4 twice - MATHEMATICS 54 Professor Constantin Teleman...

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Unformatted text preview: MATHEMATICS 54 Professor Constantin Teleman 9/8/09 Lecture 4 ASUC Lecture Notes Online is the only authorized note-taking service at UC Berkeley. Do not share, copy or illegally distribute (electronically or otherwise) these notes. Our student-run program depends on your individual subscription for its continued existence. These notes are copyrighted by the University of California and are for your personal use only. D O N O T C O P Y Sharing or copying these notes is illegal and could end note taking for this course. ANNOUNCEMENTS The linear independence part of the homework will be due by Wednesday. The linear transformation and other parts of the homework will be due by next Monday. Please check the course website for the homewordk and hanouts update. The handout I handed out today will be in the Midterm. LECTURE Linear Independence S pan of vectors Existence of solution b r Span { n a a r r ... 1 } The system b x A r r = is consistent, where A=[ 1 a r n a r ] and n a a r r ... 1 are columns of A. For homogeneous system = x A r The result is: They system has only the trivial solution = x r . The columns of A form a linear independent collection of vectors. Def: a collection n a a r r ... 1 of vectors is linearly independent if the only expression of as a linear combination of them is obtained using 0 weights. In formula: n n x a x a r r ... 1 1 + = ... 2 1 = = x x (Note: this is x A r ) Caution: the property pertains to the collection of vectors, not to the individual vectors. Example: Any collection containing { } is linearly dependent (=not linearly independent). Because 1 = , weight is 1....
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4 twice - MATHEMATICS 54 Professor Constantin Teleman...

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