Добавил:

ivanov666 Опубликованный материал нарушает ваши авторские права? Сообщите нам.

Вуз:

Казахский национальный университет им. аль-Фараби

Предмет:

[НЕСОРТИРОВАННОЕ]

Файл:

86

.pdf

Скачиваний:

Добавлен:

07.06.2023

Размер:

8.14 Mб

Скачать

☆

<<< < Предыдущая 1 2 3 4 5 6 7 8 9 1011 / 2311 12 13 14 15 16 17 18 19 20 21 22 23 > Следующая >>>

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

where

	0		0		N /2						N				N			N	N
						21								21			22	11		23
						21								21			22	11		23
P =	0		0		N	22		/2	,	P =		N	21		N	22		N	N	23	,
1						22				2			21			22		12		23
	N /2	N		/2	N							N			N			N	N
	N /2	N	22	/2	N		23					N	21		N	22		N	N	23
	21		22				23						21			22		13		23

			N			N	N	N	N
			N			11	12	11	13
						11	12	11	13
				11			2		2				N11						N21
		N		N			2	N	2				N11						N21
		N	11	N	12			N	N
P =						N				,	N	=		N	,	N		=		N .
3				2			12		2		1			12			2			22
3							12				1			12			2			22
	N					N	N
	N		11	N	13	N	N						N13						N23
			11		13	12	13	N13
				2			2	N13
				2			2

Then the improper integral (v. (2.28)):

I	5	= I	1
	5		1

where matrices

I	3	=
	3

R1,

I	I
11	12

, W
1	1

[ y

(t)R y

(t)W y(t)]dt =

(t) y

(t) y(t) y

( )

i ( i ) 1i d i

< ,

i=1

(0)

are equal:


												1				2		N /2
												1				2
						1		11			11	2			2		12		21
					1							2			2
R =			P =		1				2									N		/2
R =			P =		2			2										N		/2	,
1	11	21	1	11	2			2		12			2		12				22
1	11	21	1	11						12			2		12				22
						N	21						N	22				N23
							21							22
						2							2
						2							2

	=		22	P =
1	12		22	2

11( 1 1 ) 2 12 N21

	1 11			2 12 N21
			(			)		N
		11		1			12		21
					1	2

							W1
	1 11 2 12 N11
	1 11 2 12 N11
	1					N11 N12
=	1					N11 N12
=	2					2
	2		11	2	12	2
		1 11			N11 N13
					N11 N13
						2
						2

	1 11 ( 2 2 ) 12 N22																		N11 N23
	2 12 N22																		N12 N23						,
				11				12		(		2		2	) N		22		N N					23
		1		11				12				2		2			22			13				23
= 13 23 P3 =
		1									N N										N	11	N
		1	2 12								11		12			1 11						11			13
	2		2 12									2				1 11							2
	2											2											2
																1				N12 N13
									N							1				N12 N13						.
									N							2										.
					2	12				12						2		11					2
		1 11					N12 N13									1 11 2 12 N13
							N12 N13

		2								2

101

Lectures on the stability of the solution of an equation with differential inclusions

Here

D = SK	1

	11	,		12	,
	11			12

=1 1 1

2 2 2

N	11	,	N	12	,	N	13	,	N	21	,	N	22	,	N	23	are arbitrary parameters, matrix
	11			12			13			21			22			23
	is a feedback matrix. If the equalities are satisfied
	is a feedback matrix. If the equalities are satisfied

	11	(	2		2	)	12	N	22	=	11		12	N	21
1										1		2

11( 1 1 ) 2 12 N21 = N11 N23,

1 11 12 ( 2 2 ) N22 = N12 N23 ,

(2.42)

then matrix

=		.
	*
1	1

Now from (2.28) when performing equalities (2.42), we have

	2		( )
		i

I 5

(t)W y(t)]dt =

(

)

[ y

(t)R y(t) y

i=1

(0)

(t) y(t)

dt < .

