Decomposition of an arbitrary vector of base. Baseline decomposition

Basis (Dr. Greek. βασις, base) - a set of such vectors in the vector space, that any vector of this space may be the same in the form of a linear combination of vectors from this set - basic vectors

The basis in the space R n is called any system from n.- Linely independent vectors. Each vector of R n, which are not included in the basis, can be represented as a linear combination of basic vectors, i.e. Dispatch by base.
Let be the basis of the space R n and. Then there are such numbers λ 1, λ 2, ..., λ n, which .
The decomposition coefficients λ 1, λ 2, ..., λ n are called vector coordinates in the base. If the basis is set, the vector coefficients are definitely determined.

Comment. In each n.-Herous vector space you can choose countless different bases. In various bases, the same vector has different coordinates, but the only base in the selected basis. Example. Decay the vector by base.
Decision. . We substitute the coordinates of all vectors and perform actions on them:

Equating the coordinates, we obtain the system of equations:

I solve it: .
Thus, we get a decomposition: .
In the base, the vector has coordinates.

End of work -

This topic belongs to the section:

Vector concept. Linear operations over vectors

The vector is called a directional segment having a certain length of a certain length of which one of the limiting points. The length of the vector is called its module and is indicated by the vector module. The vector is called zero denotes if the beginning and end of it coincides with the zero vector does not have a certain ..

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L. 2-1 Basic concepts of vector algebra. Linear operations over vectors.

Baseline decomposition.

Basic concepts vector algebra

The vector is called a set of all directed segments having the same length and direction
.

Properties:

Linear operations over vectors

1.

Rule parallelogram:

FROM umbletwo vectors and called vector coming out of their general principle and being a diagonal of a parallelogram-ma built in vectors and as on the sides.

Polygon rule:

To build the amount of any number of vectors, it is necessary to put the beginning of the 2nd 3rd sample vector at the end of the 2nd, at the end of the 2nd - beginning of 3rd, etc. The vector closing the resulting broken line is the sum. Its beginning coincides with the beginning of the 1st, and the end - with the end of the latter.

Properties:

2.

Work vector number , called vector satisfying conditions:
.

Properties:

3.

Differencevectors and call vector equal and vector opposite vector .
.

- The law of the opposite element (vector).

Baseline decomposition

The sum of vectors is determined by the only way.
(only ). The inverse operation is the decomposition of the vector into several components, ambiguous:. In order to make it unambiguous, you must specify the directions by the decomposition of the vector under consideration, or, as they say, you must specify basis.

When determining the basis, it is essential to the requirement of noncomplaunarity and non-salinearity of vectors. To understand the meaning of this requirement, it is necessary to consider the concept of linear dependence and linear independence of vectors.

Arbitrary expression of the form:, called linear combinationvectors
.

Linear combination of several vectors is called trivialif all its coefficients are zero.

Vectors
called linearly dependentif there is a non-trivial linear combination of these vectors equal to zero:
(1), provided
. If equality (1) takes place only at all
at the same time equal to zero, then nonzero vectors
will be linearly independent.

Easy to prove any two collinear vectors are linearly dependent, and two non-rigid vector is linearly independent.

Proof Let's start with the first approval.

Let vectors and collinear. We show that they are linearly dependent. Indeed, if they are collinear, they differ from each other only on a numerical factor, i.e.
, hence
. Since the resulting linear combination is clearly nontrivial and equal to "0", then vectors and linearly dependent.

Consider now two non-halinar vectors and . We prove that they are linearly independent. Proof construct from nasty.

Suppose they are linearly dependent. Then there must be a nontrivial linear combination
. Let's pretend that
, then
. The resulting equality means that vectors and collinear, contrary to our initial assumption.

Similarly, you can prove: any three compartment vectors are linearly dependent, and two noncomplanar vector are linearly independent.

