Download A Review of Linear Algebra - Degree 36

Stats 443.3 & 851.3
Linear Models
Instructor:
W.H.Laverty
Office:
235 McLean Hall
Phone:
966-6096
Lectures:
Evaluation:
MWF
9:30am - 10:20am Geol 269
Lab 2:30pm – 3:30 pm Tuesday
Assignments, Term tests - 40%
Final Examination - 60%
• The lectures will be given in Power Point
Course Outline
Introduction
Review of Linear Algebra
and Matrix Analysis
Review of Probability
Theory and Statistical
Theory
Multivariate Normal
distribution
The General Linear Model
Theory and Application
Special applications of The
General Linear Model
Analysis of Variance Models, Analysis
of Covariance models
A chart illustrating Statistical Procedures
Independent variables
Dependent
Variables
Categorical
Continuous
Categorical
Multiway frequency Analysis
(Log Linear Model)
Discriminant Analysis
Continuous
ANOVA (single dep var)
MANOVA (Mult dep var)
Continuous &
Categorical
??
MULTIPLE
REGRESSION
(single dep variable)
MULTIVARIATE
MULTIPLE
REGRESSION
(multiple dependent
variable)
??
Continuous &
Categorical
Discriminant Analysis
ANACOVA
(single dep var)
MANACOVA
(Mult dep var)
??
A Review of Linear Algebra
With some Additions
Matrix Algebra
Definition
An n × m matrix, A, is a rectangular array of
elements
a1n 
 a11 a12
a

a
a
21
22
2n 

A  aij 




amn 
 am1 am 2
 
n = # of columns
m = # of rows
dimensions = n × m
Definition
A vector, v, of dimension n is an n × 1 matrix
rectangular array of elements
 v1 
v 
2

v
 
 
vn 
vectors will be column vectors (they may also be
row vectors)
A vector, v, of dimension n
 v1 
v 
2

v
 
 
vn 
can be thought a point in n dimensional space
v3
 v1 
v  v2 
 v3 
v2
v1
Matrix Operations
Addition
Let A = (aij) and B = (bij) denote two n × m
matrices Then the sum, A + B, is the matrix
 a11  b11 a12  b12
a b
a

b
21
21
22
22

A  B   aij  bij  


 am1  bm1 am 2  bm 2
a1n  b1n 
a2 n  b2 n 


amn  bmn 
The dimensions of A and B are required to be
both n × m.
Scalar Multiplication
Let A = (aij) denote an n × m matrix and let c be
any scalar. Then cA is the matrix
 ca11 ca12
 ca
ca
21
22

cA   caij  


cam1 cam 2
ca1n 
ca2 n 


camn 
Addition for vectors
v3
 v1  w1 
v  w  v2  w2 
 v3  w3 
 w1 
w   w2 
 w3 
 v1 
v  v2 
 v3 
v1
v2
Scalar Multiplication for vectors
v3
 cv1 
cv  cv2 
 cv3 
 v1 
v  v2 
 v3 
v2
v1
Matrix multiplication
Let A = (aij) denote an n × m matrix and B = (bjl)
denote an m × k matrix
Then the n × k matrix C = (cil) where
m
cil   aij b jl
j 1
is called the product of A and B and is denoted
by A∙B
In the case that A = (aij) is an n × m matrix and B
= v = (vj) is an m × 1 vector
m
Then w = A∙v = (wi) where wi 
aij v j

j 1
is an n × 1 vector
A
w3
v3
 v1 
v  v2 
 v3 
w2
v2
 w1 
w   w2   Av
 w3 
w1
v1
Definition
An n × n identity matrix, I, is the square matrix
1 0
0 1
I  In  


0 0
Note:
1. AI = A
2. IA = A.
0
0 


1
Definition (The inverse of an n × n matrix)
Let A denote the n × n matrix
a1n 
 a11 a12
a

a
a
21
22
2n 

A  aij 




ann 
 an1 an 2
 
Let B denote an n × n matrix such that
AB = BA = I,
If the matrix B exists then A is called invertible
Also B is called the inverse of A and is denoted
by A-1
Note: Let A and B be two matrices whose
inverse exists. Let C = AB. Then the inverse of
the matrix C exists and C-1 = B-1A-1.
Proof
C[B-1A-1] = [AB][B-1A-1] = A[B B-1]A-1 = A[I]A-1
= AA-1=I
The Woodbury Theorem
1
1


