How to find the basis of a given system of vectors. Linear dependence. Basis of a system of vectors Is it a basis

A linear combination of vectors is a vector
, where λ 1, ..., λ m are arbitrary coefficients.

Vector system
is called linearly dependent if there is a linear combination of it equal to , which has at least one non-zero coefficient.

Vector system
is called linearly independent if in any of its linear combinations equal to , all coefficients are zero.

The basis of the vector system
its non-empty linearly independent subsystem is called, through which any vector of the system can be expressed.

Example 2. Find the basis of a system of vectors = (1, 2, 2, 4),= (2, 3, 5, 1),= (3, 4, 8, -2),= (2, 5, 0, 3) and express the remaining vectors through the basis.

Solution: We build a matrix in which the coordinates of these vectors are arranged in columns. We bring it to a stepwise form.

~
~
~
.

The basis of this system is formed by the vectors ,,, which correspond to the leading elements of the lines, highlighted in circles. To express a vector solve the equation x 1 +x 2 + x 4 =. It reduces to a system of linear equations, the matrix of which is obtained from the original permutation of the column corresponding to , in place of the column of free members. Therefore, to solve the system, we use the resulting matrix in stepwise form, making the necessary rearrangements in it.

We consistently find:

x 1 + 4 = 3, x 1 = -1;

= -+2.

Remark 1. If it is necessary to express several vectors through the basis, then for each of them a corresponding system of linear equations is constructed. These systems will differ only in the columns of free members. Therefore, to solve them, you can create one matrix, which will have several columns of free terms. Moreover, each system is solved independently of the others.

Remark 2. To express any vector, it is sufficient to use only the basis vectors of the system that precede it. In this case, there is no need to reformat the matrix; it is enough to put a vertical line in the right place.

Exercise 2. Find the basis of the system of vectors and express the remaining vectors through the basis:

A) = (1, 3, 2, 0),= (3, 4, 2, 1),= (1, -2, -2, 1),= (3, 5, 1, 2);

b) = (2, 1, 2, 3),= (1, 2, 2, 3),= (3, -1, 2, 2),= (4, -2, 2, 2);

V) = (1, 2, 3),= (2, 4, 3),= (3, 6, 6),= (4, -2, 1);= (2, -6, -2).

1. 3. Fundamental system of solutions

A system of linear equations is called homogeneous if all its free terms are equal to zero.

The fundamental system of solutions of a homogeneous system of linear equations is the basis of the set of its solutions.

Let us be given an inhomogeneous system of linear equations. A homogeneous system associated with a given one is a system obtained from a given one by replacing all free terms by zeros.

If the inhomogeneous system is consistent and indefinite, then its arbitrary solution has the form f n +  1 f o1 + ... +  k f o k , where f n is a particular solution of the inhomogeneous system and f o1 , ... , f o k is the fundamental system solutions of the associated homogeneous system.

Example 3. Find a particular solution to the inhomogeneous system from Example 1 and fundamental system solutions of the associated homogeneous system.

Solution. Let's write the solution obtained in example 1 in vector form and decompose the resulting vector into a sum over the free parameters present in it and fixed numerical values:

= (x 1 , x 2 , x 3 , x 4) = (–2a + 7b – 2, a, –2b + 1, b) = (–2a, a, 0, 0) + (7b, 0, – 2b, b) + +(– 2, 0, 1, 0) = a(-2, 1, 0, 0) + b(7, 0, -2, 1) + (– 2, 0, 1, 0 ).

We get f n = (– 2, 0, 1, 0), f o1 = (-2, 1, 0, 0), f o2 = (7, 0, -2, 1).

Comment. The problem of finding a fundamental system of solutions to a homogeneous system is solved similarly.

Exercise 3.1 Find the fundamental system of solutions of a homogeneous system:

A)

b)

c) 2x 1 – x 2 +3x 3 = 0.

