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# lecture10 - Lecture 10 Dimensions, Independence, Basis and...

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Lecture 10 Dimensions, Independence, Basis and Complete Solution of Linear Systems Shang-Hua Teng

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Linear Independence Linear Combination { } { } i n v ,...,v ,v v = n 1 i i 2 1 is vectors of set a of α Linear Independence { } vectors of set A 2 1 n ,...,v ,v v is linearly independent if only if none of them can be expressed as a linear combination of the others
Examples 1 1 0 , 4 2 2 , 1 0 1 1 1 2 2 1 1 0 1 , ,

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Linear Independence and Null Space Theorem/Definition { } vectors of set A 2 1 n ,...,v ,v v is linearly independent if and only α 1 v 1 + 2 v 2 +…+ n v n = 0 only happens when all ’s are zero The columns of a matrix A are linearly independent when only solution to A x=0 is x = 0
2D and 3D v w u v

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How do we determine a set of vectors are independent? Make them the columns of a matrix Elimination Computing their null space
Permute Rows and Continuing Elimination (permute columns) - - - - - - = 0 1 1 1 2 1 1 3 1 1 1 1 0 2 1 1 1 1 1 1 0 0 1 1 A

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Theorem If A x = 0 has more have more unknown than equations (m > n: more columns than rows), then it has nonzero solutions. There must be free variables.
Echelon Matrices = * * * * 0 0 0 0 0 0 * * 0 0 0 0 * * * * * 0 * * * * * * A Free variables

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Reduced Row Echelon Matrix R = 1 0 0 0 0 0 0 0 0 0 * 1 0 0 0 0 * 0 * * 1 0 * 0 * * 0 1 A Free variables
Computing the Reduced Row Echelon Matrix Elimination to Echelon Matrix E 1 PA = U Divide the row of pivots by the pivots Upward Elimination E 2 E 1 PA = R

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Example: Gauss-Jordan Method for Matrix Inverse [ A I ] E 1 [ A I ] = [ U, I ] In its reduced Echelon Matrix A -1 [ A I ] = [ I A -1 ]
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## This note was uploaded on 03/08/2010 for the course CS 232 taught by Professor Bera during the Spring '09 term at BU.

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lecture10 - Lecture 10 Dimensions, Independence, Basis and...

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