A

BCD

A

A
B
C

DA
B
DA




1
1
1
1
where the inverses
1
A ,C
1
and C  DA B 
1
1
1
1
exist.
Proof:
1
1
1
1
1
1
Let H  A  A B C  DA B  DA
Then all we need to show is that
H(A + BCD) = (A + BCD) H = I.
H  A  BCD  


1
A  A B C  DA B  DA1  A  BCD 
1
1
1
1
1
 A A  A B C  DA B  DA1 A
1
1
1
1
1
 A BCD  A B C  DA B  DA BCD
1
1
1
1
1
1
 I  A B C  DA B  D
1
1
1
1
 A BCD  A B C  DA B  DA1BCD
 I  A1 BCD
1
1
1
1
1
 A B C  DA B   I  DA1BC  D
 I  A1 BCD
1
1
1
1
1
1



 A B C  DA B  C  DA B  CD
1
1
1
1
1
 I  A BCD  A BCD  I
The Woodbury theorem can be used to find the
inverse of some pattern matrices:
Example: Find the inverse of the n × n matrix
b a
a b



a a
a
1 0


a
0 1

 b  a 




b
0 0
1
1
  b  a  I    a 1 1
1 


1
0
1 1


0
1 1

a




1
1 1
A  BCD
1

1


1
where
A  b  a  I
1
1
B 
 
 
1
1
I
hence A 
ba
1
C  a
11
D  1 1
1
C  
a
1
1
1
1
 1  
1
1
C  DA B   1 1
1 
I
a
b  a   

1
1
n
b  a  an b  a  n  1
 


a b  a a b  a 
a b  a 
and
1
Thus
a b  a 
C  DA B  
b  a  n  1
1
1
1
Now using the Woodbury theorem
1
1


A

BCD

A

A
B
C

DA
B
DA




1
1
1
1
1
1
1 a b  a


1
1
1



I
I
1 1
1
I

ba
b  a   b  a  n  1
ba

1
1
1
1
a
  1 1

I
1
ba
 b  a   b  a  n  1   

1
Thus
1
b a
a b



a a
1 0
0 1
1 
ba 

0 0
c
d



d
a
a 



b
0
1 1
1 1
0 
a


  b  a   b  a  n  1  


1
1 1
d
c
d
d

d


c
1
1


1
where
a
d 
 b  a   b  a  n  1 
1
a
and c 

b  a  b  a   b  a  n  1 

1 
a
1  b  a  n  2 

1 



b  a  b  a  n  1  b  a  b  a  n  1 
Note: for n = 2
a
a
d 
 2
 b  a  b  a  b  a 2
1  b 
b
and c 
 2


b  a  b  a  b  a2
1
b a 
1
Thus 
 2

2
a
b
b

a


 b a 
 a b 


Also
b a
a b



a a
a  b a
a   a b


b  a a
 bc   n  1 ad

bd  ac  (n  2)ad




bd  ac  (n  2)ad
1
a
b a
a b
a 





b
a a
bd  ac  (n  2)ad
bc   n  1 ad
bd  ac  (n  2)ad
a  c
a   d


b  d
d
c
d
d
d 


c
bd  ac  (n  2)ad 

bd  ac  (n  2)ad 


bc   n  1 ad 
Now
a
d 
 b  a   b  a  n  1 
1  b  a  n  2 
and c 


b  a  b  a  n  1 
2


b

a
n

2
n

1
a




b
bc   n  1 ad 


b  a  b  a  n  1   b  a   b  a  n  1 

b  b  a  n  2     n  1 a 2
 b  a   b  a  n  1 
b2  ab  n  2    n  1 a 2
 2
1
2
b  ab  n  2    n  1 a
and
b  (n  2)a  a

