How to find the basis of a vector system examples. Linear dependence of vectors

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 a vector can be placed anywhere in space or on a 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 takes place on a plane. This means that the vector can be written as a coordinate line. Operations on vectors, - addition and multiplication by a number, on strings are performed componentwise. This makes it possible to expand the concept of a vector, understanding any string of numbers as a vector. For example, the solution to a system of linear equations, as well as any set of values \u200b\u200bof the variables of the system, 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) \u003d (a 1 + b 1, a 2 + b 2, ..., a n+ b n). (2)

Multiplication of a string by a number is performed according to the rule

l (a 1, a 2, ..., a n) \u003d (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, which is 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 nonzero coefficient.

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

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

x 1 + x 2 + … + x m = . (4)

If this equation has nonzero solutions, then the vector system is linearly dependent. If the zero solution is unique, then the vector system is linearly independent.

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

Then, performing the 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. The column of free members is not needed here, since the system is homogeneous.

The basis of a system of vectors (finite or infinite, in particular, of the entire linear space) is called 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 \u003d (1, 2, 2, 4), \u003d (2, 3, 5, 1), \u003d (3, 4, 8, –2), \u003d (2, 5, 0, 3) and express other vectors through the basis.

Decision... 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 \u003d. ... We bring the matrix to a stepped form:

~ ~ ~

The basis of this system of vectors is formed by vectors,,, which correspond to the leading elements of the lines, marked with circles. To express the vector, we solve the equation x 1 + x 2 + x 4 \u003d. It is reduced to a system of linear equations, the matrix of which is obtained from the initial permutation of the corresponding column to the place of the column of free terms. Therefore, when converting to a stepped view, the same transformations will be made over the matrix as above. This means that you can use the resulting matrix in a stepped form by making the necessary column permutations in it: we place the columns with circles to the left of the vertical line, and place the column corresponding to the vector to the right of the line.

We consistently find:

x 4 = 0;

x 2 = 2;

x 1 + 4 = 3, x 1 = –1;

Comment... If it is required 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.

R e n t e n t 1.4. Find the basis of the system of vectors and express the remaining vectors through the basis:

a) \u003d (1, 3, 2, 0), \u003d (3, 4, 2, 1), \u003d (1, –2, –2, 1), \u003d (3, 5, 1, 2);

b) \u003d (2, 1, 2, 3), \u003d (1, 2, 2, 3), \u003d (3, –1, 2, 2), \u003d (4, –2, 2, 2);

c) \u003d (1, 2, 3), \u003d (2, 4, 3), \u003d (3, 6, 6), \u003d (4, –2, 1); \u003d (2, –6, –2).

In a given system of vectors, a basis can usually be distinguished in 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 \u003d (1, 0,…, 0), \u003d (0, 1,…, 0),…, \u003d (0, 0,…, 1). Through this basis, any vector \u003d (a 1, a 2, ..., a n) is expressed as follows:

\u003d (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) \u003d a 1 + a 2 +… + a n .

Thus, the components in the vector row \u003d (a 1, a 2, ..., a n) Are its coefficients in the expansion in terms of 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 transformed into algebraic form and solved by means of algebra.

Lectures on algebra and geometry. Semester 1.

Lecture 9. Basis of vector space.

Abstract: a system of vectors, a linear combination of a system of vectors, coefficients of a linear combination of a system of vectors, a basis on a straight line, plane and in space, dimensions of vector spaces on a straight line, plane and in space, expansion of a vector in a basis, coordinates of a vector with respect to a basis, equality theorem two vectors, linear operations with vectors in coordinate form, orthonormal triplet of vectors, right and left triplets of vectors, orthonormal basis, the main theorem of vector algebra.

Chapter 9. Basis of vector space and decomposition of a vector in basis.

item 1. Basis on a straight line, on a plane and in space.

Definition. Any finite set of vectors is called a vector system.

Definition. Expression where
is called a linear combination of the vector system
and the numbers
are called the coefficients of this linear combination.

