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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 0 "" }{TEXT 256 24 "Linear Al
gebra Powertool" }}{PARA 18 "" 0 "" {TEXT -1 9 "Dimension" }}{PARA 4 "
" 0 "" {TEXT -1 19 "On Line Section 2.2" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 26 "Worksheet by Russell Blyth" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "r
estart: with(linalg): with(plots): with(plottools):" }}}{SECT 1 {PARA 
3 "" 0 "" {TEXT -1 7 "Outline" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "T
his worksheet covers material that corresponds to section 2.2 of the t
ext." }}{PARA 0 "" 0 "" {TEXT -1 25 "The basic objectives are:" }}
{PARA 14 "" 0 "" {TEXT -1 81 "1)  Use Gauss-Jordan elimination to find
 a basis for the column space of a matrix" }}{PARA 14 "" 0 "" {TEXT 
-1 79 "2)  Use Gauss-Jordan elimination to find a basis for the null s
pace of a matrix" }}{PARA 14 "" 0 "" {TEXT -1 67 "3)  Note a relations
hip between the dimensions of these two spaces." }}{PARA 14 "" 0 "" 
{TEXT -1 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 61 "The Dimensions \+
of the Column Space and Null Space of a Matrix" }}{EXCHG {PARA 14 "" 
0 "" {TEXT -1 153 "Consider a matrix A, as given below.  How do we fin
d a basis for the column space of the matrix? How do we find a basis f
or the null space of the matrix?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
19 "Let A be the matrix" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 
"A := matrix(5,6,[[-1,2,6,-8,-14,3],[2,4,1,-8,5,-1],[-3,1,4,-9,-10,0],
[3,-2,-1,12,-1,4],[5,7,11,-11,-19,9]]);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 22 "Compute the rank of A:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "rank(A);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 124 "How \+
does the rank of A relate to the dimensions of the vector spaces we ar
e interested in studying? Let's row reduce A next:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 23 "RedMat := gaussjord(A);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 17 "We conclude that:" }}{PARA 14 "" 0 "" {TEXT -1 
159 "1)  The rank of the matrix is 3, so the set of 6 column vectors c
annot be linearly independent.  The biggest subset of linearly indepen
dent vectors has size 3." }}{PARA 14 "" 0 "" {TEXT -1 156 "2)  The piv
ots of the reduced echelon form correspond to independent vectors.  Th
us the first three columns of vectors in A form a linearly independent
 set." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "Let's see how to write \+
the final three column vectors of A as linear combinations of the firs
t three columns:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "B:= de
lcols(A,4..6);\nv4:=col(A,4);v5:=col(A,5);v6:=col(A,6);\nlinsolve(B,v4
);\nlinsolve(B,v5);\nlinsolve(B,v6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 6 "Check:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "v1:=col(A,1
);v2:=col(A,2);v3:=col(A,3);\nevalm(2*v1-3*v2); evalm(2*v2-3*v3); eval
m(v1-v2+v3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 252 "Thus the first t
hree columns of A are a basis for the column space (why are they linea
rly independent?). The remaining three columns of A are each linear co
mbinations of the first three columns. Note that the vectors in the co
lumn space are vectors in " }{XPPEDIT 18 0 "R^5;" "6#*$%\"RG\"\"&" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "What is t
he dimension of the column space of A?" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 84 "Now find a basis for the null space o
f A, that is, for the solution space of AX = 0." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 51 "Z:=vector(5,[0,0,0,0,0]);\nnullspace:=linsolve
(A,Z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "Get a basis by " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 146 "y1:=subs(_t[1]=1,_t[2]=0,_t
[3]=0,op(nullspace));\ny2:=subs(_t[1]=0,_t[2]=1,_t[3]=0,op(nullspace))
;\ny3:=subs(_t[1]=0,_t[2]=0,_t[3]=1,op(nullspace));" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 18 "Let's check these:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 41 "evalm(A&*y1); evalm(A&*y2); evalm(A&*y3);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "So \{y1, y2, y3\} is a basis for t
he null space of A. Note that the vectors in the null space are in " }
{XPPEDIT 18 0 "R^6" "6#*$%\"RG\"\"'" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "What is the dimension o
f the null space?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 100 "What is the sum of the rank (the dimension of the column
 space) and the dimension of the null space?" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "Maple provides direct c
ommands to find bases for the column space and null space:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "colspan(A,'d'); d;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "kernel(A);" }}}{EXCHG {PARA 5 "" 0 
"" {TEXT -1 10 "Exercises:" }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 40 "1
)  Repeat all the above for the matrix:" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "E:=matrix(4,6,[[-1,2,0
,4,5,-3],[3,-7,2,0,1,4],[2,-5,2,4,6,1],[4,-9,2,-4,-4,7]]);" }}}{EXCHG 
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