# MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set.

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1 MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set.

2 Vector space A vector space is a set V equipped with two operations, addition V V (x,y) x + y V and scalar multiplication R V (r,x) rx V, that have the following properties:

3 Properties of addition and scalar multiplication 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.

4 Examples of vector spaces 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 {0}: the trivial vector space F(R): the set of all functions f : R R C(R): all continuous functions f : R 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

5 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. 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. Remarks. The zero vector in a subspace is the same as the zero vector in V. Also, the subtraction in a subspace agrees with that in V.

6 Examples of subspaces 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 n 1 x n 1 P n : polynomials of degree less than n P n is a subspace of P. Any vector space V {0}, where 0 is the zero vector in V The trivial space {0} is a subspace of V.

7 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

8 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 line or a plane in R 3 is a subspace if and only if it passes through the origin.

9 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. Proof: only if : the zero vector 0 = (0, 0,...,0) is a solution only if all equations are homogeneous. if : a system of homogeneous linear equations is equivalent to a matrix equation Ax = 0. A0 = 0 = 0 is a solution = solution set is not empty. If Ax = 0 and Ay = 0 then A(x + y) = Ax + Ay = 0. If Ax = 0 then A(rx) = r(ax) = 0.

10 Examples of subspaces of M 2,2 (R): A = diagonal matrices: b = c = 0 upper triangular matrices: c = 0 lower triangular matrices: b = 0 symmetric matrices (A T = A): b = c ( ) a b c d anti-symmetric (or skew-symmetric) matrices (A T = A): a = d = 0, c = b matrices with zero trace: a + d = 0 (trace = the sum of diagonal entries) matrices with zero determinant, ( ) ( ad ) bc ( = 0, ) do not form a subspace: + =

11 Let V be a vector space and v 1,v 2,...,v n V. Consider the set L of all linear combinations r 1 v 1 + r 2 v r n v n, where r 1, r 2,...,r n R. Theorem L is a subspace of V. Proof: First of all, L is not empty. For example, 0 = 0v 1 + 0v v n belongs to L. The set L is closed under addition since (r 1 v 1 +r 2 v 2 + +r n v n ) + (s 1 v 1 +s 2 v 2 + +s n v n ) = = (r 1 +s 1 )v 1 + (r 2 +s 2 )v (r n +s n )v n. The set L is closed under scalar multiplication since t(r 1 v 1 +r 2 v 2 + +r n v n ) = (tr 1 )v 1 +(tr 2 )v 2 + +(tr n )v n.

12 Span: implicit definition Let S be a subset of a vector space V. Definition. The span of the set S, denoted Span(S), is the smallest subspace of V that contains S. That is, Span(S) is a subspace of V; for any subspace W V one has S W = Span(S) W. Remark. The span of any set S V is well defined (it is the intersection of all subspaces of V that contain S).

13 Span: effective description Let S be a subset of a vector space V. If S = {v 1,v 2,...,v n } then Span(S) is the set of all linear combinations r 1 v 1 + r 2 v r n v n, where r 1, r 2,...,r n R. If S is an infinite set then Span(S) is the set of all linear combinations r 1 u 1 + r 2 u r k u k, where u 1,u 2,...,u k S and r 1, r 2,...,r k R (k 1). If S is the empty set then Span(S) = {0}.

14 Examples of subspaces of M 2,2 (R): ( ) ( ) The span of and consists of all matrices of the form a ( ) + b ( ) = ( ) a 0. 0 b This is the subspace of diagonal matrices. ( ) ( ) ( ) The span of,, and consists of all matrices of the form a ( ) + b ( ) + c ( ) = ( ) a c. c b This is the subspace of symmetric matrices.

15 Examples of subspaces of M 2,2 (R): ( ) 0 1 The span of is the subspace of 1 0 anti-symmetric matrices. ( ) ( ) ( ) The span of,, and is the subspace of upper triangular matrices. ( ) ( ) ( ) ( ) The span of,,, is the entire space M 2,2 (R).

16 Spanning set Definition. A subset S of a vector space V is called a spanning set for V if Span(S) = V. Examples. Vectors e 1 = (1, 0, 0), e 2 = (0, 1, 0), and e 3 = (0, 0, 1) form a spanning set for R 3 as Matrices (x, y, z) = xe 1 + ye 2 + ze 3. ( ) 1 0, 0 0 ( ) form a spanning set for M 2,2 (R) as ( a b c d ) = a ( ) + b ( , ) + c ( ) 0 0, 1 0 ( ) + d ( ) ( )

17 Problem Let v 1 = (1, 2, 0), v 2 = (3, 1, 1), and w = (4, 7, 3). Determine whether w belongs to Span(v 1,v 2 ). We have to check if there exist r 1, r 2 R such that w = r 1 v 1 + r 2 v 2. This vector equation is equivalent to a system of linear equations: 4 = r 1 + 3r 2 7 = 2r 1 + r 2 3 = 0r 1 + r 2 { r1 = 5 r 2 = 3 Thus w = 5v 1 + 3v 2 Span(v 1,v 2 ).

18 Problem Let v 1 = (2, 5) and v 2 = (1, 3). Show that {v 1,v 2 } is a spanning set for R 2. Take any vector w = (a, b) R 2. We have to check that there exist r 1, r 2 R such that { 2r1 + r w = r 1 v 1 +r 2 v 2 2 = a 5r 1 + 3r 2 = b ( ) 2 1 Coefficient matrix: C =. det C = Since the matrix C is invertible, the system has a unique solution for any a and b. Thus Span(v 1,v 2 ) = R 2.

19 Problem Let v 1 = (2, 5) and v 2 = (1, 3). Show that {v 1,v 2 } is a spanning set for R 2. Alternative solution: First let us show that vectors e 1 = (1, 0) and e 2 = (0, 1) belong to Span(v 1,v 2 ). { { 2r1 + r e 1 = r 1 v 1 +r 2 v 2 2 = 1 5r 1 + 3r 2 = 0 r1 = 3 r 2 = 5 { { 2r1 + r e 2 = r 1 v 1 +r 2 v 2 2 = 0 5r 1 + 3r 2 = 1 r1 = 1 r 2 = 2 Thus e 1 = 3v 1 5v 2 and e 2 = v 1 + 2v 2. Then for any vector w = (a, b) R 2 we have w = ae 1 + be 2 = a(3v 1 5v 2 ) + b( v 1 + 2v 2 ) = (3a b)v 1 + ( 5a + 2b)v 2.

20 Problem Let v 1 = (2, 5) and v 2 = (1, 3). Show that {v 1,v 2 } is a spanning set for R 2. Remarks on the alternative solution: Notice that R 2 is spanned by vectors e 1 = (1, 0) and e 2 = (0, 1) since (a, b) = ae 1 + be 2. This is why we have checked that vectors e 1 and e 2 belong to Span(v 1,v 2 ). Then e 1,e 2 Span(v 1,v 2 ) = Span(e 1,e 2 ) Span(v 1,v 2 ) = R 2 Span(v 1,v 2 ) = Span(v 1,v 2 ) = R 2. In general, to show that Span(S 1 ) = Span(S 2 ), it is enough to check that S 1 Span(S 2 ) and S 2 Span(S 1 ).

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