To check whether a system of vectors is linearly dependent, it is necessary to compose a linear combination of these vectors, and check whether it can be zero if at least one coefficient is equal to zero.

Case 1. A system of vectors is given by vectors

Making a linear combination

We have obtained a homogeneous system of equations. If it has a non-zero solution, then the determinant must be equal to zero. Let's compose a determinant and find its value.

The determinant is zero, therefore the vectors are linearly dependent.

Case 2. The system of vectors is defined by analytical functions:

a)
, if the identity is true, then the system is linearly dependent.

Let's make a linear combination.

It is necessary to check whether there exist a, b, c (at least one of which is not equal to zero) for which this expression equals zero.

Let's write hyperbolic functions

,
, Then

then the linear combination of vectors will take the form:

Where
, take, for example, then the linear combination is zero, therefore, the system is linearly dependent.

Answer: the system is linearly dependent.

b)
, let's make a linear combination

A linear combination of vectors must be equal to zero for any values ​​of x.

Let's check for special cases.

A linear combination of vectors is equal to zero only if all coefficients are equal to zero.

Therefore, the system is linearly independent.

Answer: the system is linearly independent.

5.3. Find some basis and determine the dimension of the linear solution space.

Let's form an extended matrix and reduce it to the form of a trapezoid using the Gaussian method.

To get some basis, let’s substitute arbitrary values:

Let's get the rest of the coordinates

Answer:

5.4. Find the coordinates of vector X in the basis, if it is given in the basis.

Finding the coordinates of a vector in a new basis comes down to solving a system of equations

Method 1. Finding using the transition matrix

Let's create a transition matrix

Let's find the vector in the new basis using the formula

Let's find the inverse matrix and perform the multiplication

,

Method 2. Finding by composing a system of equations.

Let's compose the basis vectors from the basis coefficients

,
,

Finding the vector in the new basis has the form

, Where d this is a given vector x.

The resulting equation can be solved in any way, the answer will be similar.

Answer: vector in a new basis
.

5.5. Let x = (x 1 , x 2 , x 3 ) . Are the following transformations linear?

Let's compose matrices of linear operators from the coefficients of the given vectors.

Let's check the property of linear operations for each linear operator matrix.

We find the left side by multiplying the matrix A to vector

We find the right side by multiplying the given vector by a scalar
.

We see that
This means that the transformation is not linear.

Let's check other vectors.

, the transformation is not linear.

, the transformation is linear.

Answer: Oh– not a linear transformation, In– not linear, Cx– linear.

Note. You can complete this task much easier by carefully looking at the given vectors. IN Oh we see that there are terms that do not contain elements X, which could not be obtained as a result of a linear operation. IN In there is an element X to the third power, which also could not be obtained by multiplying by a vector X.

5.6. Given x = { x 1 , x 2 , x 3 } , Ax = { x 2 – x 3 , x 1 , x 1 + x 3 } , Bx = { x 2 , 2 x 3 , x 1 } . Perform the specified operation: ( A ( B – A )) x .

Let us write down the matrices of linear operators.

Let's perform an operation on matrices

When multiplying the resulting matrix by X, we get

Answer:

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

Solution. We are looking for a general solution to the system of equations

a 1 x 1 + a 2 x 2 + a 3 x 3 = Θ

Gauss method. To do this, we write this homogeneous system in coordinates:

System Matrix

The allowed system has the form: (r A = 2, n= 3). The system is cooperative and uncertain. Its general solution ( x 2 – free variable): x 3 = 13x 2 ; 3x 1 – 2x 2 – 13x 2 = 0 => x 1 = 5x 2 => X o = . The presence of a non-zero particular solution, for example, indicates that the vectors a 1 , a 2 , a 3 linearly dependent.

Example 2.

Find out whether this system linearly dependent or linearly independent vectors:

1. a 1 = { -20, -15, - 4 }, a 2 = { –7, -2, -4 }, a 3 = { 3, –1, –2 }.

Solution. Consider a homogeneous system of equations a 1 x 1 + a 2 x 2 + a 3 x 3 = Θ

or in expanded form (by coordinates)

The system is homogeneous. If it is non-degenerate, then it has a unique solution. In the case of a homogeneous system, there is a zero (trivial) solution. This means that in this case the system of vectors is independent. If the system is degenerate, then it has non-zero solutions and, therefore, it is dependent.

