Zero vector: 0 = (0, 0,...,0) Negative of a vector: y = ( y 1, y 2,..., y n ) Vector difference: x y = x + ( y) = (x 1 y 1, x 2 y 2,...,x n y n )

3 Properties of linear operations x + y = y + x (x + y) + z = x + (y + z) x + 0 = 0 + x = x x + ( x) = ( x) + x = 0 r(x + y) = rx + ry (r + s)x = rx + sx (rs)x = r(sx) 1x = x 0x = 0 ( 1)x = x

4 Linear operations on matrices Let A = (a ij ) and B = (b ij ) be m n matrices, and r R be a scalar. Matrix sum: Scalar multiple: Zero matrix O: Negative of a matrix: Matrix difference: A + B = (a ij + b ij ) 1 i m, 1 j n ra = (ra ij ) 1 i m, 1 j n all entries are zeros A = ( a ij ) 1 i m, 1 j n A B = (a ij b ij ) 1 i m, 1 j n As far as the linear operations are concerned, the m n matrices have the same properties as mn-dimensional vectors.

5 Abstract vector space: informal description Vector space = linear space = a set V of objects (called vectors) that can be added and scaled. That is, for any u,v V and r R expressions should make sense. u + v and ru Certain restrictions apply. For instance, u + v = v + u, 2u + 3u = 5u. That is, addition and scalar multiplication in V should be like those of n-dimensional vectors.

6 Abstract vector space: definition Vector space is a set V equipped with two operations α : V V V and µ : R V V that have certain properties (listed below). The operation α is called addition. For any u,v V, the element α(u,v) is denoted u + v. The operation µ is called scalar multiplication. For any r R and u V, the element µ(r,u) is denoted ru.

7 Properties of addition and scalar multiplication (brief) A1. a + b = b + a A2. (a + b) + c = a + (b + c) A3. a + 0 = 0 + a = a A4. a + ( a) = ( a) + a = 0 A5. r(a + b) = ra + rb A6. (r + s)a = ra + sa A7. (rs)a = r(sa) A8. 1a = a

8 Properties of addition and scalar multiplication (detailed) A1. a + b = b + a for all a,b V. A2. (a + b) + c = a + (b + c) for all a,b,c V. A3. There exists an element of V, called the zero vector and denoted 0, such that a + 0 = 0 + a = a for all a V. A4. For any a V there exists an element of V, denoted a, such that a + ( a) = ( a) + a = 0. A5. r(a +b) = ra + rb for all r R and a,b V. A6. (r + s)a = ra + sa for all r, s R and a V. A7. (rs)a = r(sa) for all r, s R and a V. A8. 1a = a for all a V.

9 Associativity of addition implies that a multiple sum u 1 + u u k is well defined for any u 1,u 2,...,u k V. Subtraction in V is defined as usual: a b = a + ( b). Addition and scalar multiplication are called linear operations. Given u 1,u 2,...,u k V and r 1, r 2,...,r k R, r 1 u 1 + r 2 u r k u k is called a linear combination of u 1,u 2,...,u k.

10 Examples of vector spaces In most examples, addition and scalar multiplication are natural operations so that properties A1 A8 are easy to verify. R n : n-dimensional coordinate vectors M m,n (R): m n matrices with real entries R : infinite sequences (x 1, x 2,... ), x i R For any x = (x 1, x 2,...), y = (y 1, y 2,...) R and r R let x + y = (x 1 + y 1, x 2 + y 2,...), rx = (rx 1, rx 2,...). Then 0 = (0, 0,...) and x = ( x 1, x 2,...). {0}: the trivial vector space = 0, r0 = 0, 0 = 0.

