Math 3191 Applied Linear Algebra

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1 Math 3191 Applied Linear Algebra Lecture 11: Vector Spaces Stephen Billups University of Colorado at Denver Math 3191Applied Linear Algebra p.1/32

2 Announcements Study Guide 6 posted HWK 6 posted Math 3191Applied Linear Algebra p.2/32

3 Tonight Finish Determinants. Start Chapter 4: Vector Spaces Math 3191Applied Linear Algebra p.3/32

4 MORE THEOREMS Theorem 4 A square matrix is invertible if and only if det A 0. Theorem 5 If A is an n n matrix, then det A T = det A. Partial proof (2 [ 2 case) ] a b det = ad bc and c d [ ] T [ ] a b a c det = det = ad bc c d b d [ ] [ ] a b a c det = det. c d b d Math 3191Applied Linear Algebra p.4/32

5 3 3 case det a b c d e f g h i = a e f h i b d f g i + c d e g h det a d g b e h c f i = a e h f i b d g f i + c d g e h det a b c d e f g h i = det a d g b e h c f i. Math 3191Applied Linear Algebra p.5/32

6 Implications of Theorem 5 Theorem 3 still holds if the word row is replaced with. Math 3191Applied Linear Algebra p.6/32

7 THEOREM 6 (Multiplicative Property) For n n matrices A and B, det (AB) = (det A) (det B). EXAMPLE: Compute det A 3 if det A = 5. Solution: det A 3 = det (AAA) = (det A) (det A) (det A) = =. Math 3191Applied Linear Algebra p.7/32

8 EXAMPLE: For n n matrices A and B, show that A is singular if det B 0 and det AB = 0. Solution: Since (det A) (det B) = det AB = 0 and det B 0, then det A = 0. Therefore A is singular. Math 3191Applied Linear Algebra p.8/32

9 Sec. 4.1 Vector Spaces We can think of a vector space in general, as a collection of objects that behave as vectors do in R n. A vector space consists of 1. A nonempty collection V of objects (called vectors). 2. An operation (called vector addition) that maps any two vectors u, v V into another vector in V. (denoted u + v). 3. An operation (called scalar multiplication) that maps any real number c and a vector v V into another vector in V. (denoted cv). 4. Required Properties: Closure under vector addition and scalar multiplication. Usual properties of addition: (associativity, commutativity, additive identity, additive inverse). Usual properties of scalar multiplication: (multiplication identity, distributivity, associativity). Math 3191Applied Linear Algebra p.9/32

10 Example: Polynomials Consider the set of polynomials of degree 2 or less.: IP2 := {a 0 + a 1 t + a 2 t 2 a 0, a 1, a 2 IR}. In order to talk about this as a vector space we have to define vector addition and scalar multiplication: Vector addition: p + q is defined by (p + q)(t) = p(t) + q(t). Scalar multplication. cp is defined by (cp)(t) = c(p(t)). Closure Properties : Are p + q and cp in IP2? Yes. We can show that if p(t) = a 0 + a 1 t + a 2 t 2 and q(t) = b 0 + b 1 t + b 2 t 2, then 1. (p + q)(t) = (a 0 + b 0 ) + (a 1 + b 1 )t + (a 2 + b 2 )t 2, so p + q is a polynomial of degree cp(t) = (ca 0 ) + (ca 1 )t + (ca 2 )t 2, so cp is a polynomial of degree 2. Arithmetic Properties : All of the arithmetic properties we are used to also are satisfied for polynomial addition and scalar multiplication. Math 3191Applied Linear Algebra p.10/32

11 Vector Space Properties in Detail Closure Properties: C1 u, v V = u + v V. C2 v V, c IR = cv V. Vector Addition Properties (for all u, v, w V ) A1 There is a special vector (denoted 0) such that v + 0 = 0. (additive identity). A2 v has a negative (denoted v) satisfying v + ( v) = 0. (additive inverse). (Note: we shall often use the notation v u as a shorthand for v + ( u).) A3 (u + v) + w = u + (v + w). (associativity). A4 u + v = v + u. (commutivity). Math 3191Applied Linear Algebra p.11/32

