This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Lecture08
Lecture 08:
Vector Spaces and Subspaces #2: The Null Space N(A)
Outline: 1) Quick Review of the Column Space and Null Space 2) Finding N(A) part 1: Gaussian Elimination to Upper Echelon Form "Special Solutions" the rank r of a matrix 3) Finding N(A) part 2: GaussJordan Elimination to Reduced Row Echelon Form (RREF) Examples 4) Existence and Uniqueness: The general solution to Ax=b September 27, 2007
Review: The Column Space C(A)
Definition: The Column Space of a Matrix A is the vector subspace formed by all linear combinations of the columns of A (general mxn matrix) 1) C(A) is Ax when x assumes all values in Rn 2) C(A) is a subspace of R 3) C(A) controls the ______________________ of solutions to Ax=b 4) Ax=b will only have solutions iff b _________________________ Review: The Null Space N(A)
Definition: The Null Space of a Matrix A is the vector subspace formed by all solutions x to Ax=0 (i.e. is all linear combination of columns of A that cancel to zero) 1) N(A) is a subspace of R 2) N(A) controls the ______________________ of solutions to Ax=b 3) If A is invertible, then N(A)=______________ The Null Space: Algorithms for finding N(A)
1) Inspired Guessing A = [1 2 ; 3 6 ] 2) Gaussian Elimination to solve Ax=0 (Only small issue is that U is no longer upper triangular) The Null Space:
Algorithms for finding N(A): Gaussian Elimination to Upper Echelon Form Finding N(A): A closer look
Gaussian Elimination (including row exchanges) transforms a general mxn matrix A to U which is now "upper echelon form" A bigger example A = [1 1 2 1 ; 1 2 1 1 ] Solve Ax=0 using Gaussian Elimination 1) Identify "Pivot" columns and "Free columns" (and associated Pivot variables and free variables) 2) Identify the rank of the matrix r = number of Pivot Columns 3) Identify the number of special solutions = number of free columns = 4) Find the Special Solutions: the linear combinations of Pivot Columns required to annihilate a single free column. 5) The Null Space is formed by all linear combinations of the special solutions 1 Lecture08
The Null Space: A better way
example A = [1 1 2 1 ; 1 2 1 1 ] again September 27, 2007 GaussJordan Elimination to Reduced Row Echelon Form R=rref(A) Again A = [1 1 2 1 ; 1 2 1 1 ] 1) take A to U By Gaussian Elimination 2) Continue by GJ Elimination (eliminate up then divide by the pivots) to R 3) when done: The pivot columns will be columns of the Identity Matrix Can read the Special Solutions right out of R The Null Space: A better way
GaussJordan Elimination to Reduced Row Echelon Form R=rref(A) The Null Space: A better way
GaussJordan Elimination to Reduced Row Echelon Form R=rref(A) Example A = [1 1 2 1 ; 1 2 1 1 ] goes to R = [ 1 0 3 1 ; 0 1 1 0 ] Point: N(R)=N( ) But Rx=0 is much easier to see Last Example: A = [ 1 2 2 5 ; 2 4 8 18 ; 3 6 10 23 ] !! The Special solutions are the linear combinations of the pivot columns that annihilate each free column. The Null Space: A Trick
GaussJordan Elimination to Reduced Row Echelon Form R=rref(A) LastLast one (Easy)
Last Example: A = [ 1 2 ; 3 4 ] Last Example: A = [ 1 2 2 5 ; 2 4 8 18 ; 3 6 10 23 ] R=[1201;001 2 ;0 0 0 0 ] R= N(A)= Do it yourself Null Space:
Point: All Invertible Matrices R=I, N(A)=Z 2 Lecture08
Putting it all together: The General Solution to Ax=b
Basic Approach: 1) use GaussJordan Elimination to take [ A b ] to [ R d] 2) Find a Particular solution to Rxp=d (combination of pivot columns and no free columns that add up to d) 3) Find the special solutions to RxN=0 4) The General solution: is x=xp+xN September 27, 2007
Putting it all together: The General Solution to Ax=b
Example: A = [ 1 2 1 0 1; 2 4 1 0 0 ; 1 2 0 1 4 ] b=[111] 3 ...
View
Full
Document
This note was uploaded on 06/02/2010 for the course APMA APMA E3101 taught by Professor Spiegelman during the Fall '07 term at Columbia.
 Fall '07
 Spiegelman

Click to edit the document details