Exercise 3.2. Find a particular solution to the inhomogeneous system and a fundamental system of solutions to the associated homogeneous system:

A)

b)

In geometry, a vector is understood as a directed segment, and vectors obtained from one another by parallel translation are considered equal. All equal vectors are treated as the same vector. The origin of the vector can be placed at any point in space or plane.

If the coordinates of the ends of the vector are given in space: A(x 1 , y 1 , z 1), B(x 2 , y 2 , z 2), then

= (x 2 – x 1 , y 2 – y 1 , z 2 – z 1). (1)

A similar formula holds on the plane. This means that the vector can be written as a coordinate line. Operations on vectors, such as addition and multiplication by a number, on strings are performed componentwise. This makes it possible to expand the concept of a vector, understanding a vector as any string of numbers. For example, the solution to a system of linear equations, as well as any set of values ​​of the system's variables, can be viewed as a vector.

On strings of the same length, the addition operation is performed according to the rule

(a 1 , a 2 , … , a n) + (b 1 , b 2 , … , b n) = (a 1 + b 1 , a 2 + b 2 , … , a n+b n). (2)

Multiplying a string by a number follows the rule

l(a 1 , a 2 , … , a n) = (la 1 , la 2 , … , la n). (3)

A set of row vectors of a given length n with the indicated operations of addition of vectors and multiplication by a number forms an algebraic structure called n-dimensional linear space.

A linear combination of vectors is a vector , where λ 1 , ... , λ m– arbitrary coefficients.

A system of vectors is called linearly dependent if there is a linear combination of it equal to , in which there is at least one non-zero coefficient.

A system of vectors is called linearly independent if in any linear combination equal to , all coefficients are zero.

Thus, solving the question of the linear dependence of a system of vectors is reduced to solving the equation

x 1 + x 2 + … + x m = . (4)

If this equation has non-zero solutions, then the system of vectors is linearly dependent. If the zero solution is unique, then the system of vectors is linearly independent.

To solve system (4), for clarity, the vectors can be written not as rows, but as columns.

Then, having performed transformations on the left side, we arrive at a system of linear equations equivalent to equation (4). The main matrix of this system is formed by the coordinates of the original vectors arranged in columns. A column of free members is not needed here, since the system is homogeneous.

Basis system of vectors (finite or infinite, in particular, total linear space) is its non-empty linearly independent subsystem, through which any vector of the system can be expressed.

Example 1.5.2. Find the basis of the system of vectors = (1, 2, 2, 4), = (2, 3, 5, 1), = (3, 4, 8, –2), = (2, 5, 0, 3) and express the remaining vectors through the basis.

Solution. We build a matrix in which the coordinates of these vectors are arranged in columns. This is the matrix of the system x 1 + x 2 + x 3 + x 4 =. . We reduce the matrix to stepwise form:

~ ~ ~

The basis of this system of vectors is formed by the vectors , , , to which the leading elements of the rows, highlighted in circles, correspond. To express the vector, we solve the equation x 1 + x 2 + x 4 = . It reduces to a system of linear equations, the matrix of which is obtained from the original by rearranging the column corresponding to , in place of the column of free terms. Therefore, when reducing to a stepped form, the same transformations as above will be made on the matrix. This means that you can use the resulting matrix in a stepwise form, making the necessary rearrangements of the columns in it: we place the columns with circles to the left of the vertical bar, and the column corresponding to the vector is placed to the right of the bar.

We consistently find:

x 4 = 0;

x 2 = 2;

x 1 + 4 = 3, x 1 = –1;

Comment. If it is necessary to express several vectors through the basis, then for each of them a corresponding system of linear equations is constructed. These systems will differ only in the columns of free members. Moreover, each system is solved independently of the others.

Exercise 1.4. Find the basis of the system of vectors and express the remaining vectors through the basis:

a) = (1, 3, 2, 0), = (3, 4, 2, 1), = (1, –2, –2, 1), = (3, 5, 1, 2);

b) = (2, 1, 2, 3), = (1, 2, 2, 3), = (3, –1, 2, 2), = (4, –2, 2, 2);

c) = (1, 2, 3), = (2, 4, 3), = (3, 6, 6), = (4, –2, 1); = (2, –6, –2).