a  b  a  n  2 
bd  ac  (n  2)ad 


b  a  b  a  n  1   b  a   b  a  n  1 
0
This verifies that we have calculated the inverse
Block Matrices
Let the n × m matrix
 A11
A 
nm
n  q  A21
A12 
A22 
q
m p
p
be partitioned into sub-matrices A11, A12, A21, A22,
Similarly partition the m × k matrix
 B11
B 

mk
m  p  B21
p
l
B12 

B22 
k l
Product of Blocked Matrices
Then
 A11
A B  
 A21
A12   B11


A22   B21
 A11B11  A12 B21

 A21B11  A22 B21
B12 

B22 
A11B12  A12 B22 

A21B12  A22 B22 
The Inverse of Blocked Matrices
Let the n × n matrix
 A11
A 
nn
n  p  A21
p
p
A12 
A22 
n p
be partitioned into sub-matrices A11, A12, A21, A22,
Similarly partition the n × n matrix
 B11
B 
nn
n  p  B21
p
Suppose that B = A-1
p
B12 

B22 
n p
Product of Blocked Matrices
Then
 A11
A B  
 A21
A12   B11


A22   B21
 A11B11  A12 B21

 A21B11  A22 B21
 Ip

 0
 n p  p
0 
p n  p 

I n p 

B12 

B22 
A11B12  A12 B22 

A21B12  A22 B22 
Hence
A11B11  A12 B21  I
A11B12  A12 B22  0
1
 2
A21B11  A22 B21  0
A21B12  A22 B22  I
From (1)
3
 4
A11  A12 B21 B111  B111
From (3)
1
A22
A21  B21 B111  0 or B21 B111   A221 A21
Hence
or
1
22
1
11
A11  A12 A A21  B
B11   A11  A12 A A21 
1
22
1
1
1


 A  A A  A22  A A A  A21 A11
1
11
1
11 12
1
21 11 12
using the Woodbury Theorem
Similarly
1
B22   A22  A A A 
1
1
1
1
 A22  A22 A21  A11  A12 A22 A21  A12 A221
1
21 11 12
From
A21B11  A22 B21  0
3
1
A22
A21 B11  B21  0
and
B21   A A21B11   A A21  A11  A12 A A21 
1
22
1
22
1
22
1
similarly
B12   A A B22   A A  A22  A A A 
1
11 12
1
11 12
1
21 11 12
1
Summarizing
 A11
A 
nn
n  p  A21
A12 

A22 
p
Let
n p
p
B11

Suppose that A-1 = B 

n  p  B21
p
p
B12 

B22 
n p
then
1
1
B11   A11  A12 A A21   A  A A  A22  A A A  A21 A111
1
22
1
1
11
1
11 12
1
21 11 12
1
B22   A22  A A A   A  A A21  A11  A12 A A21  A12 A221
1
21 11 12
1
22
1
22
1
22
B12   A A B22   A A  A22  A21 A111 A12 
1
11 12
1
11 12
1
B21   A A21B11   A A21  A11  A12 A A21 
1
22
1
22
1
22
1
Example
Let
 aI
A 
p  cI
p
p
a


bI   0


dI   c
p


0
Find A-1 = B
0
0


b

0


d
b
a 0
0 d
c
 B11
 
n  p  B21
p
p
0
B12 
B22 
n p
A11  aI , A12  bI , A21  cI , A22  dI
1
B11  aI  bI  I  cI    a  bcd  I 
1
d
1
1
d
ad bc
B22   dI  cI  I  bI    d  bca  I 
1
a
1
I
a
ad bc
B12   A111 A12 B22   1a I (bI ) ad abc I   ad bbc I
1
B21   A22
A21 B11   d1 I (cI ) addbc I   ad cbc I
d

ad bc I
1
hence A   c
  ad bc I
 ad bbc I 

a
I
ad bc

I
The transpose of a matrix
Consider the n × m matrix, A
 a11 a12
a
a22
21

A   aij  


 am1 am 2
a1n 
a2 n 


amn 
then the m × n matrix,A (also denoted by AT)
 a11 a21
a
a22
12

A   a ji  


 am1 am 2
is called the transpose of A
am1 
am 2 


amn 
Symmetric Matrices
• An n × n matrix, A, is said to be symmetric if
A  A
Note:

 AB   BA
1
1 1
 AB   B A
 A
1
 A
1


The trace and the determinant of a
square matrix
Let A denote then n × n matrix
 a11 a12
a
a
21
22

A   aij  


 an1 an 2
Then
n
tr  A    aii
i 1
a1n 

a2 n 


ann 
 a11 a12
also
a
a
21
22

A  det


 an1 an 2
a1n 
a2 n 
 the determinant of A


ann 
n
  aij Aij
j 1
where Aij  cofactor of aij 
 1
i j
 the determinant of the matrix 


th
th
 after deleting i row and j col. 
 a11 a12 
det 
 a11a22  a12 a21

a21 a22 
Some properties
1.
I  1, tr  I   n
2.
AB  A B , tr  AB   tr  BA
3.
4.
1
A
1

A
 A11
A 
 A21
1

A
A

A
A
A12   22 11 12 22 A21


A22   A11 A22  A21 A111 A12

 A22 A11 if A12  0 or A21  0
Some additional Linear Algebra
Inner product of vectors
Let x and y denote two p × 1 vectors. Then.
x  y   x1 ,
 y1 
 
, x p     x1 y1 
 yp 
 
p
  xi yi
i 1
 xp yp
Note:
2

x  x  x1 
 x  the length of x
2
p
Let x and y denote two p × 1 vectors. Then.
cos  
x  y
 angle between x and y
x  x y  y
x
y

Note:
Let x and y denote two p × 1 vectors. Then.
cos  
 
x  y
if between
x  y  0 xand
 0angle
andy  2
x  x y  y
 


Thus if x   y  0, then x and y are orthogonal .
x
y
 2
Special Types of Matrices
1. Orthogonal matrices
– A matrix is orthogonal if PˊP = PPˊ = I
– In this cases P-1=Pˊ .
– Also the rows (columns) of P have length 1 and
are orthogonal to each other
Suppose P is an orthogonal matrix
then
PP  PP  I
Let x and y denote p × 1 vectors.
Let u  Px and v  Py
Then uv   Px   Py   xPPy  xy
and uu   Px   Px   xPPx  xx
Orthogonal transformation preserve length and
angles – Rotations about the origin, Reflections
Example
The following matrix P is orthogonal
1
1
P
1

1
3
2
3

3
1
2
1
6


0 
 2 6 
1
6
Special Types of Matrices
(continued)
2. Positive definite matrices
– A symmetric matrix, A, is called positive definite
 if:
2
2

x Ax  a11x1    ann xn  2a12 x1 x2   2a12 xn 1 xn  0
 
for all x  0
– A symmetric matrix, A, is called positive semi
definite if:
 
xAx  0
 
for all x  0
If the matrix A is positive definite then

 
the set of points, x , that satisfy x Ax  c where c  0
are on the surface of an n  dimensiona l ellipsoid

centered at the origin, 0.
Theorem The matrix A is positive definite if
A1  0, A2  0, A3  0,, An  0
where
 a11 a12 a13 
 a11 a12 


A1  a11  , A2  
, A3  a12 a22 a23 ,

a12 a22 
a13 a23 a33 
 a11 a12  a1n 
a a  a 
12
22
2n 

and An  A 

   


a1n a2 n  ann 
Example
.5 .25 .125
 1
 .5

1
.
5
.
25

A
 .25 .5 1
.5 


1 
.125 .25 .5
 1 .5 .25


A3  det  .5 1 .5 
.25 .5 1 
 1 .5
A2  det 

.
5
1


A  A4  0.421875  0
A3  0.5625  0
A2  0.75  0, A1  det1  1  0
Special Types of Matrices
(continued)
3. Idempotent matrices
– A symmetric matrix, E, is called idempotent if:
EE  E
– Idempotent matrices project vectors onto a linear
subspace



EEx   Ex
x

Ex
Example
Let A be any m  n matrix of rank m  n.
-1


Let E  A  A  A  A, then E is idempotent .
Proof :