We check the system for degeneracy:

= –80 – 28 + 180 – 48 + 80 – 210 = – 106 ≠ 0.

The system is non-degenerate and, thus, the vectors a 1 , a 2 , a 3 linearly independent.

Assignments. Find out whether a given system of vectors is linearly dependent or linearly independent:

1. a 1 = { -4, 2, 8 }, a 2 = { 14, -7, -28 }.

2. a 1 = { 2, -1, 3, 5 }, a 2 = { 6, -3, 3, 15 }.

3. a 1 = { -7, 5, 19 }, a 2 = { -5, 7 , -7 }, a 3 = { -8, 7, 14 }.

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

5. a 1 = { 1, 8 , -1 }, a 2 = { -2, 3, 3 }, a 3 = { 4, -11, 9 }.

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

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

8. a 1 = {-1, 7, 1 , -2}, a 2 = {2, 3 , 2, 1}, a 3 = {4, 4, 4, -3}, a 4 = {1, 6, -11, 1}.

9. Prove that a system of vectors will be linearly dependent if it contains:

a) two equal vectors;

b) two proportional vectors.

Task 1. Find out whether the system of vectors is linearly independent. The system of vectors will be specified by the matrix of the system, the columns of which consist of the coordinates of the vectors.

.

Solution. Let the linear combination equal to zero. Having written this equality in coordinates, we obtain the following system of equations:

.

Such a system of equations is called triangular. She has only one solution . Therefore, the vectors linearly independent.

Task 2. Find out whether the system of vectors is linearly independent.

.

Solution. Vectors linearly independent (see problem 1). Let us prove that the vector is a linear combination of vectors . Vector expansion coefficients are determined from the system of equations

.

This system, like a triangular one, has a unique solution.

Therefore, the system of vectors linearly dependent.

Comment. Matrices of the same type as in Problem 1 are called triangular , and in problem 2 – stepped triangular . The question of the linear dependence of a system of vectors is easily solved if the matrix composed of the coordinates of these vectors is step triangular. If the matrix does not have a special form, then using elementary string conversions , preserving linear relationships between the columns, it can be reduced to a step-triangular form.

Elementary string conversions matrices (EPS) are called following operations above the matrix:

1) rearrangement of strings;

2) multiplying a string by a non-zero number;

3) adding another string to a string, multiplied by an arbitrary number.

Task 3. Find the maximum linearly independent subsystem and calculate the rank of the system of vectors

.

Solution. Let us reduce the matrix of the system using EPS to a step-triangular form. To explain the procedure, we denote the line with the number of the matrix to be transformed by the symbol . The column after the arrow indicates the actions on the rows of the matrix being converted that must be performed to obtain the rows of the new matrix.

.

Obviously, the first two columns of the resulting matrix are linearly independent, the third column is their linear combination, and the fourth does not depend on the first two. Vectors are called basic. They form a maximal linearly independent subsystem of the system , and the rank of the system is three.

Basis, coordinates

Task 4. Find the basis and coordinates of the vectors in this basis on the set of geometric vectors whose coordinates satisfy the condition .

Solution. The set is a plane passing through the origin. An arbitrary basis on a plane consists of two non-collinear vectors. The coordinates of the vectors in the selected basis are determined by solving the corresponding system of linear equations.

There is another way to solve this problem, when you can find the basis using the coordinates.

Coordinates spaces are not coordinates on the plane, since they are related by the relation , that is, they are not independent. The independent variables and (they are called free) uniquely define a vector on the plane and, therefore, they can be chosen as coordinates in . Then the basis consists of vectors lying in and corresponding to sets of free variables And , that is .

Task 5. Find the basis and coordinates of the vectors in this basis on the set of all vectors in space whose odd coordinates are equal to each other.

Solution. Let us choose, as in the previous problem, coordinates in space.

Because , then free variables uniquely determine the vector from and are therefore coordinates. The corresponding basis consists of vectors.

Task 6. Find the basis and coordinates of the vectors in this basis on the set of all matrices of the form , Where – arbitrary numbers.

Solution. Each matrix from is uniquely representable in the form:

This relation is the expansion of the vector from with respect to the basis
with coordinates .

Task 7. Find the dimension and basis of the linear hull of a system of vectors

.

Solution. Using the EPS, we transform the matrix from the coordinates of the system vectors to a step-triangular form.

.

Columns the last matrices are linearly independent, and the columns linearly expressed through them. Therefore, the vectors form a basis , And .