12 Properties (cont.) Scalar Multiplication Properties (for all u, v V, c, d IR) S1 1u = u. (multiplicative identity). S2 c(u + v) = cu + cv. (distributivity). S3 (c + d)u = cu + du. (distributivity). S4 c(du) = (cd)u. (associativity). Math 3191Applied Linear Algebra p.12/32

13 What s the Point We will develop a lot of very useful theory about vector spaces, using only the properties of vector spaces. Much of this will be intuitive when we think about vectors in IR 2 or IR 3, but it applies to all vector spaces. As a result, much of what we learn in linear algebra, will apply to differential equations, and many other areas of applied mathematics!! Math 3191Applied Linear Algebra p.13/32

14 Vector Space Examples EXAMPLE: Let M 2 2 = {[ a c b d ] : a, b, c, d are real [ ] } In this context, note that the 0 vector is. Question: How do we define vector addition and scalar multiplication? Math 3191Applied Linear Algebra p.14/32

15 EXAMPLE: Let n 0 be an integer and let P n = the set of all polynomials of degree at most n 0. Members of P n have the form p(t) = a 0 + a 1 t + a 2 t a n t n where a 0, a 1,..., a n are real numbers and t is a real variable. The set P n is a vector space. We will just verify 3 out of the 10 axioms here. Math 3191Applied Linear Algebra p.15/32

16 proof Let p(t) = a 0 + a 1 t + + a n t n and q(t) = b 0 + b 1 t + + b n t n. Let c be a scalar. Axiom C1: The polynomial p + q is defined as follows: (p + q) (t) = p(t)+q(t). Therefore, (p + q) (t) = p(t)+q(t) = ( ) + ( ) t + + ( ) t n which is also a of degree at most. So p + q is in P n. Math 3191Applied Linear Algebra p.16/32

17 proof (cont.) Axiom A1: 0 =0 + 0t + + 0t n (zero vector in P n ) (p + 0) (t)= p(t)+0 = (a 0 + 0) + (a 1 + 0)t + + (a n + 0)t n Axiom C2: = a 0 + a 1 t + + a n t n = p(t) and so p + 0 = p (cp) (t) = cp(t) = ( ) + ( ) t + + ( ) t n which is in P n. The other 7 axioms also hold, so P n is a vector space. Math 3191Applied Linear Algebra p.17/32

18 Example Space of Continuous Functions (C(I)) Let V = C(I), the set of all continuous functions on the interval I IR. Define vector addition by (f + g)(t) := f(t) + g(t), for all t I. (In other words: if h = f + g, then h is defined by specfying its value at every t: h(t) = f(t) + g(t). Define scalar multiplication by (cf)(t) := cf(t), for all t I. Is this really a vector space? We have to check the vector space properties: Closure: Clearly f + g and cf are functions but are they continuous? Answer: Yes! (we have to use the ɛ δ defintion of continuity for the proof). Zero vector: Define the 0 function by 0(t) = 0 for all t I. Associativity: ((f+g)+h)(t) = (f+g)(t)+h(t) = f(t)+g(t)+h(t) = f(t)+(g+h)(t) = (f+(g+h))(t). etc. Math 3191Applied Linear Algebra p.18/32

19 Some Other Prominent Vector Spaces C n (I): the set of all functions on I having n continuous derivatives. C n : space of all ordered n-tuples of complex numbers. span{u 1, u 2,..., u k }: Math 3191Applied Linear Algebra p.19/32

20 Subspaces Vector spaces may be formed from subsets of other vectors spaces. These are called subspaces. A subspace of a vector space V is a subset H of V that has three properties: a. The zero vector of V is in H. b. For each u and v are in H, u + v is in H. (In this case we say H is closed under vector addition.) c. For each u in H and each scalar c, cu is in H. (In this case we say H is closed under scalar multiplication.) If the subset H satisfies these three properties, then H itself is a vector space. Math 3191Applied Linear Algebra p.20/32

