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by: Rae Kutch

41

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7

Rae Kutch
WVU
GPA 3.8

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
7
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 7 page Class Notes was uploaded by Rae Kutch on Saturday September 12, 2015. The Class Notes belongs to MATH 567 at West Virginia University taught by Staff in Fall. Since its upload, it has received 41 views. For similar materials see /class/202659/math-567-west-virginia-university in Mathematics (M) at West Virginia University.

## Popular in Mathematics (M)

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Date Created: 09/12/15
The fourth subspace Column space span of columns Ax for all possible x Row space span of rows yTA for all possible y Null space solutions of AxO all vectors orthogonal to the rows all dependencies among columns Left null space solutions of yTAO all vectors orthogonal to the columns all dependencies among the rows nullspace of AT 1 2 4 3 1 0 2 3 1 8 2 3 1 1 1 3 2 0 3 1 7 1 1 1 2 1 2 2 0 1 5 5 0 13 0 8 gtgt rref B ans 10000 0 0 02000 0 02000 0 10000 0 24000 0 14000 0 0 10000 04000 0 04000 0 0 0 0 10000 10000 0 0 0 0 0 0 0 0 0 0 0 0 RankB4dim row space dim col space Basis of Basis of row Null space basis gtgt nu11B39r39 ans 5 10000 2 2 0 2 14 2000 4000 4000 0000 4 col space 4 1 1 2 1 2 7 1 2 0 0 0 space 0 0 02000 10000 0 24000 0 10000 04000 0 0 0 worked out by hand from 4 1 0 0 4 0 1 1 1et matlab do it 02000 14000 04000 0000 0000 10000 above 2000 4000 4000 0000 Left null space basis gtgt nullB3939r39 ans 14444 03333 08889 10000 1 2 ans 10000 20000 10000 30000 20000 50000 04444 23333 21111 0 10000 10000 3 5 nullB3939r39 20000 40000 30000 10000 10000 10000 10000 70000 10000 20000 50000 0 10000 I 0000 I 0000 0000 First four columns are basis of column space Last two columns are basis of left null space Together they are a basis for R6 Characterization of column space column space Axb has a solution 4444 3333 8889 0000 04444 23333 I 1111 10000 0000 b is in the if and only if yTb0 for any y in left null space Definition of inverse If A is square we define the inverse of A if it exists as the matrix Aamp that satisfies AAampT and AdAT Theorem Aamp exists A has an inverse if and only if AT or the rank of A is n or dim col spacen AxO only if xO How to calculate A Think of solving for B that satisfies ABT 7 this is column by column in B a system of linear equations ie Acol j of Bcol j of T and we solve all these equations at once by putting all the columns of T into a superaugmented matrix A T Then we reduce as usual and assuming that AT we would obtain AIIA391 because each column on the right is one of the columns of B that we seek If you discover that the rank of A is less than n then Aamp doesn t exist eg note that if Aamp exists then AAampbb for any b which shows that Axb always has a solution If the rank of A is less than n this can t be true and so Aamp can t exist 4 subspaces in one reduction We already know how to get a basis of the column space row space and null space of the matrix A from the reduced row echelon form of A Now consider the reduction A I R B where R is the reduced row echelon form of A R will have mr rows of zeros at the bottom where r is the rank of A Then the bottom mr rows of B are a basis for the left null space Here s the reason Every row operation can be carried out by a left multiplication so A I R B means that CA TR B for some C But then we must have CAR and CTB from the latter we see that CB so that BAR Now the bottom mr rows of R are equal to the corresponding mr rows of B times A and the result is zero for each such row Thus each of the bottom mr rows of B is in the left null space The dimension of the left null space is mr and those rows of B are linearly independent because B is row equivalent to T Thus those rows are a basis of the left null space Here is our previous example of the matrix B of rank 4 We find the left null space using the procedure outlined above gtgt B eye6 ans 1 2 4 3 1 0 1 0 0 0 0 0 2 3 1 8 2 3 0 1 0 0 0 0 1 1 1 3 2 0 0 0 1 0 0 0 3 1 7 1 1 1 0 0 0 1 0 0 2 1 2 2 0 1 0 0 0 0 1 0 5 5 0 13 0 8 0 0 0 0 0 1 gtgt rref B eye6 ans 1 0 0 02000 0 02000 0 0 00690 01379 04483 00483 0 1 0 24000 0 14000 0 0 00690 01379 04483 02483 0 0 1 04000 0 04000 0 0 00345 00690 02759 00759 0 0 0 0 1 1 0 0 05172 00345 01379 01379 0 0 0 0 0 0 1 0 08621 07241 01034 01034 0 0 0 0 0 0 0 1 10690 01379 04483 04483 The rank of B is 4 the last two rows in the reduced form of B are zeroes The basis for the left nu11 space is then the last two rows in the righthand side of the augmented matrix namely 1 0 08621 07241 01034 01034 0 1 10690 01379 04483 04483 Of course it s not clear that this is any easier than simply finding the null space of the transpose of the matrix in question which is the more direct approach

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