Comment. Basis in is chosen ambiguously. For example, vectors also form a basis .

Vectors, their properties and actions with them

Vectors, actions with vectors, linear vector space.

Vectors are an ordered collection of a finite number of real numbers.

Actions: 1.Multiplying a vector by a number: lambda*vector x=(lamda*x 1, lambda*x 2 ... lambda*x n).(3.4, 0, 7)*3=(9, 12,0.21)

2. Addition of vectors (belong to the same vector space) vector x + vector y = (x 1 + y 1, x 2 + y 2, ... x n + y n,)

3. Vector 0=(0,0…0)---n E n – n-dimensional (linear space) vector x + vector 0 = vector x

Theorem. In order for a system of n vectors, an n-dimensional linear space, to be linearly dependent, it is necessary and sufficient that one of the vectors be a linear combination of the others.

Theorem. Any set of n+ 1st vectors of n-dimensional linear space of phenomena. linearly dependent.

Addition of vectors, multiplication of vectors by numbers. Subtraction of vectors.

The sum of two vectors is a vector directed from the beginning of the vector to the end of the vector, provided that the beginning coincides with the end of the vector. If vectors are given by their expansions in basis unit vectors, then when adding vectors, their corresponding coordinates are added.

Let's consider this using the example of a Cartesian coordinate system. Let

Let's show that

From Figure 3 it is clear that

The sum of any finite number of vectors can be found using the polygon rule (Fig. 4): to construct the sum of a finite number of vectors, it is enough to combine the beginning of each subsequent vector with the end of the previous one and construct a vector connecting the beginning of the first vector with the end of the last.

Properties of the vector addition operation:

In these expressions m, n are numbers.

The difference between vectors is called a vector. The second term is a vector opposite to the vector in direction, but equal to it in length.

Thus, the operation of subtracting vectors is replaced by an addition operation

A vector whose beginning is at the origin and end at point A (x1, y1, z1) is called the radius vector of point A and is denoted simply. Since its coordinates coincide with the coordinates of point A, its expansion in unit vectors has the form

A vector that starts at point A(x1, y1, z1) and ends at point B(x2, y2, z2) can be written as

where r 2 is the radius vector of point B; r 1 - radius vector of point A.

Therefore, the expansion of the vector in unit vectors has the form

Its length is equal to the distance between points A and B

MULTIPLICATION

So in the case of a plane problem, the product of a vector by a = (ax; ay) by the number b is found by the formula

a b = (ax b; ay b)

Example 1. Find the product of the vector a = (1; 2) by 3.

3 a = (3 1; 3 2) = (3; 6)

So, in the case of a spatial problem, the product of the vector a = (ax; ay; az) by the number b is found by the formula

a b = (ax b; ay b; az b)

Example 1. Find the product of the vector a = (1; 2; -5) by 2.

2 a = (2 1; 2 2; 2 (-5)) = (2; 4; -10)

Dot product of vectors and where is the angle between the vectors and ; if either, then

From the definition of the scalar product it follows that

where, for example, is the magnitude of the projection of the vector onto the direction of the vector.

Scalar squared vector:

Properties of the dot product:

Dot product in coordinates

If That

Angle between vectors

Angle between vectors - the angle between the directions of these vectors (smallest angle).

Cross product (Cross product of two vectors.) - this is a pseudovector perpendicular to a plane constructed from two factors, which is the result of the binary operation “vector multiplication” over vectors in three-dimensional Euclidean space. The product is neither commutative nor associative (it is anticommutative) and is different from the dot product of vectors. In many engineering and physics problems, you need to be able to construct a vector perpendicular to two existing ones - the vector product provides this opportunity. The cross product is useful for "measuring" the perpendicularity of vectors - the length of the cross product of two vectors is equal to the product of their lengths if they are perpendicular, and decreases to zero if the vectors are parallel or antiparallel.

The cross product is defined only in three-dimensional and seven-dimensional spaces. The result of a vector product, like a scalar product, depends on the metric of Euclidean space.

Unlike the formula for calculating scalar product vectors from coordinates in a three-dimensional rectangular coordinate system, the formula for the cross product depends on the orientation of the rectangular coordinate system or, in other words, its “chirality”

Collinearity of vectors.

Two non-zero (not equal to 0) vectors are called collinear if they lie on parallel lines or on the same line. An acceptable, but not recommended, synonym is “parallel” vectors. Collinear vectors may be identically directed (“codirectional”) or oppositely directed (in the latter case they are sometimes called “anticollinear” or “antiparallel”).