(2.43)

due to the fact that			\| y( ) \| c			< ,			\| y(0) \| c				< .			Matrix			R = R		*	0,	when
due to the fact that			\| y( ) \| c			< ,			\| y(0) \| c				< .			Matrix			R = R			0,	when
					2							2							1	1

															1			2
															1			2
									1	11				11	2		2		12
=			0,		=										2		2			0,		\| R	\| 0.
1	1	11		2																		1
1	1	11		2																		1
								1			2
								1			2
							11	2		2		12				2	12
								2		2

(2.44)

Consequently, when performing inequalities (2.44) from (2.43) it follows that the improper integral


I		=		*	(t)W y(t)dt < .
I	50	=		y	(t)W y(t)dt < .
	50				1
			0

(2.45)

[0,

If arbitrary parameters and elements of the feedback matrix		D selected so that
*	then all conditions of theorem 6 are satisfied. Consequently, in the sector
= W1 > 0,
	0 = diag( 01, 02 ), = diag( 1, 2 ), 1 > 0, 2 > 0
0 ],		are arbitrarily small

numbers, Iserman’s problem has a solution. The limit values 01, 02 are denoted from the condition of Hurwitz of the matrix A B 0 D.

For positive definiteness of the matrix W1 = W * > 0 it is necessary and sufficient that the following inequalities are performed:

102

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems


							1
						NN
	11	1111		22	1212		1111
					NN		N
11					1111		1212

22		22	1212				22
22							22

11		12	N			> 0,
11	2	12		11
								N
	11							1111
	22		22	1212					22
	22								22
							N
			22		1212			1212

N 1212

|W 11

|| 0.0.

(2.46)

The characteristic polynomial of the matrix

A B	D
0

is equal to

						1			1			01			1		1
						1									1
							01						1			01
2	( ) =\| I	3	A B	D \|=		1		2		1			02	2	1		2	=
2		3	0				02	2					02	2	02		2
							1				1				1
					3		2		a		,
				= a			a		a		,
					2			1		0

where

a2 = 3 01 1 02 2 , a1 = 3 01(2 1 1 1) 02 (2 2 2 )

01 02 ( 1 2 1 2 ), a0 = 2 01 1 02 ( 2 2 ) 01 02 ( 1 2

1 2 2 1 2 1 1 2 ).

For Hurwitz of the matrix																A B 0 D						necessary and sufficient condition is perfor-
mance of inequalities
												a0			> 0,		a1		> 0,			a2 > 0, a2a1 a0												> 0.												(2.47)
Limit values								01, 02					are denoted from equalities
a	= a	(	01	,			02	) = 2								02	(	2		2	)		01			02	(			2			2				1				1	a			2	) = 0,
0	0										01			1															1			1			2					2			1
	a = a (				01	,			02	) = 3				01	(2			1	)					02	(2			2			2	)	01		02	(			2				2	) = 0,
	1	1														1				1																		1				1

a2 = a2 ( 01, 02 ) = 3 01 1 02 2 = 0,																						a2 ( 01, 02 )a1( 01, 02 ) a0 ( 01, 02 ) = 0.

From the above results it follows:

1. For absolute stability of equilibrium state of the system (2.41) with limited nonlinearities it is necessary and sufficient that equalities (2.42), and inequalities (2.44), (2.46), (2.47) are fulfilled. (The total number of conditions is 13, and the total number

of parameters selected is 15).

	11	,
	11

	12	,
	12

N	11	,
	11

N	12	,
	12

N	13	,
	13

N	21	,
	21

N22,

N	23	,
	23

	,
1

	1	,
	1

	,
2

	,
2

2 to be

2.The effectiveness of the proposed method follows from the necessary and sufficient condition of absolute stability (2.42), (2.44), (2.45), (2.46), (2.47).

3.Constructiveness of the proposed method consists in that based on the necessary and sufficient condition of absolute stability you can select a region of absolute

103

Lectures on the stability of the solution of an equation with differential inclusions

stability in space of constructive parameters of the system

( A, B, S,	).
0

As follows

from [5, 6] by using the known methods it is not possible to select the area of absolute stability in space of constructive parameters for multidimensional regulated systems.