EE  A A  A A A A  A A
-1
-1
 A A  A  AA A  A A
-1
-1


 A A  A  A  E
-1
Example (continued)
1 1 0
A 

0 1 1 
1
1 0
1 0 
 1 1 0 
 1 1 0
-1



E  A A  A A  1 1 
1 1  



0
1
1
0
1
1
 0 1   

0 1 


1 0
1 0 2
1
 3
2 1 1 1 0 



 1 1 
 1 1  1



3
1 2 0 1 1


0 1
0 1
 23
  13
 13
1
3
2
3
1
3
 13 
1 
3 
2 
3 
 13  1 1 0

2 
0
1
1

3 
Vector subspaces of n
Let n denote all n-dimensional vectors (ndimensional Euclidean space).
Let M denote any subset of n.
n if:
Then
M
is
a
vector
subspace
of


1. 0 M

 
2. If u M and v M then u  v 
M


3. If u M then cu M
.
Example 1 of vector subspace
 
Let M  u au  a1u1  a2u2   anun  0


 a1 
a 
  2
where a 
is any n-dimensional vector.

 
 an 
Then M is a vector subspace of n.
Note: M is an (n - 1)-dimensional plane through
the origin.
Proof
 
Now M  u au  a1u1  a2u2   anun  0


1. a0  0.


2. If au  0 and av  0
    
the n au  v   au  av  0

3. If au  0
 

the n acu   cau  0


Note the vector a is orthogonal to any vector u

since au  0

Projection onto M.

Let x be any vector

The equation of the line through x
perpendicu lar to the plane
M
  
is u  x  ta
  
The point u  x  ta is the projection onto the plane

if t is chosen so that u is on the plane.

  
ax
i.e. au  ax  taa  0 and t    
aa


   ax 
and u proj  x  ta  x    a
aa
Example 2 of vector subspace




Let M  u u  b1a1  b2a 2    b p a p


 

a1 , a 2 ,, a p is any set of p n  dimensiona l vectors.
Then M is a vector subspace of n.
M is called the vector space spanned by the p
 

n -dimensional vectors: a1 , a 2 ,  , a p
M is a the plane of smallest dimension through
the origin that contains the vectors:
 

a1 , a 2 ,  , a p
Eigenvectors, Eigenvalues of a
matrix
Definition
Let A be an n × n matrix
Let x and  be such that
Ax   x with x  0
then  is called an eigenvalue of A and
and x is called an eigenvector of A and
Note:
 A  I  x  0
If A   I  0 then x   A   I  0  0
1
thus A   I  0
is the condition for an eigenvalue.
 a11   

A   I  det 
 an1


= 0
 ann   
a1n
= polynomial of degree n in .
Hence there are n possible eigenvalues 1, … , n
Thereom If the matrix A is symmetric then the
eigenvalues of A, 1, … , n,are real.
Thereom If the matrix A is positive definite then
the eigenvalues of A, 1, … , n, are
positive.
Proof A is positive definite if xAx  0 if x  0
Let  and x be an eigenvalue and
corresponding eigenvector of A.
then Ax   x
xx
and xAx   xx , or  
0
xAx
Thereom If the matrix A is symmetric and the
eigenvalues of A are 1, … , n, with
corresponding eigenvectors x1 , , xn
i.e. Axi  i xi
If i ≠ j then xix j  0
Proof: Note xj Axi  i xj xi
and xiAx j   j xix j
0   i   j  xix j
hence xix j  0
Thereom If the matrix A is symmetric with
distinct eigenvalues, 1, … , n, with
corresponding eigenvectors x1 , , xn
Assume xixi  1
then A  1 x1 x1 
  x1 ,
 n xn xn
0   x1 
1




 PDP
, xn  
 
 0
n   xn 
proof
Note xixi  1 and xix j  0 if i  j
 x1 


PP     x1 ,
 xn 
1


0
 x1x1

, xn   
 xn x1
0
I

1 
x1xn 


xn xn 
P is called an
orthogonal matrix
therefore P  P
and PP  PP  I .
thus
 x1 


I  PP   x1 , , xn     x1 x1   xn xn
 xn 
now Axi  1 xi and Axi xi  i xi xi
Ax1 x1 
A  x1 x1 
1
1
 Axn xn  1 x1 x1 
 n xn xn
 xn xn   1x1x1 
 n xn xn
A  1 x1 x1 
 n xn xn
Comment
The previous result is also true if the eigenvalues
are not distinct.
Namely if the matrix A is symmetric with
eigenvalues, 1, … , n, with corresponding
eigenvectors of unit length x1 , , xn
then A  1 x1 x1 
  x1 ,
 n xn xn
0   x1 
1