Mixed product of vectors( a, b, c)- scalar product of vector a and the vector product of vectors b and c:

(a,b,c)=a ⋅(b ×c)

sometimes called triple scalar product vectors, most likely due to the fact that the result is a scalar (more precisely, a pseudoscalar).

Geometric meaning: The modulus of the mixed product is numerically equal to the volume of the parallelepiped formed by the vectors (a,b,c) .

Properties

A mixed product is skew-symmetric with respect to all its arguments: i.e. e. rearranging any two factors changes the sign of the product. It follows that the Mixed product in the right Cartesian coordinate system (in an orthonormal basis) is equal to the determinant of a matrix composed of vectors and:

The mixed product in the left Cartesian coordinate system (in an orthonormal basis) is equal to the determinant of the matrix composed of vectors and, taken with a minus sign:

In particular,

If any two vectors are parallel, then with any third vector they form a mixed product equal to zero.

If three vectors are linearly dependent (that is, coplanar, lying in the same plane), then their mixed product is equal to zero.

Geometric meaning - The mixed product is equal in absolute value to the volume of the parallelepiped (see figure) formed by the vectors and; the sign depends on whether this triple of vectors is right-handed or left-handed.

Coplanarity of vectors.

Three vectors (or larger number) are called coplanar if they, being reduced to a common origin, lie in the same plane

Properties of coplanarity

If at least one of the three vectors is zero, then the three vectors are also considered coplanar.

A triple of vectors containing a pair of collinear vectors is coplanar.

Mixed product of coplanar vectors. This is a criterion for the coplanarity of three vectors.

Coplanar vectors are linearly dependent. This is also a criterion for coplanarity.

In 3-dimensional space, 3 non-coplanar vectors form a basis

Linearly dependent and linearly independent vectors.

Linearly dependent and independent systems vectors.Definition. The vector system is called linearly dependent, if there is at least one non-trivial linear combination of these vectors equal to the zero vector. Otherwise, i.e. if only a trivial linear combination of given vectors equals the null vector, the vectors are called linearly independent.

Theorem (linear dependence criterion). In order for a system of vectors in a linear space to be linearly dependent, it is necessary and sufficient that at least one of these vectors is a linear combination of the others.

1) If among the vectors there is at least one zero vector, then the entire system of vectors is linearly dependent.

In fact, if, for example, , then, assuming , we have a nontrivial linear combination .▲

2) If among the vectors some form linearly dependent system, then the entire system is linearly dependent.

Indeed, let the vectors , , be linearly dependent. This means that there is a non-trivial linear combination equal to the zero vector. But then, assuming , we also obtain a nontrivial linear combination equal to the zero vector.

2. Basis and dimension. Definition. System of linearly independent vectors vector space is called basis of this space if any vector from can be represented as a linear combination of vectors of this system, i.e. for each vector there are real numbers such that the equality holds. This equality is called vector decomposition according to the basis, and the numbers are called coordinates of the vector relative to the basis(or in the basis) .

Theorem (on the uniqueness of the expansion with respect to the basis). Every vector in space can be expanded into a basis in the only way, i.e. coordinates of each vector in the basis are determined unambiguously.

Linear dependence and linear independence of vectors.
Basis of vectors. Affine coordinate system

There is a cart with chocolates in the auditorium, and every visitor today will get a sweet couple - analytical geometry with linear algebra. This article will touch upon two sections of higher mathematics at once, and we will see how they coexist in one wrapper. Take a break, eat a Twix! ...damn, what a bunch of nonsense. Although, okay, I won’t score, in the end, you should have a positive attitude towards studying.

Linear dependence of vectors, linear vector independence, vector basis and other terms have not only a geometric interpretation, but, above all, an algebraic meaning. The very concept of “vector” from the point of view of linear algebra is not always the “ordinary” vector that we can depict on a plane or in space. You don’t need to look far for proof, try drawing a vector of five-dimensional space . Or the weather vector, which I just went to Gismeteo for: – temperature and atmospheric pressure respectively. The example, of course, is incorrect from the point of view of the properties of the vector space, but, nevertheless, no one forbids formalizing these parameters as a vector. Breath of autumn...