In a given system of vectors, the basis can usually be identified different ways, but all bases will have the same number of vectors. The number of vectors in the basis of a linear space is called the dimension of the space. For n-dimensional linear space n– this is the dimension of space, since this space has a standard basis = (1, 0, ... , 0), = (0, 1, ... , 0), ... , = (0, 0, ... , 1). Through this basis any vector = (a 1 , a 2 , … , a n) is expressed as follows:

= (a 1 , 0, … , 0) + (0, a 2 , … , 0) + … + (0, 0, … , a n) =

A 1 (1, 0, … , 0) + a 2 (0, 1, … , 0) + … + a n(0, 0, … ,1) = a 1 + a 2 +… + a n .

Thus, the components in the row of the vector = (a 1 , a 2 , … , a n) are its coefficients in the expansion through the standard basis.

Straight lines on a plane

The task of analytical geometry is the application of the coordinate method to geometric problems. Thus, the problem is translated into algebraic form and solved by means of algebra.

Expression of the form called linear combination of vectors A 1 , A 2 ,...,A n with odds λ 1, λ 2 ,...,λ n.

Determination of linear dependence of a system of vectors

Vector system A 1 , A 2 ,...,A n called linearly dependent, if there is a non-zero set of numbers λ 1, λ 2 ,...,λ n, in which the linear combination of vectors λ 1 *A 1 +λ 2 *A 2 +...+λ n *A n equal to the zero vector, that is, the system of equations: has a non-zero solution.
Set of numbers λ 1, λ 2 ,...,λ n is nonzero if at least one of the numbers λ 1, λ 2 ,...,λ n different from zero.

Determination of linear independence of a system of vectors

Vector system A 1 , A 2 ,...,A n called linearly independent, if the linear combination of these vectors λ 1 *A 1 +λ 2 *A 2 +...+λ n *A n equal to the zero vector only for a zero set of numbers λ 1, λ 2 ,...,λ n , that is, the system of equations: A 1 x 1 +A 2 x 2 +...+A n x n =Θ has a unique zero solution.

Check if a system of vectors is linearly dependent

Solution:

1. We compose a system of equations:

2. We solve it using the Gauss method. The Jordanano transformations of the system are given in Table 29.1. When calculating, the right-hand sides of the system are not written down since they are equal to zero and do not change during Jordan transformations.

3. From the last three rows of the table write down a resolved system equivalent to the original one system:

4. We obtain the general solution of the system:

5. Having set the value of the free variable x 3 =1 at your discretion, we obtain a particular non-zero solution X=(-3,2,1).

Answer: Thus, for a non-zero set of numbers (-3,2,1), the linear combination of vectors equals the zero vector -3A 1 +2A 2 +1A 3 =Θ. Hence, vector system linearly dependent.

Properties of vector systems

Property (1)
If a system of vectors is linearly dependent, then at least one of the vectors is expanded in terms of the others and, conversely, if at least one of the vectors of the system is expanded in terms of the others, then the system of vectors is linearly dependent.

Property (2)
If any subsystem of vectors is linearly dependent, then the entire system is linearly dependent.

Property (3)
If a system of vectors is linearly independent, then any of its subsystems is linearly independent.

Property (4)
Any system of vectors containing a zero vector is linearly dependent.

Property (5)
A system of m-dimensional vectors is always linearly dependent if the number of vectors n is greater than their dimension (n>m)

Basis of the vector system

The basis of the vector system A 1 , A 2 ,..., A n such a subsystem B 1 , B 2 ,...,B r is called(each of the vectors B 1,B 2,...,B r is one of the vectors A 1, A 2,..., A n), which satisfies the following conditions:
1. B 1 ,B 2 ,...,B r linear independent system vectors;
2. any vector A j system A 1 , A 2 ,..., A n is linearly expressed through the vectors B 1 , B 2 ,..., B r

r— the number of vectors included in the basis.

