Math 221: LINEAR ALGEBRA §5-1. Vector Space R n - Subspaces and Spanning Le Chen 1 Emory University, 2020 Fall (last updated on 08/18/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.
A rigorous defjnition of an abstract vector space will be given after we have Definitions 1. R denotes the set of real numbers, and is an example of a set of scalars. 2. R n is the set of all n-tuples of real numbers, i.e., R n = { ( x 1 , x 2 , . . . , x n ) | x i ∈ R , 1 ≤ i ≤ n } . 3. The vector space R n consists of the set R n written as column matrices, along with the (matrix) operations of addition and scalar multiplication. Unless stated otherwise, R n means the vector space R n . studied properties of the vector space R n .
A vectors is denoted by a lower case letter with an arrow written over it; for Notation example, � u , � v , and � x denote vectors. Example − 2 3 is a vector in R 5 , written � u ∈ R 5 . u = 0 . 7 � 5 π To save space on the page, the same vector � u may be written instead as a row matrix by taking the transpose of the column: � T . � � u = − 2 , 3 , 0 . 7 , 5 , π
The subset is a subspace of (verify this), as is the set itself. Any other subspace of is a subspace of . If is a subset of , we write . We are interested in nice subsets of R n , defjned as follows.
. The subset , we write is a subset of If . subspace of is a Any other subspace of itself. (verify this), as is the set is a subspace of We are interested in nice subsets of R n , defjned as follows. Definition A subset U of R n is a subspace of R n if S1. The zero vector of R n , � 0 n , is in U; S2. U is closed under addition, i.e., for all � u , � w ∈ U, � u + � w ∈ U; S3. U is closed under scalar multiplication, i.e., for all � u ∈ U and k ∈ R , k � u ∈ U.
. If , we write is a subset of We are interested in nice subsets of R n , defjned as follows. Definition A subset U of R n is a subspace of R n if S1. The zero vector of R n , � 0 n , is in U; S2. U is closed under addition, i.e., for all � u , � w ∈ U, � u + � w ∈ U; S3. U is closed under scalar multiplication, i.e., for all � u ∈ U and k ∈ R , k � u ∈ U. � � � is a subspace of R n (verify this), as is the set R n itself. The subset U = 0 n Any other subspace of R n is a proper subspace of R n .
We are interested in nice subsets of R n , defjned as follows. Definition A subset U of R n is a subspace of R n if S1. The zero vector of R n , � 0 n , is in U; S2. U is closed under addition, i.e., for all � u , � w ∈ U, � u + � w ∈ U; S3. U is closed under scalar multiplication, i.e., for all � u ∈ U and k ∈ R , k � u ∈ U. � � � is a subspace of R n (verify this), as is the set R n itself. The subset U = 0 n Any other subspace of R n is a proper subspace of R n . Notation If U is a subset of R n , we write U ⊆ R n .
Example In R 3 , the line L through the origin that is parallel to the vector − 5 x − 5 � has (vector) equation = t , t ∈ R , so d = 1 y 1 − 4 z − 4 � � t � L = d | t ∈ R . Claim. L is a subspace of R 3 . 0 3 ∈ L since 0 � ◮ First: � d = � 0 3 . u = s � v = t � ◮ Suppose � u ,� v ∈ L. Then by definition, � d and � d, for some s , t ∈ R . Thus v = s � d + t � d = ( s + t ) � u + � d . � Since s + t ∈ R , � u + � v ∈ L; i.e., L is closed under addition.
Example 4 (continued) u = t � ◮ Suppose � u ∈ L and k ∈ R ( k is a scalar). Then � d , for some t ∈ R , so u = k ( t � d ) = ( kt ) � k � d . Since kt ∈ R , k � u ∈ L ; i.e., L is closed under scalar multiplication. Therefore, L is a subspace of R 3 . Note that there is nothing special about the vector � d used in this example; the same proof works for any nonzero vector � d ∈ R 3 , so any line through the origin is a subspace of R 3 .
Example In R 3 , let M denote the plane through the origin having equation 3 x 3 x − 2 y + z = 0 ; then M has normal vector � n = − 2 . If � u = y , then 1 z u ∈ R 3 | � � � M = n • � u = 0 � , where � n • � u is the dot product of vectors � n and � u. Claim. M is a subspace of R 3 . ◮ First: � n • � 0 3 ∈ M since � 0 3 = 0 . ◮ Suppose � u ,� v ∈ M. Then by definition, � n • � u = 0 and � n • � v = 0 , so � n • ( � u + � v ) = n • � u + n • � v = 0 + 0 = 0 , and thus ( � u + � v ) ∈ M; i.e., M is closed under addition.
