Math 221: LINEAR ALGEBRA Chapter 6. Vector Spaces 6-3. Vector - PowerPoint PPT Presentation

Math 221: LINEAR ALGEBRA Chapter 6. Vector Spaces §6-3. Vector Spaces - Linear Independence Le Chen 1 Emory University, 2020 Fall (last updated on 10/09/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.

Linear Independence Definition Let V be a vector space and S = { u 1 , u 2 , . . . , u k } a subset of V. The set S is linearly independent or simply independent if the following condition holds: if s 1 u 1 + s 2 u 2 + · · · + s k u k = 0 then s 1 = s 2 = · · · = s k = 0 , i.e., the only linear combination that vanishes is the trivial one. If S is not linearly independent, then S is said to be dependent.

Linear Independence Definition Let V be a vector space and S = { u 1 , u 2 , . . . , u k } a subset of V. The set S is linearly independent or simply independent if the following condition holds: if s 1 u 1 + s 2 u 2 + · · · + s k u k = 0 then s 1 = s 2 = · · · = s k = 0 , i.e., the only linear combination that vanishes is the trivial one. If S is not linearly independent, then S is said to be dependent. Example    − 1   1   1     ,  ,  is a dependent subset of R 3 The set S = 0 1 3     1 1 5  because  − 1   1   1   0   + b  + c  = a 0 1 3 0      1 1 5 0 has nontrivial solutions, for example a = 2 , b = 3 and c = − 1 .

Problem Is the set T = { 3 x 2 − x + 2 , x 2 + x − 1 , x 2 − 3 x + 4 } an independent subset of P 2 ?

Problem Is the set T = { 3 x 2 − x + 2 , x 2 + x − 1 , x 2 − 3 x + 4 } an independent subset of P 2 ? Solution Suppose a (3 x 2 − x + 2) + b ( x 2 + x − 1) + c ( x 2 − 3 x + 4) = 0 , for some a , b , c ∈ R . Then x 2 (3 a + b + c ) + x ( − a + b − 3 c ) + (2 a − b + 4 c ) = 0 , implying that 3 a + b + c = 0 − a + b − 3 c = 0 2 a − b + 4 c = 0 .

Problem Is the set T = { 3 x 2 − x + 2 , x 2 + x − 1 , x 2 − 3 x + 4 } an independent subset of P 2 ? Solution Suppose a (3 x 2 − x + 2) + b ( x 2 + x − 1) + c ( x 2 − 3 x + 4) = 0 , for some a , b , c ∈ R . Then x 2 (3 a + b + c ) + x ( − a + b − 3 c ) + (2 a − b + 4 c ) = 0 , implying that 3 a + b + c = 0 − a + b − 3 c = 0 2 a − b + 4 c = 0 . Solving this linear system of three equations in three variables         3 1 1 0 1 0 1 0 a − t  →  ;  = − 1 1 − 3 0 0 1 − 2 0 b 2 t  , t ∈ R .     2 − 1 4 0 0 0 0 0 c t Since 1(3 x 2 − x + 2) − 2( x 2 + x − 1) − 1( x 2 − 3 x + 4) = 0 , T is a dependent subset of P 2 .

M Problem �� 1 � 0 � 1 � � �� 1 1 0 Is U = an independent subset of M 22 ? , , 0 1 1 0 1 1

M Problem �� 1 � 0 � 1 � � �� 1 1 0 Is U = an independent subset of M 22 ? , , 0 1 1 0 1 1 Solution � 1 � 0 � 1 � 0 � � � � 1 1 0 0 Suppose a + b + c = for some 0 1 1 0 1 1 0 0 a , b , c ∈ R .

M Problem �� 1 � 0 � 1 � � �� 1 1 0 Is U = an independent subset of M 22 ? , , 0 1 1 0 1 1 Solution � 1 � 0 � 1 � 0 � � � � 1 1 0 0 Suppose a + b + c = for some 0 1 1 0 1 1 0 0 a , b , c ∈ R . Then a + c = 0 , a + b = 0 , b + c = 0 , a + c = 0 .

Problem �� 1 � 0 � 1 � � �� 1 1 0 Is U = an independent subset of M 22 ? , , 0 1 1 0 1 1 Solution � 1 � 0 � 1 � 0 � � � � 1 1 0 0 Suppose a + b + c = for some 0 1 1 0 1 1 0 0 a , b , c ∈ R . Then a + c = 0 , a + b = 0 , b + c = 0 , a + c = 0 . This system of four equations in three variables has unique solution a = b = c = 0 , and therefore U is an independent subset of M 22 .

Any set of polynomials with distinct degrees is independent. For example, is an independent subset of . Example e n } (the standard basis of R n ) is an As we saw earlier, { � e 1 ,� e 2 , . . . ,� independent subset of R n .

For example, Any set of polynomials with distinct degrees is independent. is an independent subset of . Example e n } (the standard basis of R n ) is an As we saw earlier, { � e 1 ,� e 2 , . . . ,� independent subset of R n . Example (An independent subset of P n ) Consider { 1 , x , x 2 , . . . , x n } , and suppose that a 0 · 1 + a 1 x + a 2 x 2 + · · · + a n x n = 0 for some a 0 , a 1 , . . . , a n ∈ R . Then a 0 = a 1 = · · · = a n = 0 , and thus { 1 , x , x 2 , . . . , x n } is an independent subset of P n .

