Math 211 Math 211 Lecture #19 Nullspaces and Subspaces October - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #19 Nullspaces and Subspaces October 10, 2001

2 Structure of the Solution Set Structure of the Solution Set Let x p be a particular solution to A x p = b . Theorem: 1. If A x h = 0 then x = x p + x h also satisfies A x = b . 2. If A x = b , then there is a vector x h such that A x h = 0 and x = x p + x h . • The solution set for A x = b is known if we know one particular solution x p and the solution set for the homogeneous system A x h = 0 . Return

3 Solution Set of a Homogeneous System Solution Set of a Homogeneous System Our goal is to understand such sets better. In particular we want answers to the following questions. • What are the properties of these solution sets? • Is there a convenient way to describe them? Return Solution set

4 Nullspace of a Matrix Nullspace of a Matrix The nullspace of a matrix A is the set null( A ) = { x | A x = 0 } . • The nullspace of A is the same as the solution set for the homogeneous system A x = 0 . Return

5 Properties of a Nullspace Properties of a Nullspace Let A be a matrix. Proposition: 1. If x and y are in null( A ) , then x + y is in null( A ) . 2. If a is a scalar and x is in null( A ) , then a x is in null( A ) . Return

6 Subspaces of R n Subspaces of R n A nonempty subset V of R n that has the Definition: properties 1. if x and y are vectors in V , x + y is in V , 2. if a is a scalar, and x is in V , then a x is in V , is called a subspace of R n . • A nullspace of a matrix is a subspace. Return

7 Examples of Subspaces Examples of Subspaces • The nullspace of a matrix is a subspace. • A line through 0 is a subspace. V = { t v | t ∈ R } . • A plane through 0 is a subspace. V = { a v + b w | a, b ∈ R } . • { 0 } and R n are subspaces of R n . � These are called the trivial subspaces. Return

8 Linear Combinations Linear Combinations Any linear combination of vectors in a Proposition: subspace V is also in V . • Subspaces of R n have the same kind of linear structure as R n itself. • In particular nullspaces of matrices have the same kind of linear structure as R n . Return

9 Example 1 Example 1 4 3 − 1   A = − 3 − 2 1   1 2 1 The nullspace of A is null( A ) = { a v | a ∈ R } , where v = (1 , − 1 , 1) T . Return

10 Example 2 Example 2 4 3 − 1 6   B = − 3 − 2 1 − 4   1 2 1 4 • null( B ) = { a v + b w | a, b ∈ R } , where v = (1 , − 1 , 1 , 0) T and w = (0 , − 2 , 0 , 1) T . • null( B ) consists of all linear combinations of v and w . Return

11 The Span of a Set of Vectors The Span of a Set of Vectors In every example the subspace has been the set of all linear combinations of a few vectors. The span of a set of vectors is the set of all Definition: linear combinations of those vectors. The span of the vectors v 1 , v 2 , . . . , and v k is denoted by span( v 1 , v 2 , . . . , v k ) . If v 1 , v 2 , . . . , and v k are all vectors in Proposition: R n , then V = span( v 1 , v 2 , . . . , v k ) is a subspace of R n . null( A ) null( B ) Return Examples Nullspace

12 When is w in span( v 1 , v 2 , . . . , v k ) ? When is w in span( v 1 , v 2 , . . . , v k ) ? ⇔ w is a linear combination of v 1 , v 2 , . . . , and v k ? ⇔ There are constants a 1 , a 2 , . . . , a k such that a 1 v 1 + a 2 v 2 + . . . + a k v k = w . There is a solution a = ( a 1 , a 2 , · · · , a k ) T to the system ⇔ V a = w , where V = [ v 1 , v 2 , · · · . v k ] . Return Span

13 Examples Examples Let v 1 = (1 , 2) T , v 2 = (1 , 0) T , and v 3 = (2 , 0) T . • span( v 1 , v 2 ) = R 2 . (Proof?) • span( v 1 , v 3 ) = R 2 . (Proof?) • span( v 2 , v 3 ) = span( v 2 ) . (Proof?) � span( v 2 , v 3 ) = { t v 2 | t ∈ R } . � v 2 and v 3 have the same direction. Return Span