Absolute stability. Now consider equation (2.41) with given																																																	1	= x ,		2
																																																	1		1	2
				1		0	0		Using transformation																					A = KAK 1,																				D = SK 1,
				1		0	0																																		B = KB,
where matrix S =								.																																	B = KB,
						1	0
				1		1	0

y	= y		y		3	(				1	),	y	2		= y					y			2		y			3				(	2	),			y	= y				y				2	y				,
1	1				3		1			1			2					1					2					3		2			2				3				1					2				3
									1	= y				,				2		= y							,	( )								,
									1			1						2								2								0
												1				0				0									1					0										0
where matrix S = (S1, S 2 ) =																						,				S1			=								,		S 2 =								.
												0			1					0											0														0
																															0					1									0

For this example improper integral

I	=	*	( (t)) (t)dt =												[ y		*		(t) y								(t) y				*	(t)					y(t) y					*	(t)						y(t)]dt =
I	=		( (t)) (t)dt =												[ y				(t) y								(t) y					(t)			2		y(t) y						(t)					3	y(t)]dt =
1							1																1												2													3
	0														0
																	2				( )
																	2			1
															=						i ( i ) 1i d i ,
																i=1					(0)
																				1
where
		1					0			0																11				12				0										0					0		0


=				0			1			0		,				2	=						0								12			0			,			3	=				0				0		0 .
	1															2															12									3
				0			0			0																								0											0				0		0
				0			0			0														11						12				0											0				0		0
																								11						12

= x1 x2 ,

we get

(2.48)

Improper integral

			I		=						( (t))						( (t))									( (t)) (t)]dt =
			I		=		[				( (t))						( (t))									( (t)) (t)]dt =
										*						1									*
				2											2	0														2
						0

			=		[ y		*									*	(t)									*	(t)			y(t)]dt 0,
			=		[ y			(t) y(t) y									(t)			2	y(t) y						(t)		3	y(t)]dt 0,
													1							2									3
					0
where
		1									0			0								2 1									2 1						0
			21		01																		21	01				21				21		02
	= 0								22			1		0	,			2	=					0						2	22		1			22	0 ,
1									22				02					2													22		02			22
			0								0			0								2 1								2			22	1			0
																								21			01						22		02
							1							1								1		22						1			21				1
					21 01								22 02				22 02													21 01					22		02
					21 01								22 02				22 02							2						21 01			2		22		02
																								2									2
													22																						22
									1				22									1											1		22
3	=			22 02										2				22 02					22							22 02						2		,
					1				21						1							1		22							1					1
		21 01								2			22 02				22 02							2						21 01			22 02
										2														2

104

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

Improper integral

[ y

P y]dt = 0,

P y y

where matrices

1, P2 , P3

are defined by formulas shown above.

Improper integral

= I

[ y

(t)W y(t)]dt

(t)R y(t) y

(t) y(t) y

( )

i ( i ) 1i d i

< ,

= diag ( 11, 12 ),

2 = diag ( 21, 22 ) > 0.

i=1

(0)

where

									1							1				0										N /2
									1											0										N /2
													21			01															21
													21			01															21
					R		=					0					1										1		N					/2	,
					R		=					0					1					22					02		N		22			/2	,
					0																	22					02				22
											N			/2					N				/2							N
											N	21		/2					N	22			/2							N		23
		N			2							21								22					2							23					N	N
		N			2					1															2						1						N	N
	11			21				21		01					21						12							22			02						11		23
	11			21				21		01					21						12							22			02						11		23
=					N														N							2								1			N	N		,
=					N		21											12	N					22		2			22				02			22	N	N	23	,
0							21											12						22					22				02			22	12		23
			N			2					1														2						1						N	N
		11	N		21	2					01										12				2			22			02						N	N	23
		11			21				21		01										12							22			02						13		23

