key theorems key ideaskey algorithms linked to examples next
DESCRIPTION
V is a vector space of dimension n. S = { v 1, v 2, v 3,..., v n } then S is INDEPENDENT if and only if S SPANS V. Return to outlineTRANSCRIPT
KEY THEOREMS
KEY IDEAS KEY ALGORITHMS
LINKED TO EXAMPLES
next
key theorems key ideas key algorithms n vectorsin an n dimensionalvector space
VECTOR SPACEindependentspan
Solve system equations
basis Find dot productcoordinates Take matrix times vector
dimensiondomain,null space, range of alinear mapping
LINEAR MAPPING Write matrix equationdomainnull spacerange
Find matrix for lin mapTake product of matrices
detA 0 matrix for Find inverse of matrixcompositioninverse
Find determinant of matrix
similarity ,eigenstuff
EIGENSTUFF Find eigenstuffsimilarity Similar diagonal matrix
V is a vector space of dimension n.
S = { v1 , v2 , v3 , . . . , vn } then
S is INDEPENDENT if and only if S SPANS V.
Return to outline
If T is a LINEAR MAPPING then:
the dimension of the DOMAIN of T = the dimension of the NULL SPACE of T + the dimension of the RANGE of T
Return to outline
0
A and B are SIMILAR matrices if and only if there exists a matrix P such that:
B = P –1 A PIf A is the matrix for T relative
tothe standard basis
then B is the matrix for T relative to
the columns of PIf B is diagonal then
the diagonal entries of B are eigenvalues andthe columns of P are eigenvectors
Return to outline
3221424274263
zyxwzyxwzyxw
next
3221424274263
zyxwzyxwzyxw
next321214214274263
Reduces to:
3221424274263
zyxwzyxwzyxw
next321214214274263
000002210010021
Reduces to:
3221424274263
zyxwzyxwzyxw
next
000002210010021
zyxwzyxwzyxw
3221424274263
zyxwzyxwzyxw
next
zyxw
zyxwzyxwzyxw
2221
000002210010021
3221424274263
zyxwzyxwzyxw
nextzzzy
xxxw
122
121
zyxw
2221
3221424274263
zyxwzyxwzyxw
1200
0012
02
01
122
121
zx
zzzy
xxxw
return to outline
( 3 )-1 4 2
( 2 ) 5 3-2
• =
next
( 3 )-1 4 2
( 2 ) 5 3-2
• =
6 + -5 + 12 + -4 = 9
return to outline
( 3 -1 2 ) ( 1 ) = ( ) 2 1 -1 43
next
( 3 -1 2 ) ( 1 ) = ( 5 ) 2 1 -1 43
nextdot product of row 1 of matrix with vector
= entry 1 of answer
( 3 -1 2 ) ( 1 ) = ( 5 ) 2 1 -1 4 33
dot product of row 2 of matrix with vector
= entry 2 of answer return to outline
3221424274263
zyxwzyxwzyxw
347
212121424263
zyxw
System of linear equations:
Equivalent matrix equation:
return to outline
A toy maker manufactures bears and dolls.It takes 4 hours and costs $3 to make 1 bear.It takes 2 hours and costs $5 to make 1 doll.
cost totalrequired timetotal
dolls #bears #
TFind the matrix for T
next
A toy maker manufactures bears and dolls.It takes 4 hours and costs $3 to make 1 bear.It takes 2 hours and costs $5 to make 1 doll.
cost totalrequired timetotal
dolls #bears #
TFind the matrix for T
columnfirst 34
bear 1 make cost to bear 1 make torequired time
01
T
column second52
doll 1 make cost to doll 1 make torequired time
10
T
5324
for matrix the T
return to outline
( 3 -1 2 ) ( 1 2 ) = ( ) 2 1 -1 4 13 1
next
A B
( 3 -1 2 ) ( 1 2 ) = ( 5 ) 2 1 -1 4 13 1
dot product of row 1 of A with column 1 of B
= entry in row 1 column 1 of AB next
A B
( 3 -1 2 ) ( 1 2 ) = ( 5 7 ) 2 1 -1 4 13 1
dot product of row 1 of A with column 2 of B
= entry in row 1 column 2 of AB next
A B
( 3 -1 2 ) ( 1 2 ) = ( 5 7 ) 2 1 -1 4 1 33 1
dot product of row 2 of A with column 1 of B
= entry in row 2 column 1 of AB next
A B
( 3 -1 2 ) ( 1 2 ) = ( 5 7 ) 2 1 -1 4 1 3 43 1
dot product of row 2 of A with column 2 of B
= entry in row 2 column 2 of AB return to outline
100425010312001213
111100527010102001
Reduces to
next
100425010312001213
111100527010102001
Reduces to
A
A-1
return to outline
next
82
122A
To find eigenvalues for A, solve for :
08682
122det)det( 2
AI
next
82
122A
To find eigenvalues for A, solve for :
08682
122det)det( 2
AI
The eigenvalues are 2 and 4
next
82
122A
To find eigenvalues for A, solve for :
08682
122det)det( 2
AI
The eigenvalues are 2 and 4
An eigenvector belonging to 2 is in the null space of 2I - A
next
82
122A
To find eigenvalues for A, solve for :
08682
122det)det( 2
AI
The eigenvalues are 2 and 4
An eigenvector belonging to 2 is in the null space of 2I - A
62
124 2I - A
next
82
122A
To find eigenvalues for A, solve for :
08682
122det)det( 2
AI
The eigenvalues are 2 and 4
An eigenvector belonging to 2 is in the null space of 2I - A
62
124 2I - A
an eigenvector belonging to 2 is any nonzero multiple of
13
next
82
122A
To find eigenvalues for A, solve for :
08682
122det)det( 2
AI
The eigenvalues are 2 and 4
eigenvectors are:
13
12
next
82
122A
The eigenvalues are 2 and 4
eigenvectors are:
13
12
A is similar to the diagonal matrix B
4002
82
122A
The eigenvalues are 2 and 4
eigenvectors are:
13
12
B = P –1 A P
4002
=
1
1123
1123
82122
return to outline
bcaddbca
det
next
bcaddbca
det
cba
ifchebgda
ifchebgda
columnor row a choose
det
next
bcaddbca
det
cba
ifchebgda
ifchebgda
det
ifhe
a det
next
bcaddbca
det
ifhe
a
cba
ifchebgda
ifchebgda
det
det
ifgd
b det
next
bcaddbca
det
ifgd
bifhe
a
cba
ifchebgda
ifchebgda
detdet
det
hegd
c det
Return to outline
Return to outline
A is an nn matrix
detA 0 iff
A is nonsingular (invertible) iffThe columns of A are a basis for Rn iffThe null space of A contains only the zero vector iffA is the matrix for a 1-1 linear transformation