Linear dependence and independence Linear dependence 1 Definition - PowerPoint PPT Presentation

L INEAR A LGEBRA Berkant Ustao˘ glu CRYPTOLOUNGE . NET

Linear dependence and independence

Linear dependence 1 Definition (linear (in)dependence) Let { � v 1 , � v 2 , . . . , � v k } be a set of vectors. If v k = � a 1 � v 1 + a 2 � v 2 + · · · + a k � ⇒ a 1 = a 2 = · · · = a k = 0 0 then the vectors � v 1 , � v 2 , . . . , � v k are called linearly independent otherwise they are linearly dependent .

C 4 example 2           1 0 0 0 0 0 1 0 0 0            x 1 +  x 2 +  x 3 +  x 4 =           0 0 1 0 0       0 0 0 1 0

Linear dependence 3 Theorem The standard basis vectors are linearly independent, in other words the columns and rows of I are linearly independent.

C 2 example 4 � 4 � 1 � 2 � � � Are , and linearly dependent? 3 1 2 � 4 � 1 � 2 � 0 � � � � x 1 + x 2 + x 3 = 3 1 2 0 ◮ x 1 = x 2 = x 3 = 0 is solution ◮ x 1 = 0 , x 2 = 2 and x 3 = − 1 is another solution

C 2 example 5 � 4 � 1 � � Are and linearly dependent? 3 1 � 4 � 1 � 0 � � � x 1 + x 2 = 3 1 0 1. x 1 = 0 and x 2 = 0 is the only solution

CVS example 6 � 4 � 1 � � Are and linearly dependent? 3 1 � 4 � 1 � 4 � � � � � � � α ⊙ ⊕ β ⊙ = 3 1 3 look at � ( 4 α − 4 α + 4 ) + ( β − 4 β + 4 ) − 4 � 4 � � = ( 3 α − 3 α + 3 ) + ( β − 3 β + 3 ) − 3 3 ◮ α = 0 and β = 0 is a solution ◮ α = 7 and β = 0 is a solution

CVS example 7 � 4 � Is linearly dependent? 3 � 4 � 4 � � α ⊙ = 3 3 look at � 4 α − 4 α + 4 � 4 � � = 3 α − 3 α + 3 3 ◮ α = 0 is a solution ◮ α = 7 is a solution

C 2 example 8 � 4 � Is linearly dependent? 3 � 4 � 4 α � 0 � � � α = = 3 α 3 0 ◮ α = 0 is the only solution

Functions 9 ◮ Are the functions p 0 ( x ) = x 0 , p 1 ( x ) = x 1 and p 2 ( x ) = x 2 linearly dependent or independent? ◮ Are the functions 5 x 0 , sin 2 x and 3 cos 2 x linearly dependent or independent?

Linear dependence 10 Theorem Let { � v 1 , � v 2 , . . . , � v k } be a collection of vectors. If k = 1 the system of vectors is linearly dependent if and only if v 1 = � � 0 .

Linear dependence 11 Theorem Let { � v 1 , � v 2 , . . . , � v k } be a collection of vectors. If for some v i = � 1 ≤ i ≤ k we have that � 0 then the system of vectors is linearly dependent.

Linear dependence 12 Theorem Let { � v 1 , � v 2 , . . . , � v k } be a collection of vectors. If for some 1 ≤ i � = j ≤ k we have that � v i = � v j then the system of vectors is linearly dependent.

Linear dependence 13 Theorem Let { � v 1 , � v 2 , . . . , � v k } be a collection of linearly dependent vectors and k > 1 . Then there is an index i such that � v i can be written as a linear combination of the remaining vectors.

Linear dependence 14 Theorem Let { � v 1 , � v 2 , . . . , � v k } be a collection of vectors. If a subset of { � v 1 , � v 2 , . . . , � v k } is linearly dependent then { � v 1 , � v 2 , . . . , � v k } is also linearly dependent.

Linear independence 15 Theorem Let { � v 1 , � v 2 , . . . , � v k } be linearly independent, then any subset of { � v 1 , � v 2 , . . . , � v k } is also linearly independent.

Span 16 Theorem S ⊂ V and � u ∈ V , then � S ∪ � u � = � S � ⇐ ⇒ � u ∈ � S � Corollary If � u ∈ S , then � S \ � ⇒ � u � = � S � ⇐ u ∈ � S �

Span 17 Theorem Let S = { � v 1 ,� v 2 , . . . ,� v k } . Then S is linearly independent if and only if ∀ i ∈ [ 1 , . . . , k ] , � S \ � v i � � � S �

Span 18 Theorem Let S = { � v 1 ,� v 2 , . . . ,� v k } . Then there is a set B that is linearly independent and � S � = � B �

A sufficient condition 19 Theorem � � b 1 , � � b 2 , . . . , � Let { � a 1 , � a 2 , . . . , � a k } and b s be two sets of vectors. Suppose that for each 1 ≤ i ≤ k we have that � a i is � � b 1 , � � b 2 , . . . , � a linear combination of b s . Suppose also k > s then { � a 1 , � a 2 , . . . , � a k } is linearly dependent.

� b s independent 20 � 1 � 0 � 0 � � � � � 2 0 0 0 0 = − 3 + 3 2 − 3 3 0 0 1 0 0 1 � 2 � 1 � 0 � 0 � � � � 0 0 0 0 = 2 + 4 + 4 4 4 0 0 1 0 0 1 � 1 � 1 � 0 � 0 � � � � 0 0 0 0 = + 2 + 2 2 2 0 0 1 0 0 1 � 1 � 1 � 0 � 0 � � � � 0 0 0 0 = + 3 + 3 1 0 0 1 0 0 1

� b s dependent 21 � 8 � 1 � 0 � 2 � � � � 8 1 0 2 = − 3 + 3 2 3 3 0 0 1 1 2 2 � 3 � 1 � 0 � 2 � � � � 3 1 0 2 = + 4 + 6 6 0 0 1 1 2 2 � 5 � 1 � 0 � 2 � � � � 5 1 0 2 = + 3 + 2 7 7 0 0 1 1 2 2 � 3 � 1 � 0 � 2 � � � � 3 1 0 2 = + 2 + 4 4 0 0 1 1 2 2

k ≤ s 22 � 1 � 0 � 0 � � � � � 2 2 1 0 0 = − 3 + 1 2 − 3 1 0 0 1 0 0 1 � 0 � 1 � 0 � 0 � � � � 0 1 0 0 = + 0 + 3 0 0 3 0 0 1 0 0 1 � 1 � 0 � 0 � � � � � 0 0 1 0 0 = − 2 + 0 0 − 2 0 0 0 1 0 0 1

k ≤ s 23 � 2 � 1 � 1 � 0 � � � � 2 1 1 0 = 2 − 0 + 2 2 0 0 0 0 0 1 0 � 4 � 1 � 2 � 0 � � � � 4 1 2 0 = 0 + 2 − 2 4 0 0 0 3 0 1 0