Let L, P and S be a line, plane and space of points, respectively, and
... Then
- vector spaces of vectors as directed segments on the straight line L, on the plane P and in the space S, respectively.

any nonzero vector
, i.e. any nonzero vector collinear on the line L:
and
.

Basis designation
:
- basis
.

Definition. The basis of the vector space
is any ordered pair of noncollinear vectors of the space
.

where
,
- basis
.

Definition. The basis of the vector space
is called any ordered triplet of non-coplanar vectors (i.e., not lying in the same plane) of the space
.

- basis
.

Comment. The basis of a vector space cannot contain a zero vector: in space
by definition, in space
two vectors will be collinear if at least one of them is zero, in space
three vectors will be coplanar, that is, they will lie in the same plane if at least one of the three vectors is zero.

item 2. Decomposition of a vector in basis.

Definition. Let be - an arbitrary vector,
- an arbitrary system of vectors. If equality holds

then they say that the vector presented as a linear combination of this vector system. If the given system of vectors
is a basis of the vector space, then equality (1) is called the expansion of the vector on the basis
... Linear combination coefficients
in this case are called the coordinates of the vector on the basis
.

Theorem. (On the expansion of a vector in terms of a basis.)

Any vector of a vector space can be decomposed in its basis and, moreover, in a unique way.

Evidence. 1) Let L be an arbitrary line (or axis) and
- basis
... Take an arbitrary vector
... Since both vectors and collinear with the same line L, then
... We will use the collinearity theorem for two vectors. Because
, then there is (exists) such a number
, what
and thus we got the vector expansion on the basis
vector space
.

Now let us prove the uniqueness of such a decomposition. Suppose the opposite. Let there be two expansions of the vector on the basis
vector space
:

and
where
... Then
and using the law of distributivity, we get:

Because
, then from the last equality it follows that
, ch.d.

2) Now let P be an arbitrary plane and
- basis
... Let be
an arbitrary vector of this plane. Let us postpone all three vectors from any one point of this plane. Let's build 4 lines. Let's draw a straight line on which the vector lies , straight
on which the vector lies ... Through the end of the vector draw a straight line parallel to the vector and a straight line parallel to the vector ... These 4 lines carve a parallelogram. See below fig. 3. According to the parallelogram rule
and
,
,
- basis ,
- basis
.

Now, by what has already been proved in the first part of this proof, there exist numbers
, what

and
... From here we get:

and the possibility of expansion in a basis is proved.

Now let us prove the uniqueness of the expansion in terms of the basis. Suppose the opposite. Let there be two expansions of the vector on the basis
vector space
:
and
... We get equality

Where follows
... If
then
, and since
then
and the expansion coefficients are:
,
... Let now
... Then
where
... By the collinearity theorem for two vectors, this implies that
... This contradicts the hypothesis of the theorem. Consequently,
and
, ch.d.

3) Let
- basis
let it go
arbitrary vector. Let's carry out the following constructions.

Set aside all three basis vectors
and vector from one point and build 6 planes: the plane in which the basis vectors lie
, plane
and plane
; further through the end of the vector draw three planes parallel to the three planes just constructed. These 6 planes carve a parallelepiped:

By the rule of addition of vectors, we obtain the equality:

. (1)

By construction
... Hence, by the collinearity theorem for two vectors, it follows that there exists a number
, such that
... Similarly,
and
where
... Now, substituting these equalities into (1), we get:

and the possibility of expansion in a basis is proved.

Let us prove the uniqueness of such a decomposition. Suppose the opposite. Let there be two expansions of the vector on the basis
:

And. Then

Note that by hypothesis the vectors
non-coplanar, therefore, they are pairwise non-collinear.