																																					N			N
										N									1								1										N		11	N					12						1			22
										N																														2

													11				21		01					22			02																						22		02		2
													11				21		01					22			02																						22		02
				W =						N		11	N			12						1							22											N								1
				W =								11				12					22	1							22											N			12				22	1				22
					0								2								22		02					2															12				22			02		22
													2															2									N			N
										N13									1				21											1			N			N									22 02
							N11			N13									1				21											1					13						12				22 02					22
																																							13						12						1			22
										2							21 01						2							22	02									2														2
										2													2
													N		11		N			13								1						21							1
															11					13								1						21							1
																	2								21			01					2					22			02
																	2																2
																	N13 N12															1				22
																					2								22			02					2							.
																					2								22			02					2
																		N										1								1
																				13					21		01						22			02
																				13					21		01						22			02

		If the equalities are satisfied
N		=		2			1		,		N					2							1			= N					N						,						2							1	= N			N		,
N	21	=	12	2	22		02		,	11	N			21		2				21			01			= N					N				23		,			12			2			22				02	= N		12	N	23	,
	21		12		22		02			11				21						21			01						11						23					12						22				02			12		23
then the matrix						0			*	,	improper integral
then the matrix						0		= 0		,	improper integral
																																							2	1		( )
					I 4 = [ y* (t)R0 y(t) y* (t)W0 y(t)]dt																																					i ( i ) 1i d i
									0																													i=1					(0)
																																									1

d 1

0 dt [ 2 y* (t) 0 y(t)]dt < .

(2.49)

105

Lectures on the stability of the solution of an equation with differential inclusions

		*	0		when inequalities are satisfied
Matrix R0 = R0			0		when inequalities are satisfied
									1						1			0
		1				1		0,	1									0				0,
		1						0,				21			01							0,
												21			01
				21		01					0					1				1					(2.50)
											0					1			22	02
																			22	02
											N		2								N		2
\| R \|= ( 1		1 )( 1						1 )N			N		21	( 1				1 )			N		22 ( 1		1 ) 0.
\| R \|= ( 1	21	1 )( 1					22	1 )N		23			21	( 1			22	1 )					22 ( 1	21	1 ) 0.
0	21	01					22	02		23	4						22	02				4		21	01
											4											4

When equalities (2.49) and inequalities (2.50) are satisfied, the improper integral

Matrices

W	= W	*	> 0

0	0


I		=		*	(t)W y(t)]dt < .
I	40	=	[ y		(t)W y(t)]dt < .
	40				0
			0

then and only then

(2.51)

N		1		1	> 0,
	21	01	22	02
11

N					1			1			N	N						1
N			21		1		22	1			11		12			22		1			22
	11		21	01			22		02			2				22	02			2
												2								2		,	\| W
N	N					1									1								0
																							0
11		12			22					22	N			22				22	> 0
	2				22	02			2		12			22	02			22
	2								2

(2.52)

The characteristic polynomial of the matrix A B 0 D is equal to
										1	01		0		1
										1			0		1

	3	( ) =\| I			3	A B	0	D0 \|=		1			1	02	1	=
	3				3		0							02
										1			1		1
						= b			b b ,
								3	2
								2	1			0
where b2 = 3 01			02 ,			b2 = 3 2 01 2 02 01 02 ,						b0	= 2 01 02.		It is easy to prove
that matrix A B				D is a Hurwitz matrix.
			0
Thus, when conditions (2.49), (2.50), (2.52) are satisfied, from evaluation (2.51) it
follows that	lim y(t) = 0					(v. proof of Theorem 5).
		t

So, the following statements are true:

1. For absolute stability of equilibrium state of multidimensional regulated systems with limited nonlinearities it is sufficient that equality (2.49), and inequalities (2.50), (2.52) are satisfied. (The total number of conditions is 9, and the total number of

parameters 11, 12, 22 > 0, 21 > 0, N11, N12 , N13 , N21, N22, N23 to be selected is 10).

2. Constructiveness of the proposed method is that based on relations (2.49), (2.50), (2.52) we can find the area of absolute stability in space of constructive parameters of the system which is essential for application problems.

106

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

3. The effectiveness of the proposed method follows from using properties of the

improper integral	I	3	.	Notice, that in the absence of the improper integral	I3	the total

number of selected parameters is 4, conditions is 10.

	11	,
	11

	12	,
	12

	22	> 0,
	22

	21	> 0,
	21

and the number of

Lecture 18.