No, I'm not going to bore you with theory, linear vector spaces, the task is to understand definitions and theorems. The new terms (linear dependence, independence, linear combination, basis, etc.) apply to all vectors from an algebraic point of view, but geometric examples will be given. Thus, everything is simple, accessible and clear. In addition to problems of analytical geometry, we will also consider some typical algebra problems. To master the material, it is advisable to familiarize yourself with the lessons Vectors for dummies And How to calculate the determinant?

Linear dependence and independence of plane vectors.
Plane basis and affine coordinate system

Let's consider the plane of your computer desk (just a table, bedside table, floor, ceiling, whatever you like). The task will be next steps:

1) Select plane basis. Roughly speaking, a tabletop has a length and a width, so it is intuitive that two vectors will be required to construct the basis. One vector is clearly not enough, three vectors are too much.

2) Based on the selected basis set coordinate system(coordinate grid) to assign coordinates to all objects on the table.

Don't be surprised, at first the explanations will be on the fingers. Moreover, on yours. Please place index finger left hand on the edge of the tabletop so that he looks at the monitor. This will be a vector. Now place little finger right hand on the edge of the table in the same way - so that it is directed at the monitor screen. This will be a vector. Smile, you look great! What can we say about vectors? Data vectors collinear, which means linear expressed through each other:
, well, or vice versa: , where is some number different from zero.

You can see a picture of this action in class. Vectors for dummies, where I explained the rule for multiplying a vector by a number.

Will your fingers set the basis on the plane of the computer desk? Obviously not. Collinear vectors travel back and forth across alone direction, and a plane has length and width.

Such vectors are called linearly dependent.

Reference: The words “linear”, “linearly” denote the fact that in mathematical equations and expressions there are no squares, cubes, other powers, logarithms, sines, etc. There are only linear (1st degree) expressions and dependencies.

Two plane vectors linearly dependent if and only if they are collinear.

Cross your fingers on the table so that there is any angle between them other than 0 or 180 degrees. Two plane vectorslinear Not dependent if and only if they are not collinear. So, the basis is obtained. There is no need to be embarrassed that the basis turned out to be “skewed” with non-perpendicular vectors of different lengths. Very soon we will see that not only an angle of 90 degrees is suitable for its construction, and not only unit vectors of equal length

Any plane vector the only way is expanded according to the basis:
, where are real numbers. The numbers are called vector coordinates in this basis.

It is also said that vectorpresented as linear combination basis vectors. That is, the expression is called vector decompositionby basis or linear combination basis vectors.

For example, we can say that the vector is decomposed along an orthonormal basis of the plane, or we can say that it is represented as a linear combination of vectors.

Let's formulate definition of basis formally: The basis of the plane is called a pair of linearly independent (non-collinear) vectors, , while any a plane vector is a linear combination of basis vectors.

An essential point of the definition is the fact that the vectors are taken in a certain order. Bases – these are two completely different bases! As they say, you cannot replace the little finger of your left hand in place of the little finger of your right hand.

We have figured out the basis, but it is not enough to set a coordinate grid and assign coordinates to each item on your computer desk. Why isn't it enough? The vectors are free and wander throughout the entire plane. So how do you assign coordinates to those little dirty spots on the table left over from a wild weekend? A starting point is needed. And such a landmark is a point familiar to everyone - the origin of coordinates. Let's understand the coordinate system:

I'll start with the “school” system. Already in the introductory lesson Vectors for dummies I highlighted some differences between the rectangular coordinate system and the orthonormal basis. Here standard picture:

When they talk about rectangular coordinate system, then most often they mean the origin, coordinate axes and scale along the axes. Try typing “rectangular coordinate system” into a search engine, and you will see that many sources will tell you about coordinate axes familiar from the 5th-6th grade and how to plot points on a plane.

On the other hand, it seems that a rectangular coordinate system can be defined in terms of an orthonormal basis. And that's almost true. The wording sounds as follows:

origin, And orthonormal the basis is set Cartesian rectangular plane coordinate system . That is, the rectangular coordinate system definitely is defined by a single point and two unit orthogonal vectors. That is why you see the drawing that I gave above - in geometric problems, both vectors and coordinate axes are often (but not always) drawn.

I think everyone understands that using a point (origin) and an orthonormal basis ANY POINT on the plane and ANY VECTOR on the plane coordinates can be assigned. Figuratively speaking, “everything on a plane can be numbered.”