11 Functional vector spaces F(R): the set of all functions f : R R Given functions f, g F(R) and a scalar r R, let (f + g)(x) = f (x) + g(x) and (rf )(x) = rf (x) for all x R. Zero vector: o(x) = 0. Negative: ( f )(x) = f (x). C(R): all continuous functions f : R R Linear operations are inherited from F(R). We only need to check that f, g C(R) = f +g, rf C(R), the zero function is continuous, and f C(R) = f C(R). C 1 (R): all continuously differentiable functions f : R R C (R): all smooth functions f : R R P: all polynomials p(x) = a 0 + a 1 x + + a n x n

12 Some general observations The zero vector is unique. If z 1 and z 2 are zeros then z 1 = z 1 + z 2 = z 2. For any a V, the negative a is unique. Suppose b and b are negatives of a. Then b = b + 0 = b + (a + b) = (b + a) + b = 0 + b = b. 0a = 0 for any a V. Indeed, 0a + a = 0a + 1a = (0 + 1)a = 1a = a. Then 0a + a = a = 0a + a a = a a = 0a = 0. ( 1)a = a for any a V. Indeed, a + ( 1)a = ( 1)a + a = ( 1)a + 1a = ( 1 + 1)a = 0a = 0.

13 Counterexample: dumb scaling Consider the set V = R n with the standard addition and a nonstandard scalar multiplication: r a = 0 for any a R n and r R. Properties A1 A4 hold because they do not involve scalar multiplication. A5. r (a + b) = r a + r b 0 = A6. (r + s) a = r a + s a 0 = A7. (rs) a = r (s a) 0 = 0 A8. 1 a = a 0 = a A8 is the only property that fails. As a consequence, property A8 does not follow from properties A1 A7.

14 Subspaces of vector spaces Definition. A vector space V 0 is a subspace of a vector space V if V 0 V and the linear operations on V 0 agree with the linear operations on V. Examples. F(R): all functions f : R R C(R): all continuous functions f : R R C(R) is a subspace of F(R). P: polynomials p(x) = a 0 + a 1 x + + a k x k P n : polynomials of degree less than n P n is a subspace of P.

15 Subspaces of vector spaces Counterexamples. R n : n-dimensional coordinate vectors Q n : vectors with rational coordinates Q n is not a subspace of R n. 2(1, 1,...,1) / Q n = Q n is not a vector space (scaling is not well defined). P: polynomials p(x) = a 0 + a 1 x + + a n x n P n: polynomials of degree n (n > 0) P n is not a subspace of P. x n + (x n + 1) = 1 / Pn = Pn is not a vector space (addition is not well defined).

16 If S is a subset of a vector space V then S inherits from V addition and scalar multiplication. However S need not be closed under these operations. Proposition A subset S of a vector space V is a subspace of V if and only if S is nonempty and closed under linear operations, i.e., x,y S = x + y S, x S = rx S for all r R. Proof: only if is obvious. if : properties like associative, commutative, or distributive law hold for S because they hold for V. We only need to verify properties A3 and A4. Take any x S (note that S is nonempty). Then 0 = 0x S. Also, x = ( 1)x S.

17 Example. V = R 2. The line x y = 0 is a subspace of R 2. The line consists of all vectors of the form (t, t), t R. (t, t) + (s, s) = (t + s, t + s) = closed under addition r(t, t) = (rt, rt) = closed under scaling The parabola y = x 2 is not a subspace of R 2. It is enough to find one explicit counterexample. Counterexample 1: (1, 1) + ( 1, 1) = (0, 2). (1, 1) and ( 1, 1) lie on the parabola while (0, 2) does not = not closed under addition Counterexample 2: 2(1, 1) = (2, 2). (1, 1) lies on the parabola while (2, 2) does not = not closed under scaling

18 Example. V = R 3. The plane z = 0 is a subspace of R 3. The plane z = 1 is not a subspace of R 3. The line t(1, 1, 0), t R is a subspace of R 3 and a subspace of the plane z = 0. The line (1, 1, 1) + t(1, 1, 0), t R is not a subspace of R 3 as it lies in the plane x + y + z = 3, which does not contain 0. In general, a straight line or a plane in R 3 is a subspace if and only if it passes through the origin.

19 System of linear equations: a 11 x 1 + a 12 x a 1n x n = b 1 a 21 x 1 + a 22 x a 2n x n = b 2 a m1 x 1 + a m2 x a mn x n = b m Any solution (x 1, x 2,...,x n ) is an element of R n. Theorem The solution set of the system is a subspace of R n if and only if all b i = 0.