Absolute stability and Iserman’s problem of multidimensional systems in a simple critical case. Problem statement. Nonsingular transformation

Problem statement. Equations of motion of multidimensional regulated systems in a simple critical case have a form:

x = Ax B ( ),

= ( ),

= Dx E ,

x(0) = x	,	(0) =	,	t I = [0, ),
0		0

(2.53)

where

are constant matrices of orders

n n,

n m,

m m,

m n,

m m , respectively,	A	is a Hurwitz matrix, i.e.

eigenvalues of the matrix			A.
Function

Re	( A) < 0,
j

j = 1, n,

	( A)
j

are

( )				= { ( ) = ( ( ),...,										( )) C(R								m		, R		m		) \| ( ) =
																						m				m
			0					1	1			m			m													i	i
0	(	)				2	,	( ) = (					(			),...,					(				)),			(0) = 0,
0	(	)		i		i	,	( ) = (					(			),...,				m	(		m		)),			(0) = 0,
i	i			i	0i	i					1			1			)R		,	m			m					< },
						, = (					,...,						)R	m	,		0 <							< },
																		m
										1					m												*

	i		(	)
i	i	i	i

\| ( ) \|	,
*

(2.54)

where = diag ( 1,..., m ) > 0,	i	> 0,	i = 1, m	are arbitrarily small numbers.
Notice, that

( ) = { ( ) C(R

, R

) | 0

( )

i = 1, m,

( ) = 0,

| ( ) |

0 <

< }.

(2.55)

Practical systems of automatic control relate to the systems with limited resources and for such systems function ( ) satisfies the conditions (2.54), (2.55).

As follows from inclusions (2.54), (2.55) equations of motion (2.53) can be represented as


z = A1z B1 ( ),	= S1z,	z(0) = z0 , \| z0 \|< , t I, ( ) 1,			(2.56)
where A1, B1, S1 matrices of orders (n m) (n m),			(n m) m,	m (n m) , respecti-
vely:

107

Lectures on the stability of the solution of an equation with differential inclusions

A B D

B E

= A ( ),

z =

A =

B =

S = D,

E .

The equilibrium states of the system (2.53), (2.54) are determined from solutions

of algebraic equations

A z

B (

) = 0,

= S z

If the matrix

A = A ( ) is a

1 *

Hurwitz matrix, the function ( ) 1 turns to zero only when

= 0,

then the system

(2.53),

(2.54)

has

the

only

equilibrium state

= (x , ) = 0.

Consequently

x = 0,

= 0, where

Notice, that trivial solutions

x(t) 0,

(t) 0,

t I

(2.53), (2.54) correspond to the

equilibrium state. We study the stability of generally unperturbed movement

x(t) 0,

(t) 0,

t I

in the system (2.53), (2.54) for all

( ) 0 (or

z(t) 0,

I , ,

( ) 1 ).

We assume with a sufficiently small neighborhood of the point

= 0,

function

( ) can be approximated by the linear function

( ) = ,

where = diag ( 1,..., m ).

Consequently, when | |< ,

> 0

is a sufficiently small number, perturbed motion

equation (2.56) can be written as:

z = A1z B1 S1z = A1( )z,

z(0) = z0 ,

| z0

|< ,

t I,

(2.57)

where

A ( ) = A B S ,

= diag(

,...,

)

is a limit diagonal matrix

determined by

Hurwitz

the

matrix

A1( ).

the

matrix

A1( ) = A1( ) B1 S,

0 is a Hurwitz matrix, then there is a number

> 0 such as

| x(t) |< 1 when

| z

moreover

lim z(t) = 0. Thus, when the matrix

A ( ),

is a Hur-

witz matrix, trivial solution

z(t) 0,

t I , (x(t) 0,

(t) 0,

t I )

of the system (2.57)

is asymmetrically stable by Lyapunov when t .