 PDP
, xn  
 
 0
n   xn 
An algorithm
for computing eigenvectors, eigenvalues of positive
definite matrices
• Generally to compute eigenvalues of a matrix
we need to first solve the equation for all
values of .
– |A – I| = 0 (a polynomial of degree n in )

• Then solve the equation for the eigenvector , x ,


Ax  x
Recall that if A is positive definite then
A  1 x1 x1 
 n xn xn
 

where x1 , x2 ,, xn are the orthogonal eigenvecto rs
 
 

of length 1. i.e. xi xi  1 and xi x j  0 if i  j 


and 1  2    n  0 are the eigenvalue s
It can be shown that



2
2
2
2
A  1 x1  x1  2 x2  x2    n xn  xn



m
m
m
m
and that A  1 x1  x1  2 x2  x2    n xn  xn
m
     m  






m
m  
n
2
 1  x1  x1    x2  x2      xn  xn   1 x1  x1


 1 
 1 
Thus for large values of m
 
m
A  a constant x1  x1
The algorithim
1. Compute powers of A - A2 , A4 , A8 , A16 , ...
2. Rescale (so that largest element is 1 (say))
3. Continue until there is no change, The
 
m
resulting matrix will be A  cx1  x1



 
m
4. Find b so that A  b  b   cx1  x1
5. Find

x1 


1 
  b and 1 using Ax1  1 x1
b  b

To find x2 and 2 Note :
 
 
 
A  1 x1  x1  2 x2  x2    n xn  xn
6. Repeat steps 1 to 5 with the above matrix

to find x2 and 2
7. Continue to find



x3 and 3 , x4 and 4 ,, xn and n
Example
A=
eigenvalue
eignvctr
5
4
2
4
10
1
1
12.54461
0.496986
0.849957
0.174869
2
1
2
2
3.589204
0.677344
-0.50594
0.534074
3
0.866182
0.542412
-0.14698
-0.82716
Differentiation with respect to a
vector, matrix
Differentiation with respect to a vector
Let x denote a p × 1 vector. Let f  x  denote a
function of the components of x .
 df  x  


 dx1 
df  x  




dx
 df  x  
 dx 
p 

Rules
1. Suppose
f  x   ax  a1x1 
 an xn
 f  x  


 x1   a1 
df  x  
 

then

 a


dx
 f  x    a p 
 x 
p 

2. Suppose
f  x   xAx  a11x12 
2a12 x1 x2  2a13 x1 x3 
 a pp x2p 
 2a p 1, p x p 1 x p
 f  x  


 x1 
df  x  
  2 Ax
then



dx
 f  x  
 x 
p 

i.e.
f  x 
xi
 2ai1 x1 
 2aii xi 
 2aip x p
Example
1. Determine when f  x   xAx  bx  c
is a maximum or minimum.
solution
df  x 
dx
 2 Ax  b  0 or x   12 A1b
2. Determine when f  x   xAx is a maximum if
xx  1. Assume A is a positive definite matrix.
solution
let g  x   xAx   1  xx 
 is the Lagrange multiplier.
dg  x 
dx
 2 Ax  2 x  0
or Ax   x
This shows that x is an eigenvector of A.
and f  x   xAx   xx  
Thus x is the eigenvector of A associated with
the largest eigenvalue, .
Differentiation with respect to a matrix
Let X denote a q × p matrix. Let f (X) denote a
function of the components of X then:
 f  X 

x11

df  X   f  X   



 xij  
dX

  f  X 


x
q1

f  X  

x1 p 


f  X  

x pp 
Example
Let X denote a p × p matrix. Let f (X) = ln |X|
then
d ln X
dX
X
Solution
X  xi1 X i1 
1
 xij X ij 
 xip X ip
Note Xij are cofactors
 ln X
xij
1

