Linear Algebra
Vector:
1
T2
1 2 1 2 'n n
n
x
x
x x x x x x x
x
- linear space nR
15-algebra(2)
2
A set of vectors is said to be linearly
dependent if there exist real numbers ,
not all zeros, such that
If (*) holds only for , then this
set of vectors is said to be linearly independent.
mxxx ,, 21
)( .02211 mmxxx
m ,, 21
021 m
2 :R
1x
2x
21 xx
1x
2x
3x5-algebra(2)
3
In general, in the space, there are vectors
that are linearly independent, but any
vectors must be linear dependent.
nR n
1n
The dimension of a linear space is the maximum
number of linearly independent vectors in the
space. So, the space is an dimensional
linear space.
nR n
5-algebra(2)
4
A set of linearly independent vectors in is called
a basis if every vector in can be expressed as a
unique linear combination of the vectors in the set.
nR
nR
Any set of linearly independent vectors is a
basis for .
n
nR
Proof: Let be linearly independent. nqqq ,, 21
For any , are linearly dependent. nRx xqqq n ,,, 21
121 ,,, nn There are , not all zero, s.t.,
012211 xqqq nnn
5-algebra(2)
5
012211 xqqq nnn
Observe that .
(otherwise, would be linearly dependent)
01 n
nqqq ,, 21
n
n
n
nn
qqqx
1
2
1
2
1
1
1
Uniqueness? Let:
nn
nn
qqq
qqqx
2211
2211
nnn qqq )()()(0 222111
nn ,, 2211? 5-algebra(2)
6
Determination of the coordinates ni i ,,2,1,
nnqqqx 2211
xqqq
n
n
2
1
21
xqqq n
n
1
21
2
1
5-algebra(2)
7
For an orthonormal basis,
1
0
0
,,
0
1
0
,
0
0
1
21
n
iii
xIxxiii n
n
1
21
2
1
5-algebra(2)
8
Example:
2
3
1
Rx
2
2
,
1
3
21 qq
2
1
3
1
4
3
4
1
2
1
2
1
3
1
21
23
1
2
1
21 2qqx
5-algebra(2)
9
Norms of vectors (length or magnitude)
Any real-valued function of , , can be a norm if x x
xx ,01) and iff ; 0x 0x
Rxx ,2) ;
212121 ,, xxxxxx 3) .
1x
2x
21 xx
Triangular
inequality
5-algebra(2)
10
Example of norms:
(1 norm)
pn
i
p
ip
i
i
n
i
i
n
i
i
xx
xx
xxxxx
xx
1
1
1
2
2
1
1
max
:'
(2 norm or Euclidean norm)
(∞ norm)
(p norm)
5-algebra(2)
11
Orthonormalization
• is normalized if ;
• and are orthogonal if
1' xxx
1x 2x 0' 21 xx
A set of vectors is said to be
orthonormal if
,,,2,1, mixi
ji
ji
xx ji
1
0
'
5-algebra(2)
12
Schmidt orthonormalization procedure:
Given linear independent,,,, 21 meee
1
1
111 ,
u
u
qeu
2
2
212122 ,)'(
u
u
qqeqeu
1e
2e
1q
2u
3
3 3 1 3 1 2 3 2 3
3
( ' ) ( ' ) ,
u
u e q e q q e q q
u
m
m
mmmmmmmm
u
u
qqeqqeqqeqeu ,)'()'()'( 112211
5-algebra(2)
13
Example:
2
1
,
1
1
21 ee
2
2
2
2
2
1
2
1
,
1
1
1
1
111
u
u
qeu
2
2 2 1 2 1 2
2
2 1 2
1 3 2 2 2 2
( ' ) ,
2 12 2 2
22 2
u
u e q e q q
u
5-algebra(2)
m equations with n unknowns
14
11,,, nmnm RxRyRAyAx
Linear algebraic equations
Let naaaA 21
yAx
yxaxaxa nn 2211
is a linear combination of the columns of y A
5-algebra(2)
15
- Range space of : All possible linear combinations
of columns of . The rank of is the dimension of
the range space.
A
A A
Rank( ) = the largest number of linearly independent
columns of
= the largest number of linearly independent
rows of
= size of the largest square submatrix whose
determinant is non-zero
A
A
A
min ,m n
5-algebra(2)
16
Example:
4321:
0202
4321
2110
aaaaA
1a and are linearly independent2a
24213 2, aaaaa
Any 3 columns are linearly dependent
2)(:)(Rank AA
5-algebra(2)
17
- A vector is a null vector of if . x A 0Ax
AThe null space of consists of all its null vectors.