Two cases are possible:
or
.

a) Let
, then equality (3) implies:

. (4)

It follows from equality (4) that the vector decomposes on the basis
, i.e. vector lies in the plane of vectors
and therefore the vectors
coplanar, which contradicts the condition.

b) The case remains
, i.e.
... Then from equality (3) we obtain either

Because
Is the basis of the space of vectors lying in the plane, and we have already proved the uniqueness of the expansion in terms of the basis of vectors of the plane, then it follows from equality (5) that
and
, ch.d.

The theorem is proved.

Consequence.

1) There is a one-to-one correspondence between the set of vectors of the vector space
and the set of real numbers R.

2) There is a one-to-one correspondence between the set of vectors of the vector space
and cartesian square

3) There is a one-to-one correspondence between the set of vectors of the vector space
and cartesian cube
the set of real numbers R.

Evidence. Let us prove the third statement. The first two are proved similarly.

Select and fix in space
some basis
and arrange the mapping
according to the following rule:

those. each vector is associated with an ordered set of its coordinates.

Since for a fixed basis each vector has a single set of coordinates, the correspondence given by rule (6) is indeed a mapping.

It follows from the proof of the theorem that different vectors have different coordinates with respect to the same basis, i.e. mapping (6) is an injection.

Let be
an arbitrary ordered set of real numbers.

Consider a vector
... By construction, this vector has coordinates
... Consequently, mapping (6) is a surjection.

A mapping that is both injective and surjective is bijective, i.e. one-to-one, p.t.d.

The corollary is proved.

Theorem. (On the equality of two vectors.)

Two vectors are equal if and only if their coordinates are equal relative to the same basis.

The proof immediately follows from the previous corollary.

item 3. Dimension of vector space.

Definition. The number of vectors in the basis of a vector space is called its dimension.

Designation:
Is the dimension of the vector space V.

Thus, in accordance with this and previous definitions, we have:

1)
Is the vector space of vectors of the line L.

- basis
,
,
,
- vector expansion
on the basis
,
- vector coordinate on the basis
.

2)
Is the vector space of vectors of the plane P.

- basis
,
,
,
- vector expansion
on the basis
,
- vector coordinates on the basis
.

3)
- vector space of vectors in the space of points S.

- basis
,
,
- vector expansion
on the basis
,
- vector coordinates on the basis
.

Comment. If
then
and you can choose the basis
space
so that
- basis
and
- basis
... Then
and
, .

Thus, any vector of the straight line L, plane P and space S can be expanded in the basis
:

Designation. By virtue of the theorem on the equality of vectors, we can identify any vector with an ordered triple of real numbers and write:

This is possible only when the basis
fixed and there is no danger of confusion.

Definition. Writing a vector in the form of an ordered triple of real numbers is called the coordinate form of writing a vector:
.

item 4. Linear operations with vectors in coordinate notation.

Let be
- space basis
and
- two of its arbitrary vectors. Let be
and
- writing these vectors in coordinate form. Let, further,
- an arbitrary real number. In this notation, the following theorem holds.

Theorem. (About linear operations on vectors in coordinate form.)

2)
.

In other words, in order to add two vectors, you need to add their corresponding coordinates, and to multiply a vector by a number, you need to multiply each coordinate of a given vector by a given number.

Evidence. Since by the condition of the theorem, then using the axioms of the vector space, which obey the operations of addition of vectors and multiplication of a vector by a number, we obtain:

This implies .

The second equality is proved similarly.

The theorem is proved.

p. 5. Orthogonal vectors. Orthonormal basis.

Definition. Two vectors are called orthogonal if the angle between them is equal to the right angle, i.e.
.

Designation:
- vectors and orthogonal.

Definition. Three vectors
is called orthogonal if these vectors are pairwise orthogonal to each other, i.e.
,
.

Definition. Three vectors
is called orthonormal if it is orthogonal and the lengths of all vectors are equal to one:
.

Comment. It follows from the definition that an orthogonal and, therefore, an orthonormal triple of vectors is non-coplanar.

Definition. Ordered noncoplanar triplet of vectors
, plotted from one point, is called right (right-oriented), if, when viewed from the end of the third vector onto the plane in which the first two vectors lie and , the shortest rotation of the first vector to the second going counterclockwise. Otherwise, the triplet of vectors is called left (left-oriented).