21 EXAMPLE: Let H = 82 >< 6 4 >: a 0 b >= : a and b are real. Show that H is a subspace of R 3. >; Solution: Verify properties a, b and c of the definition of a subspace. a. The zero vector of R 3 is in H (let a = and b = ). b. Adding two vectors in H always produces another vector whose second entry is and therefore the sum of two vectors in H is also in H. (H is closed under addition) c. Multiplying a vector in H by a scalar produces another vector in H (H is closed under scalar multiplication). Since properties a, b, and c hold, V is a subspace of R 3. Math 3191Applied Linear Algebra p.21/32

22 Note: Vectors (a, 0, b) in H look and act like the points (a, b) in R 2. Math 3191Applied Linear Algebra p.22/32

23 EXAMPLE: Is H =? {[ x x + 1 ] : x is real I.e., does H satisfy properties a, b and c? } a subspace of 3 x x 1 Graphical Depiction of H Math 3191Applied Linear Algebra p.23/32

24 Solution All three properties must hold in order for H to be a subspace of R 2. Property (a) is not true because. Therefore H is not a subspace of R 2. x x 1 Math 3191Applied Linear Algebra p.24/32

25 Another way to show that H is not a subspace of R 2 : Let u = and v = , then u + v = 4 5 and so u + v = , which is in H. So property (b) fails and so H is not a subspace of R 2. 3 x x 1 Math 3191Applied Linear Algebra p.25/32

26 A Shortcut for Determining Subspaces THEOREM 1 If v 1,..., v p are in a vector space V, then Span{v 1,..., v p } is a subspace of V. Proof: In order to verify this, check properties a, b and c of definition of a subspace. a. 0 is in Span{v 1,..., v p } since 0 = v 1 + v v p Math 3191Applied Linear Algebra p.26/32

27 b. To show that Span{v 1,..., v p } closed under vector addition, we choose two arbitrary vectors in Span{v 1,..., v p } : u =a 1 v 1 + a 2 v a p v p and v =b 1 v 1 + b 2 v b p v p. Then u + v = (a 1 v 1 + a 2 v a p v p ) + (b 1 v 1 + b 2 v b p v p ) = ( v 1 + v 1 ) + ( v 2 + v 2 ) + + ( v p + v p ) = (a 1 + b 1 ) v 1 + (a 2 + b 2 ) v (a p + b p ) v p. So u + v is in Span{v 1,..., v p }. Math 3191Applied Linear Algebra p.27/32

28 proof (cont.) c. To show that Span{v 1,..., v p } closed under scalar multiplication, choose an arbitrary number c and an arbitrary vector in Span{v 1,..., v p } : Then v =b 1 v 1 + b 2 v b p v p. cv =c (b 1 v 1 + b 2 v b p v p ) = v 1 + v v p So cv is in Span{v 1,..., v p }. Since properties a, b and c hold, Span{v 1,..., v p } is a subspace of V. Math 3191Applied Linear Algebra p.28/32

29 Recap 1. To show that H is a subspace of a vector space, use Theorem To show that a set is not a subspace of a vector space, provide a specific example showing that at least one of the axioms a, b or c (from the definition of a subspace) is violated. Math 3191Applied Linear Algebra p.29/32

30 EXAMPLE: Is V = {(a + 2b, 2a 3b) : a and b are real} a subspace of R 2? Why or why not? Solution: Write vectors in V in column form: [ ] a + 2b 2a 3b [ = = 1 2 ] [ a 2a ] + + [ [ 2b 3b 2 3 So V =Span{v 1, v 2 } and therefore V is a subspace of by Theorem 1. ] ] Math 3191Applied Linear Algebra p.30/32

31 Is H = a + 2b a + 1 EXAMPLE: : a and b are real a subspace of R3? a Why or why not? Solution: 0 is not in H since a = b = 0 or any other combination of values for a and b does not produce the zero vector. So property fails to hold and therefore H is not a subspace of R 3. Math 3191Applied Linear Algebra p.31/32

32 EXAMPLE: [ 2a Is the set H of all matrices of the form 3a + b subspace of M 2 2? Explain. Solution: Since [ ] [ ] [ 2a b 2a 0 0 b = + 3a + b 3b 3a 0 b 3b [ ] [ ] Therefore H =Span subspace of M 2 2. = a {[ b ], [ ]} b 3b ] ] a and so H is a Math 3191Applied Linear Algebra p.32/32