Definition

The

trivial solution

x(t) 0,

(t) 0,

t I

of the

system

(2.53),

(2.54) is called absolutely stable if matrices A1 = A1( ), are Hurwitz matrices (asymptotic stability by Lyapunov

( ) 1

the solution of the differential equation

A1( ), when (2.56)

.);

has the

,		0	>	0
		0		0

2) for all property

lim z(t;0, z0 , ) =
t
0 ,	\| x0 \|< ,	\|

0	\|<

\| z	0	\|<	(or	lim	x(t;0, x , ) = 0,	lim	(t;0, x	,	, ) = 0,	,
	0			lim	0	lim	0	0
				t		t
).

Definition 5. Conditions of absolute stability of the system (2.53), (2.54) are

called ratios, binding constructive parameters of the system ( A,	B, ,	D, E, 0 ),
0 = diag ( 01,..., 0m ), under whichthe equilibrium state ( x* = 0,	* = 0)	is absolutely
stable.

108

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

Problem 5. Find the condition of absolute stability of the equilibrium state of the

system (2.53), (2.54).

Function

(t) = S1z(t),

t I is a control formed by the principle of feedback, and

the matrix

of order

m (m n)

is called a feedback matrix. Iserman’s problem con-

sists in choosing a feedback matrix

so that from asymptotic stability of the trivial

solution

z(t) 0,

of the linear system

(2.57) for any ,

0 ,

0 = 0 2

there follows absolute stability of the trivial solution

x(t) 0,

(t) 0,

t I

of the sys-

tem (2.53), (2.54), where

= diag(

,...,

) is a limit diagonal matrix of Hurwitz of

the matrix

( ),

2 > 0

is a diagonal matrix of order m m with arbitrarily small

positive elements.

Definition 6. We assume that in the sector [ ,

] Iserman’s problem has a solu-

tion, if: 1) there exists matrix S1 of order

m (m n)

such as

0 = 0

; 2) for any

( ) = ,

solution of the system (2.56) is asymptotically stable; 3) for

any ( ) 1

trivial solution of the system (2.53), (2.54), (2.55) is absolutely stable.

Problem 6. Find the sector [ ,

2 ], where Iserman’s problem has a solution.

Note that Iserman’s problem does not always have a solution. The problem con-

sists of finding such a feedback matrix

of order

m (m n),

for that in the sector

[ , 0 ], where 0 = 0 2

Iserman’s problem had a solution. Sector

[ ,

0 = 0 2

means that

[ i , 0i ],

= 0i 2i ,

where

= diag ( 1,..., m ),

= diag ( 01,..., 0m ),

= diag(

,...,

= diag(

,...,

> 0, 2 > 0

Nonsingular transformation. For ease of verification of the proposed condition of absolute stability it is advisable to convert the original equation (2.56). Let the


matrix B1 =\|\| b1,...,bm \|\|,	where bi ,	i =			are column vectors (m n) 1.
			1, m
Lemma 7. Let the vectors i		R	n m	, i = 1, m such that:

	*	*	= 0, i = 1, m, i j,
	i bi = 1,	i bj				(2.58)

where bi bj , i j,

(*)

is a transpose sign. Then along the solution of the system

(2.56) the following identities hold


z(t) = A z(t)							(	(t)), t I = [0, ), i = 1, m.				(2.59)
i				i 1		i	i
If in addition rank	B	*	= m		and Gram determinant is the following
	B		= m
	1
				< 1, 1 >			< 1, 2 >		...	< 1, m >
( 1,..., m ) =					...			...	...	...	0,	(2.60)
( 1,..., m ) =					...			...	...	...	0,
				< m , 1 >			< m , 2 >		...	< m , m >

109

Lectures on the stability of the solution of an equation with differential inclusions

then vectors i ,

i = 1, m

scalar product of vectors

exist and they are linearly independent, where

	,		j	,	i, j = 1, m.
i			j

	,	j
i		j

is a

left

Proof. As

z(t) = A1z(t

	*	= (		,...,		),	i = 1, m, then by multiplying the identity on the
	i		i1		im n

) B1 i ( i (t)),					t I		*
							on i	we have

i*z(t) = i* A1z(t) i*B1 i ( i (t)), t I, i = 1, m,

where i* that ratio

	*
B = (		b ,...,
1	i	1

(2.58) can be

*	Hence taking into account (2.58), we get (2.59). Notice,
i bm ).

written as

...