Any set of polynomials with distinct degrees is independent. For example, Example e n } (the standard basis of R n ) is an As we saw earlier, { � e 1 ,� e 2 , . . . ,� independent subset of R n . Example (An independent subset of P n ) Consider { 1 , x , x 2 , . . . , x n } , and suppose that a 0 · 1 + a 1 x + a 2 x 2 + · · · + a n x n = 0 for some a 0 , a 1 , . . . , a n ∈ R . Then a 0 = a 1 = · · · = a n = 0 , and thus { 1 , x , x 2 , . . . , x n } is an independent subset of P n . Polynomials with distinct degrees { 2 x 4 − x 3 + 5 , − 3 x 3 + 2 x 2 + 2 , 4 x 2 + x − 3 , 2 x − 1 , 3 } is an independent subset of P 4 .

Any set of polynomials with distinct degrees is independent. For example, Example e n } (the standard basis of R n ) is an As we saw earlier, { � e 1 ,� e 2 , . . . ,� independent subset of R n . Example (An independent subset of P n ) Consider { 1 , x , x 2 , . . . , x n } , and suppose that a 0 · 1 + a 1 x + a 2 x 2 + · · · + a n x n = 0 for some a 0 , a 1 , . . . , a n ∈ R . Then a 0 = a 1 = · · · = a n = 0 , and thus { 1 , x , x 2 , . . . , x n } is an independent subset of P n . Polynomials with distinct degrees { 2 x 4 − x 3 + 5 , − 3 x 3 + 2 x 2 + 2 , 4 x 2 + x − 3 , 2 x − 1 , 3 } is an independent subset of P 4 . How would you prove this?

. M M , constitutes an independent subset of , zeros elsewhere, and matrices that have a ‘1’ in position In general, the set of Example               1 0 0 1 0 0 0 0 0 0 0 0   U = 0 0 0 0 1 0 0 1 0 0 0 0   ,   ,   ,   ,   ,   0 0 0 0 0 0 0 0 1 0 0 1   is an independent subset of M 32 .

Example               1 0 0 1 0 0 0 0 0 0 0 0   U = 0 0 0 0 1 0 0 1 0 0 0 0   ,   ,   ,   ,   ,   0 0 0 0 0 0 0 0 1 0 0 1   is an independent subset of M 32 . An independent subset of M mn In general, the set of mn m × n matrices that have a ‘1’ in position ( i , j ) and zeros elsewhere, 1 ≤ i ≤ m , 1 ≤ j ≤ n , constitutes an independent subset of M mn .

Example Let V be a vector space. 1. If v is a nonzero vector of V, then { v } is an independent subset of V. Proof. Suppose that k v = 0 for some k ∈ R . Since v � = 0 , it must be that k = 0 , and therefore { v } is an independent set. 2. The zero vector of V, 0 is never an element of an independent subset of V. Proof. Suppose S = { 0 , v 2 , v 3 , . . . , v k } is a subset of V. Then 1( 0 ) + 0( v 2 ) + 0( v 3 ) + · · · + 0( v k ) = 0 . Since the coefficient of 0 (on the left-hand side) is ‘ 1 ’, we have a nontrivial vanishing linear combination of the vectors of S. Therefore S is dependent.

u v u w v 5w u v w u v w Problem Let V be a vector space and let { u , v , w } be an independent subset of V. Is S = { u + v , 2 u + w , v − 5w } an independent subset of V? Justify your answer.

Problem Let V be a vector space and let { u , v , w } be an independent subset of V. Is S = { u + v , 2 u + w , v − 5w } an independent subset of V? Justify your answer. Solution Suppose that a linear combination of the vectors of S is equal to zero, i.e., a ( u + v ) + b (2 u + w ) + c ( v − 5w ) = 0 for some a , b , c ∈ R . Then ( a + 2 b ) u + ( a + c ) v + ( b − 5 c ) w = 0 . Since { u , v , w } is independent, a + 2 b = 0 a + c = 0 b − 5 c = 0 . Solving for a , b and c, we find that the system has unique solution a = b = c = 0 . Therefore, S is linearly independent.

it equal to the Use the standard approach: take a linear combination of the matrices and set zero matrix. Two key points: Since 0 , the matrices are all nonzero. Since 0 , the matrices are all zero. Problem Suppose that A is an n × n matrix with the property that A k = 0 but A k − 1 � = 0 . Prove that B = { I , A , A 2 , . . . , A k − 1 } is an independent subset of M nn .

Use the standard approach: take a linear combination of the matrices and set Problem Suppose that A is an n × n matrix with the property that A k = 0 but A k − 1 � = 0 . Prove that B = { I , A , A 2 , . . . , A k − 1 } is an independent subset of M nn . Hint it equal to the n × n zero matrix. Two key points: ◮ Since A k − 1 � = 0 , the matrices A , A 2 , . . . , A k − 2 are all nonzero. ◮ Since A k = 0 , the matrices A k +1 , A k +2 , A k +3 , . . . are all zero.

Again, the proof of the corresponding result for generalizes to an arbitrary vector space . Theorem Let V be a vector space and let U = { v 1 , v 2 , . . . , v k } ⊆ V be an independent set. If v is in span ( U ) , then v has a unique representation as a linear combination of elements of U.