AA
Nullity ( ) = maximum number of linear independent
null vectors
= number of columns of - rank( )
A
0202
4321
2110
A Nullity( ) = 4 - 2 = 2 A
5-algebra(2)
18
0202
4321
2110
A
0Ax
1
0
2
0
,
0
1
1
1
21 nn
and are linearly independent and form a
basis of the null space, that is:
Null space =
1n 2n
Rnn 212211 ,:
5-algebra(2)
19
Theorem: Given nmRAyAx ,
1. Solution exists iff , that is, x range( )y A
yAA ,
2. Solution exists for any iffx y
mA (full row rank)
5-algebra(2)
20
Theorem: Given . Assume that solution
exists. Let .
nmRAyAx ,
px )(Ank
2. : then,0k
1. : solution is unique; 0k
,2211 kkp nnnxx
where is a basis of Null( ).A knnn ,,, 21
5-algebra(2)
21
- Determinant and inverse of a square matrix
1
1
,
,)(det
j
ijij
n
i
ijij
iCa
jCaA
ji
nj
ij
j
ij
a
a
a
C
)1(
1
: Inverse of 1A A
IAAAA 11
Inverse of exists iffA .0)(det A
5-algebra(2)
22
T
1 Adj( )
det( ) det( )
ijCA
A
A A
Example:
1121
1222
12212211
1
2221
12111 1
aa
aa
aaaaaa
aa
A
A shorthand method for computing 1A
1
4
3
4
1
2
1
2
1
10
01
4
2
4
2
4
3
4
1
01
10
10
4
3
4
1
21
10
10
31
21
40
10
01
21
23
AI
IA
5-algebra(2)
23
- Similarity transformation
nRxAxx ,Consider
0)(det, QQzxLet (state transformation)
zAAQzQAxQxQz :111
AQQA 1 (similarity transformation)
AA and are said to be similar (to each other)
5-algebra(2)
24
- Eigenvalues and eigenvectors
x Ax
x
Ax
0, xxAx
0,0)( xxAI
0)(:)(det AI Eigenvalues
For a given eigenvalue :
0, xxAx Eigenvectors
5-algebra(2)
25
- Each eigenvalue corresponds to one or more linearly
independent eigenvectors
- Eigenvectors corresponding to different eigenvalues are
linearly independent
- If eigenvalues of are all distinct: A n ,,, 21
linearly independent eigenvectors:n nqqq ,,, 21
nnn qAqqAqqAq ,,, 222111
n
nn qqqqqqA
2
1
2121
Q Q5-algebra(2)
26
1
2
1
QQA
n
QAQ
2
1
1
Diagonalization!
- If eigenvalues of are not all distinct: A
Example:
0)()(det, 41
4 AIRA
1 (4 repeated eigenvalues)
5-algebra(2)
27
0)( 1 qAI
Suppose or 3)( 1 AI 134)(Nullity 1 AI
One (linearly independent) eigenvector, 1q
121 )( qqAI 2q
231 )( qqAI 3q
341 )( qqAI 4q
are generalized eigenvectors,
linearly independent, and:
4321 ,,, qqqq
4134
3123
2112
111
qqAq
qqAq
qqAq
qAq
5-algebra(2)
28
4134
3123
2112
111
qqAq
qqAq
qqAq
qAq
1
1
1
1
43214321
1
1
1
qqqqqqqqA
Q Q
JQAQ
:
1
1
1
1
1
1
1
1
(Jordan block of size 4)
5-algebra(2)
29
0)( 1 qAI
0)()(det, 41
44 AIRA
Suppose or 2)( 1 AI 224)(Nullity 1 AI
Two linearly independent eigenvectors, 11, pq
Two chains of generalized eigenvectors
1
1
1
1
21212121
1
0
1
ppqqppqqA
2
11
J
J
QAQ
5-algebra(2)
30
1
1
1
1
13211321
0
1
1
pqqqpqqqA
2
11
J
J
QAQ
- Example (an application of Jordan block)
JQAQ 1
QA
Q
QAQQAQJ detdet
det
1
detdetdetdetdet 11
i
n
i
JA
1
detdet
A matrix is nonsingular iff it has no zero eigenvalue!
Or:
5-algebra(2)
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