Here, Fig. 6 shows the right triplet of vectors
... The following figure 7 shows the left triplet of vectors
:

Definition. Basis
vector space
is called orthonormal if
orthonormal triplet of vectors.

Designation. In what follows, we will use the right orthonormal basis
, see the following figure.

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

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

Decision... Consider a homogeneous system of linear equations

and 1 x 1 + and 2 x 2 + and 3 x 3 + and 4 x 4 + and 5 x 5 = 0

or in expanded form.

We will solve this system by the Gauss method, without changing the positions of the rows and columns, and, in addition, choosing the main element not in the upper left corner, but over the entire row. 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

and 1 1 x 1 + and 2 1 x 2 + and 3 1 x 3 + and 4 1 x 4 + and 5 1 x 5 = 0 ,

where and 1 1 = , and 2 1 = , and 3 1 = , and 4 1 = , and 5 1 = . (1)

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

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

From (1) we have:

and 2 1 = and 3 1 (-1) + and 4 1 0 + and 1 1 1 \u003d\u003e and 2 1 = and 1 1 – and 3 1 .

and 5 1 = and 3 1 0 + and 4 1 1 + and 1 1 2 \u003d\u003e and 5 1 = 2and 1 1 + and 4 1 .

Vectors and 2 and and 5 decompose in basis and 1 , and 3 , and 4 with the same coefficients as vectors and 2 1 and and 5 1 diagonal and 1 1 , and 3 1 , and 4 1 (those odds x i). Consequently,

and 2 = and 1 – and 3 , and 5 = 2and 1 + and 4 .

Tasks. 1. Find the basis of the system of vectors and vectors not included in the basis, expand in 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 }.

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 , in which there is at least one nonzero 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 the vector system = (1, 2, 2, 4),= (2, 3, 5, 1),= (3, 4, 8, -2),\u003d (2, 5, 0, 3) and express the remaining vectors in terms of the basis.

Solution. Build a matrix in which the coordinates of these vectors are arranged in columns. We bring it to a stepped form.

~
~
~
.

The basis of this system is formed by vectors ,,, which correspond to the leading elements of the lines, enclosed by circles. 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 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 a stepped form, making the necessary permutations in it.

We consistently find:

x 1 + 4 \u003d 3, x 1 \u003d -1;

= -+2.

Remark 1. If it is required to express several vectors in terms of 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 compose one matrix in which there will be several columns of free members. 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 in front of it. In this case, there is no need to reshape 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 in terms of the basis:

and) = (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);

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

1. 3. Fundamental decision system

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

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

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

If an inhomogeneous system is consistent and indefinite, then its arbitrary solution has the form f н +  1 f о1 + ... +  kf о k, where f о is a particular solution of the inhomogeneous system and f о1, ..., f о k is the fundamental system solutions of the associated homogeneous system.

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

Solution. Let us write the solution obtained in Example 1 in vector form and expand the resulting vector into a sum in terms of the free parameters available in it and fixed numerical values:

\u003d (x 1, x 2, x 3, x 4) \u003d (–2a + 7b - 2, a, –2b + 1, b) \u003d (–2a, a, 0, 0) + (7b, 0, - 2b, b) + + (- 2, 0, 1, 0) \u003d a (-2, 1, 0, 0) + b (7, 0, -2, 1) + (- 2, 0, 1, 0 ).

We get f n \u003d (- 2, 0, 1, 0), f о1 \u003d (-2, 1, 0, 0), f о2 \u003d (7, 0, -2, 1).

Comment. The problem of finding a fundamental system of solutions to a homogeneous system is solved in a similar way.

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

and)

b)

c) 2x 1 - x 2 + 3x 3 \u003d 0.

R e n t in g 3.2. Find a particular solution to an inhomogeneous system and a fundamental system of solutions to the associated homogeneous system:

and)

b)