Returning to the concept of the basis and the task of the decomposition of the vector in a specific basis, it can be said that the basis on the plane and in space is formed from the set of linearly independent vectors.Such a concept of basis is common, because It applies to the space of any number of measurements.

Expression of the form:
, is called the decomposition of the vector by vectors ,…,.

If we consider the basis in three-dimensional space, then the decomposition of the vector basisus
will be
where
-coordinates of the vector.

The task of expanding an arbitrary vector in some basis is very important to the following statement: any vectormay be the only way unfolded in this base
. In other words, coordinates
for any vector regarding Basisa
definitively.

The introduction of the basis in space and on the plane allows you to put in line with each vector an ordered triple (pair) numbers is its coordinates. This very important result that allows you to establish a link between geometric objects and numbers, makes it possible to analytically describe and explore the position and movement of physical objects.

The totality of the point and base is called Coordinate system.

If the vectors forming the basis are single and pairly perpendicular, then the coordinate system is called rectangulara Base orthonormal.

L. 2-2 works of vectors

Baseline decomposition

Consider vector
as defined by its coordinates:
.

- Components of the vector in the directions of basic vectors
.

Expression of type
called decomposition of vector basisus
.

Similarly, you can decompose basisus
vector
:

.

Cosine Corners formed by the Vector with basic ortes
called guides cosinees

;
;
.

Scalar product of vectors.

Scalar product of two vectors and called the number equal to the product of the modules of these vectors on the cosine of the corner between them

The scalar product of two vectors can be viewed as a product of a module of one of these vectors on the orthogonal projection of another vector to the direction of the first
.

Properties:

If the coordinates of the vectors are known
and
, after performing the expansion of the vectors of the basis
:
and
, Find

because
,
T.

.

.

The condition perpendicularity of vectors:
.

Conditions of collinearity rectors:
.

Vector artwork vectors

or

Vector product vector on vector called such vector
which satisfies the conditions:

Properties:

The considered algebraic properties make it possible to find an analytical expression for vector art through the coordinates of the components of the vectors in the orthonormal basis.

Given:
and
.

because .
,
,
,
,
,
T.

. This formula can be recorded in short, in the form of a third-order determinant:

.

Mixed vectors

Mixed product of three vectors ,and called a number equal to vector work
multiplied scalar to vector .

True the following equality:
, so mixed work is recorded
.

As follows from the definition, the result of a mixed product of three vectors is the number. This number has a visual geometric meaning:

Module of mixed work
equal to the amount of parallelepiped, built on the above general to the beginning vectors ,and .

Properties of mixed work:

If vectors ,,specified in the orthonormal basis
its coordinates, the calculation of the mixed work is carried out by the formula

.

Indeed, if
T.

;
;
, then
.

If vectors ,,complian, then vector work
perpendicular to the vector . And vice versa if
, then the volume of parallelepiped equal to zero., And this is possible only when the compartment vectors (linearly dependent).

Thus, three vectors are compartment, then and only if their mixed work is zero.

Base spacethey call such a system of vectors in which all other space vectors can be represented as a linear combination of vectors included in the basis.
In practice, this is all implemented simply enough. The basis, as a rule, is checked on a plane or in space, and for this you need to find the determinant of the second, third order matrix is \u200b\u200bcomposed of the coordinates of the vectors. Below are schematically recorded conditions under which vectors form basis

To stripe vector B for basic vectors
E, E ..., E [N] It is necessary to find the coefficients x, ..., x [n] at which the linear combination of vectors E, E ..., E [N] is equal to the vector b:
x1 * E + ... + x [n] * e [n] \u003d b.
To do this, the vector equation should be converted to the system linear equations and find solutions. It is also fairly easy to implement.
Found coefficients x, ..., x [n] are called Coordinates of the vector B in the base E, E ..., E [N].
Let's go to practical side Topics.