Are coordinate vectors required to be unit? No, they can have an arbitrary non-zero length. Consider a point and two orthogonal vectors of arbitrary non-zero length:

Such a basis is called orthogonal. The origin of coordinates with vectors is defined by a coordinate grid, and any point on the plane, any vector has its coordinates in a given basis. For example, or. The obvious inconvenience is that the coordinate vectors V general case have different lengths other than unity. If the lengths are equal to one, then the usual orthonormal basis is obtained.

! Note : in the orthogonal basis, as well as below in the affine bases of plane and space, units along the axes are considered CONDITIONAL. For example, one unit along the x-axis contains 4 cm, one unit along the ordinate axis contains 2 cm. This information is enough to, if necessary, convert “non-standard” coordinates into “our usual centimeters”.

And the second question, which has actually already been answered, is whether the angle between the basis vectors must be equal to 90 degrees? No! As the definition states, the basis vectors must be only non-collinear. Accordingly, the angle can be anything except 0 and 180 degrees.

A point on the plane called origin, And non-collinear vectors, , set affine plane coordinate system :

Sometimes such a coordinate system is called oblique system. As examples, the drawing shows points and vectors:

As you understand, the affine coordinate system is even less convenient; the formulas for the lengths of vectors and segments, which we discussed in the second part of the lesson, do not work in it Vectors for dummies, many delicious formulas related to scalar product of vectors. But the rules for adding vectors and multiplying a vector by a number, formulas for dividing a segment in this respect, as well as some other types of problems that we will consider soon, are valid.

And the conclusion is that the most convenient special case of an affine coordinate system is the Cartesian rectangular system. That’s why you most often have to see her, my dear one. ...However, everything in this life is relative - there are many situations in which an oblique angle (or some other one, for example, polar) coordinate system. And humanoids might like such systems =)

Let's move on to the practical part. All tasks this lesson valid both for the rectangular coordinate system and for the general affine case. There is nothing complicated here; all the material is accessible even to a schoolchild.

How to determine collinearity of plane vectors?

Typical thing. In order for two plane vectors were collinear, it is necessary and sufficient that their corresponding coordinates be proportional Essentially, this is a coordinate-by-coordinate detailing of the obvious relationship.

Example 1

a) Check if the vectors are collinear .
b) Do the vectors form a basis? ?

Solution:
a) Let us find out whether there is for vectors proportionality coefficient, such that the equalities are satisfied:

I’ll definitely tell you about the “foppish” type of application of this rule, which works quite well in practice. The idea is to immediately make up the proportion and see if it is correct:

Let's make a proportion from the ratios of the corresponding coordinates of the vectors:

Let's shorten:
, thus the corresponding coordinates are proportional, therefore,

The relationship could be made the other way around; this is an equivalent option:

For self-test, you can use the fact that collinear vectors are linearly expressed through each other. In this case, the equalities take place . Their validity can be easily verified through elementary operations with vectors:

b) Two plane vectors form a basis if they are not collinear (linearly independent). We examine vectors for collinearity . Let's create a system:

From the first equation it follows that , from the second equation it follows that , which means the system is inconsistent(no solutions). Thus, the corresponding coordinates of the vectors are not proportional.

Conclusion: the vectors are linearly independent and form a basis.

A simplified version of the solution looks like this:

Let's make a proportion from the corresponding coordinates of the vectors :
, which means that these vectors are linearly independent and form a basis.

Usually this option is not rejected by reviewers, but a problem arises in cases where some coordinates are equal to zero. Like this: . Or like this: . Or like this: . How to work through proportion here? (indeed, you cannot divide by zero). It is for this reason that I called the simplified solution “foppish”.

Answer: a) , b) form.

A small creative example for your own solution:

Example 2

At what value of the parameter are the vectors will they be collinear?

In the sample solution, the parameter is found through the proportion.

There is an elegant algebraic way to check vectors for collinearity. Let’s systematize our knowledge and add it as the fifth point:

For two plane vectors the following statements are equivalent:

2) the vectors form a basis;
3) the vectors are not collinear;

+ 5) the determinant composed of the coordinates of these vectors is nonzero.

Respectively, the following opposite statements are equivalent:
1) vectors are linearly dependent;
2) vectors do not form a basis;
3) the vectors are collinear;
4) vectors can be linearly expressed through each other;
+ 5) the determinant composed of the coordinates of these vectors is equal to zero.

I really, really hope that at the moment you already understand all the terms and statements you come across.

Let's take a closer look at the new, fifth point: two plane vectors are collinear if and only if the determinant composed of the coordinates of the given vectors is equal to zero:. For use of this characteristic Naturally, you need to be able to find determinants.