X ij = (i,j)th element of X-1
X
Example
Let X and A denote p × p matrices.
Let f (X) = tr (AX) then
Solution
p
d tr  AX 
p
dX
tr  AX    aik xki
k 1 k 1
tr  AX 
xij
 a ji
 A
Differentiation of a matrix of functions
Let U = (uij) denote a q × p matrix of functions
of x then:
 du11

dx

dU  duij 


dx  dx  
duq1

 dx
du1 p 

dx 

duqp 

dx 
Rules:
1.
2.
3.
d  aU 
dx
dU
a
dx
d U  V 
dx
d UV 
dx
dU dV


dx dx
dU
dV

V U
dx
dx
4.

d U
1
dx
Proof:
  U
1
dU 1
U
dx
U 1U  I
1
dU
1 dU
U U
 0
dx
dx p p
1
dU
1 dU
U  U
dx
dx
1
dU
1 dU
 U
U 1
dx
dx
5.
dtrAU
 dU 
 tr  A

dx
 dx 
p
p
Proof: tr  AU    aik uki
i 1 k 1
tr  AU 
x
6.
dtrAU
dx
1
uki
 dU 
  aik
 tr  A

x
 dx 
i 1 k 1
p
p

1 dU
1 
 tr  AU
U 
dx


dtrAX 1
 ij  1
  tr E X AX 1
dxij

7.
E
 kl 
 kl 
 kl 
 (eij ) where eij

1 i  k , j  l

0 otherwise
Proof:
dtrAX
dxij
8.
1


1 dX
1
 tr   AX
X    tr AX 1 E  ij  X 1


dx
ij


 ij  1
  tr E X AX 1



dtrAX 1
  X 1 AX 1
dX

The Generalized Inverse of a
matrix
Recall
B (denoted by A-1) is called the inverse of A if
AB = BA = I
• A-1 does not exist for all matrices A
• A-1 exists only if A is a square matrix and
|A| ≠ 0
• If A-1 exists then the system of linear
equations Ax  b has a unique solution
1
xA b
Definition
B (denoted by A-) is called the generalized inverse (Moore –
Penrose inverse) of A if
1. ABA = A
2. BAB = B
3. (AB)' = AB
4. (BA)' = BA
Note: A- is unique
Proof: Let B1 and B2 satisfying
1. ABiA = A
2. BiABi = Bi
3. (ABi)' = ABi
4. (BiA)' = BiA
Hence
B1 = B1AB1 = B1AB2AB1 = B1 (AB2)'(AB1) '
= B1B2'A'B1'A'= B1B2'A' = B1AB2 = B1AB2AB2
= (B1A)(B2A)B2 = (B1A)'(B2A)'B2 = A'B1'A'B2'B2
= A'B2'B2= (B2A)'B2 = B2AB2 = B2
The general solution of a system of Equations
Ax  b
The general solution
x  A b   I  A A z

b   I  where
A A  z is arbitrary

Suppose a solution exists
Ax0  b
Let
x  Ab   I  A A  z
then Ax  A  Ab   I  A A z 



  AA b   A  AA A z 

 AA Ax0  Ax0  b
Calculation of the Moore-Penrose g-inverse
Let A be a p×q matrix of rank q < p, then A   AA  A

Proof
thus
also
 
 
A A A A


1
 


A  A  A A

1
A A  I
AA A  AI  A and A AA  IA  A
A A  I is symmetric
 
and AA  A A A

1
A is symmetric
1
Let B be a p×q matrix of rank p < q, then B  B  BB 

Proof
thus
 
 
BB  B  B BB


1
 

 BB

BB

1
I
BB  B  IB  B and B  BB   B  I  B 
BB  I is symmetric
also
 
and B B  B BB

1
B is symmetric
1
Let C be a p×q matrix of rank k < min(p,q),
then C = AB where A is a p×k matrix of rank k and B is a k×q
matrix of rank k
then C  B  BB 

Proof
1
 AA A
1
   A A

CC   AB  B BB

is symmetric, as well as




1
1
 

A  A A A

1
 
 


  
 
A
1
1
1


C C   B BB
A A A AB  B BB B


1


Also CC C   A A A A AB  AB  C


1
1
1




and C CC    B BB B   B BB
A A A

  1

1
 B BB
A A A  C 