Baseline decomposition by vector vectors

Task 1. Check if Vectors A1, A2 Base on the plane

1) A1 (3; 5), A2 (4; 2)
Solution: We make a determinant from the coordinates of the vectors and calculate it

The determinant is not zero, hence vectors are linearly independent, and therefore form a basis.

2) A1 (2; -3), A2 (5; -1)
Solution: Calculate the determinant composed of vectors

The determinant is 13 (not equal to zero) - it follows that vectors A1, A2 is a basis on the plane.

---=================---

Consider typical examples from the Maup program on the discipline "Higher Mathematics".

Task 2. Show that vectors A1, A2, A3 form the basis of the three-dimensional vector space, and decompose the vector B for this basis (when solving a linear system algebraic equations Use the crawler method).
1) a1 (3; 1; 5), a2 (3; 2; 8), a3 (0; 1; 2), b (-3; 1; 2).
Solution: First consider the system of vectors A1, A2, A3 and check the determinant of the matrix A

Built in vectors other than zero. The matrix contains one zero element, so the determinant is more expedient to calculate as a schedule on the first column or third line.

In the recruitment of computations, they received that the determinant is different from zero, therefore vectors A1, A2, A3 linearly independent.
According to the definition, the vectors form the basis in R3. We write the schedule of the vector B by base

The vectors are equal when their respective coordinates are equal.
Therefore, from the vector equation, we obtain a system of linear equations

I will solve the slown Cramer method . To do this, write a system of equations in the form

The main determinant of the Slava is always equal to the determinant composed of the Basis vectors.

Therefore, in practice it is not calculated twice. To find auxiliary determinants, we put a column of free members in place of each column of the main determinant. Determinents calculate according to the rule of triangles



We substitute the found identifiers in the Cramer formula



So, the decomposition of the vector B on the basis has the form B \u003d -4a1 + 3a2-A3. The coordinates of the vector B in the basis A1, A2, A3 will be (-4.3, 1).

2) A1 (1; -5; 2), a2 (2; 3; 0), a3 (1; -1; 1), b (3; 5; 1).
Solution: Checking vectors on the basis - we make a determinant from the coordinates of the vectors and calculate it

The determinant is not equal to zero, therefore vectors form a basis in space. It remains to find the schedule of the vector B through this basis. To do this, write a vector equation

and convert to the system of linear equations

Record matrix equation

Next for Cramer formulas we find auxiliary determinants



Using Cramer formulas



So, the specified vector B has a schedule in two vector of the basis B \u003d -2A1 + 5A3, and its coordinates in the base are equal to B (-2,0, 5).

In vector calculation and its applications, the task of decomposition is of great importance, consisting in the presentation of this vector as a sum of several vectors called the components of this

vector. This task having in general countless decisions becomes quite defined if you specify some elements of the components of the vectors.

2. Examples of decomposition.

Consider a few very often encountered cases of decomposition.

1. To decompose this vector with two components of the vector of which one, for example A, is set in size and direction.

The task is reduced to the determination of the difference in two vectors. Indeed, if the vectors are the components of the vector C, then equality should be performed

Hence the second component vector is determined.

2. Ensure this vector with two components, of which one must lie in a given plane and the second must lie on a given direct A.

To determine the components of the vectors, we transfer the vector with so that its beginning coincides with the intersection point of the specified direct with the plane (point O - see Fig. 18). From the end of the vector C (point C) will spend direct

intersections with a plane (B - point of intersection), and then from a point with spend straight parallel

Vectors and will be desired, i.e., it is natural that the specified decomposition is possible if straight A and the plane are not parallel.

3. There are three compartment vectors A, B and C, and the vectors are not collinear. It is required to decompose vector with vectors

We give all three set vector to one point O. Then, by virtue of their compartment, they will be located in the same plane. On this vector with both diagonally, we construct parallelograms, the sides of which are parallel to the lines of the vectors (Fig. 19). This construct is always possible (if only the vectors are not collinear) and the only one. From fig. 19 shows that