Theorem 29.1 On the unit basis of a system of vectors.

If a system of m-dimensional vectors contains m different unit vectors E 1 E 2 ,..., E m , then they form the basis of the system.

Algorithm for finding the basis of a system of vectors

In order to find the basis of the system of vectors A 1 ,A 2 ,...,A n it is necessary:

• Create a homogeneous system of equations corresponding to the system of vectors A 1 x 1 +A 2 x 2 +...+A n x n =Θ
• Bring this system

Find the basis of the system of vectors and vectors not included in the basis, expand them according to the basis:

A 1 = {5, 2, -3, 1}, A 2 = {4, 1, -2, 3}, A 3 = {1, 1, -1, -2}, A 4 = {3, 4, -1, 2}, A 5 = {13, 8, -7, 4}.

Solution. Consider a homogeneous system of linear equations

A 1 X 1 + A 2 X 2 + A 3 X 3 + A 4 X 4 + A 5 X 5 = 0

or in expanded form.

We will solve this system by the Gaussian method, without swapping rows and columns, and, in addition, choosing main element not in the upper left corner, but along the entire line. The challenge is to select the diagonal part of the transformed system of vectors.

~ ~

~ ~ ~ .

The allowed system of vectors, equivalent to the original one, has the form

A 1 1 X 1 + A 2 1 X 2 + A 3 1 X 3 + A 4 1 X 4 + A 5 1 X 5 = 0 ,

Where A 1 1 = , A 2 1 = , A 3 1 = , A 4 1 = , A 5 1 = . (1)

Vectors A 1 1 , A 3 1 , A 4 1 form a diagonal system. Therefore, the vectors A 1 , A 3 , A 4 form the basis of the vector system A 1 , A 2 , A 3 , A 4 , A 5 .

Let us now expand the vectors A 2 And A 5 on basis A 1 , A 3 , A 4 . To do this, we first expand the corresponding vectors A 2 1 And A 5 1 diagonal system A 1 1 , A 3 1 , A 4 1, bearing in mind that the coefficients of the expansion of a vector along the diagonal system are its coordinates x i.

From (1) we have:

A 2 1 = A 3 1 · (-1) + A 4 1 0 + A 1 1 ·1 => A 2 1 = A 1 1 – A 3 1 .

A 5 1 = A 3 1 0 + A 4 1 1 + A 1 1 ·2 => A 5 1 = 2A 1 1 + A 4 1 .

Vectors A 2 And A 5 are expanded in basis A 1 , A 3 , A 4 with the same coefficients as vectors A 2 1 And A 5 1 diagonal system A 1 1 , A 3 1 , A 4 1 (those coefficients x i). Hence,

A 2 = A 1 – A 3 , A 5 = 2A 1 + A 4 .

Tasks. 1.Find the basis of the system of vectors and vectors not included in the basis, expand them according to the basis:

1. a 1 = { 1, 2, 1 }, a 2 = { 2, 1, 3 }, a 3 = { 1, 5, 0 }, a 4 = { 2, -2, 4 }.

2. a 1 = { 1, 1, 2 }, a 2 = { 0, 1, 2 }, a 3 = { 2, 1, -4 }, a 4 = { 1, 1, 0 }.

3. a 1 = { 1, -2, 3 }, a 2 = { 0, 1, -1 }, a 3 = { 1, 3, 0 }, a 4 = { 0, -7, 3 }, a 5 = { 1, 1, 1 }.

4. a 1 = { 1, 2, -2 }, a 2 = { 0, -1, 4 }, a 3 = { 2, -3, 3 }.

2. Find all bases of the vector system:

1. a 1 = { 1, 1, 2 }, a 2 = { 3, 1, 2 }, a 3 = { 1, 2, 1 }, a 4 = { 2, 1, 2 }.

2. a 1 = { 1, 1, 1 }, a 2 = { -3, -5, 5 }, a 3 = { 3, 4, -1 }, a 4 = { 1, -1, 4 }.