As in the previous example, there is nothing special about the plane chosen Example 6 (continued) ◮ Suppose � u ∈ M and k ∈ R . Then � n • � u = 0 , so n • ( k � u ) = k ( � n • � u ) = k (0) = 0 , � and thus k � u ∈ M ; i.e., M is closed under scalar multiplication. Therefore, M is a subspace of R 3 . for this example; any plane through the origin is a subspace of R 3 .
. The zero vector of . Therefore, , so and In this case, . with is the vector Problem � a � � b � a subspace of R 4 ? Is U = a , b , c , d ∈ R and 2 a − b = c + 2 d � c � � d � Justify your answer.
Problem � a � � b � a subspace of R 4 ? Is U = a , b , c , d ∈ R and 2 a − b = c + 2 d � c � � d � Justify your answer. Solution 1 a b The zero vector of R 4 is the vector with a = b = c = d = 0 . c d In this case, 2 a − b = 2(0) + 0 = 0 and c + 2 d = 0 + 2(0) = 0 , so 2 a − b = c + 2 d . Therefore, � 0 4 ∈ U .
Suppose and and Solution (continued) a 1 a 2 b 1 b 2 � v 1 = � v 2 = are in U . c 1 c 2 d 1 d 2 Then 2 a 1 − b 1 = c 1 + 2 d 1 and 2 a 2 − b 2 = c 2 + 2 d 2 . Now a 1 a 2 a 1 + a 2 b 1 b 2 b 1 + b 2 � v 1 + � v 2 = + = , c 1 c 2 c 1 + c 2 d 1 d 2 d 1 + d 2 2( a 1 + a 2 ) − ( b 1 + b 2 ) = (2 a 1 − b 1 ) + (2 a 2 − b 2 ) = ( c 1 + 2 d 1 ) + ( c 2 + 2 d 2 ) = ( c 1 + c 2 ) + 2( d 1 + d 2 ) . v 1 + � v 2 ∈ U . Therefore, �
and Finally, suppose and Solution (continued) a b v = � ∈ U k ∈ R . c d Then 2 a − b = c + 2 d . Now a ka b kb k � v = k = , c kc d kd 2 ka − kb = k (2 a − b ) = k ( c + 2 d ) = kc + 2 kd . Therefore, k � v ∈ U . It follows from the Subspace Test that U is a subspace of R 4 .
scalar multiplication.) Note that is not closed under addition, or not closed under (You could also show that . is not a subspace of , and thus Problem � 1 � � a subspace of R 3 ? Justify your answer. Is U = s s , t ∈ R � � t �
scalar multiplication.) Problem � 1 � � a subspace of R 3 ? Justify your answer. Is U = s s , t ∈ R � � t � Solution Note that � 0 3 �∈ U , and thus U is not a subspace of R 3 . (You could also show that U is not closed under addition, or not closed under
. if and only if with equality if and only if . Similarly, implies , and if and only if . This means if and only if ; thus . Therefore Since contains only , the zero vector, i.e., . As we already observed, is a subspace of , and therefore is a subspace of , Problem � r � r 2 + s 2 = 0 � a subspace of R 3 ? Is U = 0 r , s ∈ R and � � s � Justify your answer.
Problem � r � r 2 + s 2 = 0 � a subspace of R 3 ? Is U = 0 r , s ∈ R and � � s � Justify your answer. Solution Since r ∈ R , r 2 ≥ 0 with equality if and only if r = 0 . Similarly, s ∈ R implies s 2 ≥ 0 , and s 2 = 0 if and only if s = 0 . This means r 2 + s 2 = 0 if and only if r 2 = s 2 = 0 ; thus r 2 + s 2 = 0 if and only if r = s = 0 . Therefore U contains only � 0 3 , the zero vector, i.e., U = { � 0 3 } . As we already observed, { � 0 n } is a subspace of R n , and therefore U is a subspace of R 3 .
Definitions Let A be an m × n matrix. The null space of A is defined as x ∈ R n | A � x = � null ( A ) = { � 0 m } , and the image space of A is defined as x ∈ R n } . im ( A ) = { A � x | � x ∈ R m , so im ( A ) ⊆ R m while x ∈ R n , A � Note. Since A is m × n and � null ( A ) ⊆ R n .
. . is a subspace of Therefore, null . null and thus , so . Then and null Let null and thus , so and . Then null Let . null , Since Problem Prove that if A is an m × n matrix, then null ( A ) is a subspace of R n .
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