Let's decide Example 1 in the second way:

a) Let us calculate the determinant made up of the coordinates of the vectors :
, which means that these vectors are collinear.

b) Two plane vectors form a basis if they are not collinear (linearly independent). Let's calculate the determinant made up of vector coordinates :
, which means the vectors are linearly independent and form a basis.

Answer: a) , b) form.

It looks much more compact and prettier than a solution with proportions.

With the help of the material considered, it is possible to establish not only the collinearity of vectors, but also to prove the parallelism of segments and straight lines. Let's consider a couple of problems with specific geometric shapes.

Example 3

The vertices of a quadrilateral are given. Prove that a quadrilateral is a parallelogram.

Proof: There is no need to create a drawing in the problem, since the solution will be purely analytical. Let's recall the definition of a parallelogram:
Parallelogram A quadrilateral whose opposite sides are parallel in pairs is called.

Thus, it is necessary to prove:
1) parallelism of opposite sides and;
2) parallelism of opposite sides and.

We prove:

1) Find the vectors:

2) Find the vectors:

The result is the same vector (“according to school” – equal vectors). Collinearity is quite obvious, but it is better to formalize the decision clearly, with arrangement. Let's calculate the determinant made up of vector coordinates:
, which means that these vectors are collinear, and .

Conclusion: The opposite sides of a quadrilateral are parallel in pairs, which means it is a parallelogram by definition. Q.E.D.

More good and different figures:

Example 4

The vertices of a quadrilateral are given. Prove that a quadrilateral is a trapezoid.

For a more rigorous formulation of the proof, it is better, of course, to get the definition of a trapezoid, but it is enough to simply remember what it looks like.

This is a task for you to solve on your own. Complete solution at the end of the lesson.

And now it’s time to slowly move from the plane into space:

How to determine collinearity of space vectors?

The rule is very similar. In order for two space vectors to be collinear, it is necessary and sufficient that their corresponding coordinates be proportional.

Example 5

Find out whether the following space vectors are collinear:

A) ;
b)
V)

Solution:
a) Let’s check whether there is a coefficient of proportionality for the corresponding coordinates of the vectors:

The system has no solution, which means the vectors are not collinear.

“Simplified” is formalized by checking the proportion. In this case:
– the corresponding coordinates are not proportional, which means the vectors are not collinear.

Answer: the vectors are not collinear.

b-c) These are points for independent decision. Try it out in two ways.

There is a method for checking spatial vectors for collinearity through a third-order determinant, this method covered in the article Vector product of vectors.

Similar to the plane case, the considered tools can be used to study the parallelism of spatial segments and straight lines.

Welcome to the second section:

Linear dependence and independence of vectors in three-dimensional space.
Spatial basis and affine coordinate system

Many of the patterns that we examined on the plane will also be valid for space. I tried to minimize the theory notes, since the lion's share of the information has already been chewed. However, I recommend that you read the introductory part carefully, as new terms and concepts will appear.

Now, instead of the plane of the computer desk, we explore three-dimensional space. First, let's create its basis. Someone is now indoors, someone is outdoors, but in any case, we cannot escape three dimensions: width, length and height. Therefore, to construct a basis, three spatial vectors will be required. One or two vectors are not enough, the fourth is superfluous.

And again we warm up on our fingers. Please raise your hand up and spread it out different sides thumb, index and middle finger . These will be vectors, they look in different directions, they have different lengths and have different angles among themselves. Congratulations, the basis of three-dimensional space is ready! By the way, there is no need to demonstrate this to teachers, no matter how hard you twist your fingers, but there is no escape from definitions =)

Next, let's ask important issue, do any three vectors form a basis of three-dimensional space? Please press three fingers firmly onto the top of the computer desk. What happened? Three vectors are located in the same plane, and, roughly speaking, we have lost one of the dimensions - height. Such vectors are coplanar and, it is quite obvious that the basis of three-dimensional space is not created.

It should be noted that coplanar vectors do not have to lie in the same plane, they can be in parallel planes (just don’t do this with your fingers, only Salvador Dali did this =)).

Definition: vectors are called coplanar, if there is a plane to which they are parallel. It is logical to add here that if such a plane does not exist, then the vectors will not be coplanar.