= 0,

i1 11

in m

1n m

........

i1bi1

i2bi2 ... in mbin m = 1,

........

...

= 0, i =1, m,

i1 m1

in m

mn m

k 2

, k = 1, m.

...

kn m

= m then this system of equations has a solution

i ,

From the

If rank B1

i = 1, m.

condition (2.60) it follows that vectors

i ,

i = 1, m

are linearly independent. The lem-

ma is proved.

,...,

n m

Lemma 8. Let the vectors

from

such that

b = 0,

i = 1, m,

k = 1, n

, i j. Then along the solution of the system (2.56) the

identities hold

, bi bj

z(t) =

A z(t), k = 1, n, t I.

(2.61)

If in addition rank B* = m and

Gram determinant is the following

< 01, 01

> ...

< 01, 0n >

( 01,..., 0n ) =

...

(2.62)

< 0n , 01 > ... < 0n , 0n >

then vectors 01,..., 0n exist and are linearly independent.

110

<<< < Предыдущая 1 2 3 4 5 6 7 8 9 1011 / 2311 12 13 14 15 16 17 18 19 20 21 22 23 > Следующая >>>

Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]

#
07.06.20237.09 Mб881.pdf
#
07.06.20237.1 Mб182.pdf
#
07.06.20237.5 Mб183.pdf
#
07.06.20237.55 Mб284.pdf
#
07.06.20238.12 Mб785.pdf
#
07.06.20238.14 Mб186.pdf
#
07.06.20238.65 Mб1287.pdf
#
07.06.20238.69 Mб388.pdf
#
07.06.20239.46 Mб189.pdf
#
07.06.2023704.58 Кб19.pdf
#
07.06.20239.48 Mб290.pdf

			N			N	N	N	N
			N			11	12	11	13
						11	12	11	13
				11			2		2				N11						N21
		N		N			2	N	2				N11						N21
		N	11	N	12			N	N
P =						N				,	N	=		N	,	N		=		N .
3				2			12		2		1			12			2			22
3							12				1			12			2			22
	N					N	N
	N		11	N	13	N	N						N13						N23
			11		13	12	13	N13
				2			2	N13
				2			2

N					1			1			N	N						1
N			21		1		22	1			11		12			22		1			22
	11		21	01			22		02			2				22	02			2
												2								2		,	\| W
N	N					1									1								0
																							0
11		12			22					22	N			22				22	> 0
	2				22	02			2		12			22	02			22
	2								2

			N			N	N	N	N
			N			11	12	11	13
						11	12	11	13
				11			2		2				N11						N21
		N		N			2	N	2				N11						N21
		N	11	N	12			N	N
P =						N				,	N	=		N	,	N		=		N .
3				2			12		2		1			12			2			22
3							12				1			12			2			22
	N					N	N
	N		11	N	13	N	N						N13						N23
			11		13	12	13	N13
				2			2	N13
				2			2

N					1			1			N	N						1
N			21		1		22	1			11		12			22		1			22
	11		21	01			22		02			2				22	02			2
												2								2		,	\| W
N	N					1									1								0
																							0
11		12			22					22	N			22				22	> 0
	2				22	02			2		12			22	02			22
	2								2

			N			N	N	N	N
			N			11	12	11	13
						11	12	11	13
				11			2		2				N11						N21
		N		N			2	N	2				N11						N21
		N	11	N	12			N	N
P =						N				,	N	=		N	,	N		=		N .
3				2			12		2		1			12			2			22
3							12				1			12			2			22
	N					N	N
	N		11	N	13	N	N						N13						N23
			11		13	12	13	N13
				2			2	N13
				2			2

N					1			1			N	N						1
N			21		1		22	1			11		12			22		1			22
	11		21	01			22		02			2				22	02			2
												2								2		,	\| W
N	N					1									1								0
																							0
11		12			22					22	N			22				22	> 0
	2				22	02			2		12			22	02			22
	2								2