14 Linear Independence of Two Vectors Linear Independence of Two Vectors We need a condition that will keep unneeded vectors out of a spanning list. We will work toward a general definition. • Two vectors are linearly dependent if one is a scalar multiple of the other. � v 2 and v 3 are linearly dependent. � v 1 and v 2 are linearly independent. Return

15 Linear Independence of Three Vectors Linear Independence of Three Vectors • Three vectors v 1 , v 2 , and v 3 are linearly dependent if one is a linear combination of the other two. � Example: v 1 = (1 , 0 , 0) T , v 2 = (0 , 1 , 0) T , and v 3 = (1 , 2 , 0) T v 3 = v 1 + 2 v 2 . � Notice that v 1 + 2 v 2 − v 3 = 0 . Return

16 Linear Independence Linear Independence • Three vectors are linearly dependent if there is a non-trivial linear combination of them which equals the zero vector. � Non-trivial means that at least one of the coefficients is not 0. • A set of vectors is linearly dependent if there is a non-trivial linear combination of them which equals the zero vector. Return

17 Linear Independence Linear Independence The vectors v 1 , v 2 , . . . , and v k are linearly Definition: independent if the only linear combination of them which is equal to the zero vector is the one with all of the coefficients equal to 0. • In symbols, c 1 v 1 + c 2 v 2 + · · · + c k v k = 0 ⇒ c 1 = c 2 = · · · = c k = 0 . Return Two vectors Three vectors More vectors

18 Basis of a Subspace Basis of a Subspace A set of vectors v 1 , v 2 , . . . , and v k form a Definition: basis of a subspace V if 1. V = span( v 1 , v 2 , . . . , v k ) 2. v 1 , v 2 , . . . , and v k are linearly independent. Return Span

19 Examples of Bases Examples of Bases • The vector v = (1 , − 1 , 1) T is a basis for null( A ) . � null( A ) is the subspace of R 3 with basis v . • The vectors v = (1 , − 1 , 1 , 0) T and w = (0 , − 2 , 0 , 1) T form a basis for null( B ) . � null( B ) is the subspace of R 4 with basis { v , w } . Return Examples Nullspace

20 Basis of a Subspace Basis of a Subspace Let V be a subspace of R n . Proposition: 1. If V � = { 0 } , then V has a basis. 2. Every basis of V has the same number of elements. The dimension of a subspace V is the Definition: number of elements in a basis of V . Return Examples Examples Nullspace

21 Linear Independence? Linear Independence? How do we decide if a set of vectors is linearly independent? 1 − 1 5       − 2 − 3 0       v 1 =  , v 2 =  , v 3 =       0 2 − 4     2 0 6 Return

22 c 1 v 1 + c 2 v 2 + c 3 v 3 = 0 ⇔ [ v 1 , v 2 , v 3 ] c = 0 ⇔ c ∈ null([ v 1 , v 2 , v 3 ]) . • c = ( − 3 , 2 , 1) T ∈ null([ v 1 , v 2 , v 3 ]) , ⇒ − 3 v 1 + 2 v 2 + v 3 = 0 . • v 1 , v 2 , v 3 are linearly dependent. Return Example

23 1 − 1 5       − 2 − 3 0       v 1 =  , v 2 =  , v 3 =       0 2 − 4     2 0 3 • null([ v 1 , v 2 , v 3 ]) = { 0 } . • v 1 , v 2 , v 3 are linearly independent. Return Method

24 Suppose that v 1 , v 2 , . . . , and v k are Proposition: vectors in R n . Set V = [ v 1 , v 2 , · · · , v k ] . 1. If null( V ) = { 0 } , then v 1 , v 2 , . . . , and v k are linearly independent. 2. If c = ( c 1 , c 2 , . . . , c k ) T is a nonzero vector in null( V ) , then c 1 v 1 + c 2 v 2 + · · · + c k v k = 0 , so the vectors are linearly dependent.