Linear Independence Recall: The vectors { v 1 , . . . , v n } - PowerPoint PPT Presentation

Linear Independence Recall: ⇀ ⇀ The vectors { v 1 , . . . , v n } generate a coordinate system ⇀ ⇀ for Span { v n } v 1 , . . . ,

We want an efficient coordinate system. ⇀ ⇀ ⇀ ⇀ E.g. Suppose, as above, that b = v 1 + 2 v 2 + 3 v 3 . ⇀ ⇀ ⇀ Suppose also that v 3 = v 1 + v 2 .

Lemma: ⇀ ⇀ ⇀ If w = r 1 v 1 + r 2 v 2 ⇀ ⇀ ⇀ ⇀ ⇀ Then Span { w } = Span { v 2 } v 1 , v 2 , v 1 ,

⇀ “each v i contributes something new” ⇐ ⇒ ⇀ No v i is in the span of the other vectors. Define: { ⇀ ⇀ v n } is linearly independent v 1 , . . . , if and only if

Fun With Negations ⇀ ⇀ { v n } is linearly dependent v 1 , . . . , if and only if if and only if

 − 1     − 3  2 ⇀ ⇀ ⇀ E.g. Let v 1 = 0 v 2 = 2 v 3 = 4  ,  ,     0 2 3 Is { ⇀ ⇀ ⇀ v 3 } independent? v 1 , v 2 ,

      3 0 6 ⇀ ⇀ ⇀ E.g. Let v 1 = 3 v 2 = 5 v 3 = − 4  ,  ,     − 6 − 4 − 4 Is { ⇀ ⇀ ⇀ v 3 } independent? v 1 , v 2 ,

What does a solution mean ?

Two Observations about Dependence 1. If there is a non-trivial dependence relation, then one vector is in the span of the others

2. If one vector is in the span of the others, then there is a non-trivial dependence relation

Theorem 7: The geometric meaning of dependence ⇀ ⇀ { v n } is Dependent v 1 , . . . , if and only if

⇀ ⇀ Warning: Not all the v i will be generated by other v j . � 0 � � 1 � � 2 � E.g. Consider { , } , , 1 0 0