Three coplanar vectors are always linearly dependent, that is, they are linearly expressed through each other. For simplicity, let's again imagine that they lie in the same plane. Firstly, vectors are not only coplanar, they can also be collinear, then any vector can be expressed through any vector. In the second case, if, for example, the vectors are not collinear, then the third vector is expressed through them in a unique way: (and why is easy to guess from the materials in the previous section).

Fair and converse statement:three non-coplanar vectors are always linearly independent, that is, they are in no way expressed through each other. And, obviously, only such vectors can form the basis of three-dimensional space.

Definition: The basis of three-dimensional space is called a triple of linearly independent (non-coplanar) vectors, taken in a certain order, and any vector of space the only way is decomposed over a given basis, where are the coordinates of the vector in this basis

Let me remind you that we can also say that the vector is represented in the form linear combination basis vectors.

The concept of a coordinate system is introduced in exactly the same way as for the plane case; one point and any three linearly independent vectors are sufficient:

origin, And non-coplanar vectors, taken in a certain order, set affine coordinate system of three-dimensional space :

Of course, the coordinate grid is “oblique” and inconvenient, but, nevertheless, the constructed coordinate system allows us definitely determine the coordinates of any vector and the coordinates of any point in space. Similar to a plane, some formulas that I have already mentioned will not work in the affine coordinate system of space.

The most familiar and convenient special case of an affine coordinate system, as everyone guesses, is rectangular space coordinate system:

A point in space called origin, And orthonormal the basis is set Cartesian rectangular space coordinate system . Familiar picture:

Before moving on to practical tasks, let’s again systematize the information:

For three space vectors the following statements are equivalent:
1) the vectors are linearly independent;
2) the vectors form a basis;
3) the vectors are not coplanar;
4) vectors cannot be linearly expressed through each other;
5) the determinant, composed of the coordinates of these vectors, is different from zero.

I think the opposite statements are understandable.

Linear dependence/independence of space vectors is traditionally checked using a determinant (point 5). The remaining practical tasks will be of a pronounced algebraic nature. It's time to hang up the geometry stick and wield the baseball bat of linear algebra:

Three vectors of space are coplanar if and only if the determinant composed of the coordinates of the given vectors is equal to zero: .

I would like to draw your attention to a small technical nuance: the coordinates of vectors can be written not only in columns, but also in rows (the value of the determinant will not change because of this - see properties of determinants). But it is much better in columns, since it is more beneficial for solving some practical problems.

For those readers who have a little forgotten the methods of calculating determinants, or maybe have little understanding of them at all, I recommend one of my oldest lessons: How to calculate the determinant?

Example 6

Check whether the following vectors form the basis of three-dimensional space:

Solution: In fact, the entire solution comes down to calculating the determinant.

a) Let us calculate the determinant made up of the coordinates of the vectors (the determinant is revealed in the first line):

, which means that the vectors are linearly independent (not coplanar) and form the basis of three-dimensional space.

Answer: these vectors form a basis

b) This is a point for independent decision. Full solution and answer at the end of the lesson.

There are also creative tasks:

Example 7

At what value of the parameter will the vectors be coplanar?

Solution: Vectors are coplanar if and only if the determinant composed of the coordinates of these vectors is equal to zero:

Essentially, you need to solve an equation with a determinant. We swoop down on zeros like kites on jerboas - it’s best to open the determinant in the second line and immediately get rid of the minuses:

We carry out further simplifications and reduce the matter to the simplest linear equation:

Answer: at

It’s easy to check here; to do this, you need to substitute the resulting value into the original determinant and make sure that , opening it again.

In conclusion, let's look at one more typical task, which is more algebraic in nature and is traditionally included in the course of linear algebra. It is so common that it deserves its own topic:

Prove that 3 vectors form the basis of three-dimensional space
and find the coordinates of the 4th vector in this basis

Example 8

Vectors are given. Show that vectors form a basis in three-dimensional space and find the coordinates of the vector in this basis.

Solution: First, let's deal with the condition. By condition, four vectors are given, and, as you can see, they already have coordinates in some basis. What this basis is is not of interest to us. And the following thing is of interest: three vectors may well form a new basis. And the first stage completely coincides with the solution of Example 6; it is necessary to check whether the vectors are truly linearly independent:

Let's calculate the determinant made up of vector coordinates:

, which means that the vectors are linearly independent and form the basis of three-dimensional space.

! Important : vector coordinates Necessarily write down into columns determinant, not in strings. Otherwise, there will be confusion in the further solution algorithm.