Lecture 2
Matrix Series
We consider matrix series


k 1
Ak  lim
N 
N
A
k
k 1
,
(1)
where Ak is likes matrices.
If exist limit (1) then a matrix series is to converge (be convergent) and the
limits of matrix series is called “sum series”.
Necessary condition of matrix series convergence
Theorem 1. If matrix series (1) is to converge (be convergent), then
lim Ak  0.
x
k
Proof. Let S k   A j . If matrix series (1) is to converge, then  S  lim Sk
N 
j 1
Ak  Sk  Sk 1, 
lim Ak  lim Sk  lim Sk 1  S  S  0 .
k 
k 
k 
The matrix series (1) be absolute convergent if

A
k 1
(2)
k
is to converge.
Theorem 2. If matrix seriesis is to absolute converge, then is to converge.

  (k ) 
(k )


Proof. Let Ak   aij  , k  1, 2.... The  Ak   aij  (2) matrix
k 1
 k 1

seriesis is to converge. By definition, every number series

a
k 1
(k )
ij
(i  1,2,..., m, j  1,2,..., n) is to converge. By theorem of series

theory, all series
a
k 1
(k )
ij
(i  1, 2 , . .m. , ,j
1, 2n, . is
. . .to
, absolute
)
converge,
N
that is  S  lim S N  lim  Ak , else matrix seriesis (1) is to converge.
N 
Theorem 3. Let
N 
k 1
Sufficient Condition
A is canonical norm of matrix A . If number series


k 1
is to converge, then
converge.
Ak
(3)
matrix seriesis (1) is to converge and is to absolute
(k )
Proof. Let Ak   aij  ,
k  1, 2,... We consider the number series
1

a
(i  1, 2,...,m , j  1, 2,...,n .)
(k )
ij
k 1
(4)
aij( k )  Ak since, every of series (4) is to converge and is to absolute converge.
  (k ) 
A


k
 aij  is to absolute
k 1
 k 1


From this it follows that (by defination), series
converge.
Degree Matrix Series

A X
1) right
k
k 0
k
,
(5)

X
2) left
k 0
k
(5')
Ak ,
where X is n -dimensional square matrix.
j 1, n
Cases: 1) Ak  aij( k ) i 1,m (row vector),
2) Ak  aij( k ) i 1,n (column vector).
j 1, m
Theorem 4. If r is radius of convergence of degree series


k 0
where Ak
Ak x k ,
(6)
is canonical norm, then degree matrix series (5) and (5')
x r
(7)
to fulfill a condition (7) is to converge.
Specifically, if the condition x  r is fulfilled, then degree matrix series

a
k 0
k
(k  0,1,...)
(k  0,1,...)
X k with number coefficients ak
is convergent, where r is

radius of convergence of degree series  ak x k .
k 0
Proof.
Ак X k  Ak  X
k

since, then it follows from (7) that series  Ak X k
k 0
is convergent. It follows that, by Theorem 5, degree matrix series (5) is
convergent. Just as a convergence formula (5/) is proofed. Second derivation of
Theorem: a k is number, then ak  ak Ak  a11( k )  is numbers .

Theorem 5.

A  AX  AX 2  ...  AX k  ... ,
A  XA  X A  ....  X A  ... ,
2
k
(8)
(8')
where X is quadrat matrix. If condition
X 1
(9)
is fulfilled, then the geometric series (8) and (8') is convergent and

 AX k  A( E  X ) 1 ,
k 0

X
k
A  ( E  X ) 1 A .
k 0
2
Proof. By Theorem 4 if condition (9) is satisfiable then the geometric series (8)

is convergent and  S   AX k . Consider the equality:
k 0
A( E  X  X 2  ....  X k )( E  X )  A( E  X k 1 ) .
(10)
k 1
In (10) since k   and және condition (9) is satisfiable then X  0 , k  
is satisfiable and we have:
S ( E  X )  AE  A
(11)

Specifically, in the equality (11) if A  E then S1 ( E  X )  E , where S1   X k .
k 0
 det( E  X )  0  matrix
So det S1  det( E  X )  det E  1 and det S1  
E  X 
1
is non-singular that is  ( E  X ) . The equality (11) on the right side

of the two parts on matrix ( E  X ) 1 is multiply: S   AX k  A( E  X ) 1 .
k 0

Specifically,  X k A  ( E  X ) 1 A , if X  1 is satisfiable.
k 0
Corollary 1. If X  1 then

( E  X ) 1   X k , and if

E  1 then
k 0

(E  X )   X
1
k
1

.
1 X
Remark. If X  1 then (residual member of norm of the matrix series (8))
k 0
Rk  A( E  X ) 1  A( E  X  ...  X k )  A X k 1  X k  2  ... 

A  X
k 1
 X
k 2

A X
 ... 
1 X
k 1
.
A X
R  ( E  X ) A  ( E  X  ...  X ) A 
1 X
1
/
k
k
Specifically,
(8'):
k 1
.

Example.
Xn
.
e 
n 0 n!
X
(12)
Adjunction
Theorem. Let X is squarte matrix and 1 , 2 ,..., n is eigenvalues of matrix
X . The inequality  j  X ( j  1,2,..., n) is correct.
Proof. Let X   and consider the matrix Y 
1
 X , where   0 .
 
1

and
the
series
X 
1
 
 
E  Y  Y 2  ...  Y k ... is convergent. By Theorem of matrix geometric series the
matrix geometric series E  X  X 2  ....  X k  ... if and only if is to converge if
Then
Y 
3
all eigenvalues i  i ( x), (i  1,2,..., n) the condition
 j  1, ( j  1, 2,...,n )
is satisfiable. If series E  Y  Y 2  ...  Y k ... is non-convergt then for k  
X k  0 not to hold. Then the eigenvalues 1 , 2 ,..., n of matrix Y the
inequality  j  1 ( j  1, n)
Y
1
X
 
 j 
is
satisfiable.
1
  j ; ( j  1,2,..., n) 
 
Bat
j    
1
 j   j  1. Otherwise,  is any number   j    X
 
since
 j  1,2,..., n  .
Hamilton-Kelli equality
Theorem. Let X is any squarte matrix. If     n  p1n1  ....  p n , where
 ( )  det( E  X ), then  ( x)  X n  p1 X n1  p2 X n2  ....  pn E  0 .
4