How to determine whether vectors are linearly dependent or independent. Linear dependence of vectors

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 consist of the following actions:

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 left index finger on the edge of the tabletop so that he looks at the monitor. This will be a vector. Now place right little finger 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, , wherein 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 that are left over after 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's the 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 completely defined in terms of an orthonormal basis. And that's almost true. The wording is 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 in general have different lengths other than unity. If the lengths are equal to unity, 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 relation, 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 problems in this lesson are 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” version of applying 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 by now you already understand all the terms and statements you have encountered.

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:. To apply this feature, of course, 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 remember 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. Full 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 is 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 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 in different directions thumb, index and middle finger. These will be vectors, they look in different directions, have different lengths and have different angles between 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 an important question: 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 us 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).

The converse is also true: 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’s calculate the determinant made up of vector coordinates (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, we will consider another typical problem, which is more algebraic in nature and is traditionally included in a linear algebra course. 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.

The vector system is called linearly dependent, if there are numbers among which at least one is different from zero, such that the equality https://pandia.ru/text/78/624/images/image004_77.gif" width="57" height="24 src=" >.

If this equality is satisfied only in the case when all , then the system of vectors is called linearly independent.

Theorem. The vector system will linearly dependent if and only if at least one of its vectors is a linear combination of the others.

Example 1. Polynomial is a linear combination of polynomials https://pandia.ru/text/78/624/images/image010_46.gif" width="88 height=24" height="24">. The polynomials constitute a linearly independent system, since the polynomial https: //pandia.ru/text/78/624/images/image012_44.gif" width="129" height="24">.

Example 2. The matrix system, , https://pandia.ru/text/78/624/images/image016_37.gif" width="51" height="48 src="> is linearly independent, since a linear combination is equal to the zero matrix only in in the case when https://pandia.ru/text/78/624/images/image019_27.gif" width="69" height="21">, , https://pandia.ru/text/78/624 /images/image022_26.gif" width="40" height="21"> linearly dependent.

Solution.

Let's make a linear combination of these vectors https://pandia.ru/text/78/624/images/image023_29.gif" width="97" height="24">=0..gif" width="360" height=" 22">.

Equating the same coordinates of equal vectors, we get https://pandia.ru/text/78/624/images/image027_24.gif" width="289" height="69">

Finally we get

And

The system has a unique trivial solution, so a linear combination of these vectors is equal to zero only in the case when all coefficients are equal to zero. Therefore, this system of vectors is linearly independent.

Example 4. The vectors are linearly independent. What will the vector systems be like?

a).;

b).?

Solution.

a). Let's make a linear combination and equate it to zero

Using the properties of operations with vectors in linear space, we rewrite the last equality in the form

Since the vectors are linearly independent, the coefficients for must be equal to zero, i.e..gif" width="12" height="23 src=">

The resulting system of equations has a unique trivial solution .

Since equality (*) executed only when https://pandia.ru/text/78/624/images/image031_26.gif" width="115 height=20" height="20"> – linearly independent;

b). Let's make an equality https://pandia.ru/text/78/624/images/image039_17.gif" width="265" height="24 src="> (**)

Applying similar reasoning, we obtain

Solving the system of equations by the Gauss method, we obtain

or

The latter system has an infinite number of solutions https://pandia.ru/text/78/624/images/image044_14.gif" width="149" height="24 src=">. Thus, there is a non-zero set of coefficients for which holds the equality (**) . Therefore, the system of vectors – linearly dependent.

Example 5 A system of vectors is linearly independent, and a system of vectors is linearly dependent..gif" width="80" height="24">.gif" width="149 height=24" height="24"> (***)

In equality (***) . Indeed, at , the system would be linearly dependent.

From the relation (***) we get or Let's denote .

We get

Problems for independent solution (in the classroom)

1. A system containing a zero vector is linearly dependent.

2. System consisting of one vector A, is linearly dependent if and only if, a=0.

3. A system consisting of two vectors is linearly dependent if and only if the vectors are proportional (that is, one of them is obtained from the other by multiplying by a number).

4. If you add a vector to a linearly dependent system, you get a linearly dependent system.

5. If a vector is removed from a linearly independent system, then the resulting system of vectors is linearly independent.

6. If the system S is linearly independent, but becomes linearly dependent when adding a vector b, then the vector b linearly expressed through system vectors S.

c). System of matrices , , in the space of second-order matrices.

10. Let the system of vectors a,b,c vector space is linearly independent. Prove the linear independence of the following vector systems:

a).a+b, b, c.

b).a+https://pandia.ru/text/78/624/images/image062_13.gif" width="15" height="19">– arbitrary number

c).a+b, a+c, b+c.

11. Let a,b,c– three vectors on the plane from which a triangle can be formed. Will these vectors be linearly dependent?

12. Two vectors are given a1=(1, 2, 3, 4),a2=(0, 0, 0, 1). Find two more four-dimensional vectors a3 anda4 so that the system a1,a2,a3,a4 was linearly independent .

In this article we will cover:

• what are collinear vectors;
• what are the conditions for collinearity of vectors;
• what properties of collinear vectors exist;
• what is the linear dependence of collinear vectors.

Collinear vectors are vectors that are parallel to one line or lie on one line.

Example 1

Conditions for collinearity of vectors

Two vectors are collinear if any of the following conditions are true:

• condition 1 . Vectors a and b are collinear if there is a number λ such that a = λ b;
• condition 2 . Vectors a and b are collinear with equal coordinate ratios:

a = (a 1 ; a 2) , b = (b 1 ; b 2) ⇒ a ∥ b ⇔ a 1 b 1 = a 2 b 2

• condition 3 . Vectors a and b are collinear provided that the cross product and the zero vector are equal:

a ∥ b ⇔ a, b = 0

Note 1

Condition 2 not applicable if one of the vector coordinates is zero.

Note 2

Condition 3 applies only to those vectors that are specified in space.

Examples of problems to study the collinearity of vectors

We examine the vectors a = (1; 3) and b = (2; 1) for collinearity.

How to solve?

In this case, it is necessary to use the 2nd collinearity condition. For given vectors it looks like this:

The equality is false. From this we can conclude that vectors a and b are non-collinear.

Answer : a | | b

Example 2

What value m of the vector a = (1; 2) and b = (- 1; m) is necessary for the vectors to be collinear?

How to solve?

Using the second collinearity condition, vectors will be collinear if their coordinates are proportional:

This shows that m = - 2.

Answer: m = - 2 .

Criteria for linear dependence and linear independence of vector systems

A system of vectors in a vector space is linearly dependent only if one of the vectors of the system can be expressed in terms of the remaining vectors of this system.

Proof

Let the system e 1 , e 2 , . . . , e n is linearly dependent. Let us write a linear combination of this system equal to the zero vector:

a 1 e 1 + a 2 e 2 + . . . + a n e n = 0

in which at least one of the combination coefficients is not equal to zero.

Let a k ≠ 0 k ∈ 1 , 2 , . . . , n.

We divide both sides of the equality by a non-zero coefficient:

a k - 1 (a k - 1 a 1) e 1 + (a k - 1 a k) e k + . . . + (a k - 1 a n) e n = 0

Let's denote:

A k - 1 a m , where m ∈ 1 , 2 , . . . , k - 1 , k + 1 , n

In this case:

β 1 e 1 + . . . + β k - 1 e k - 1 + β k + 1 e k + 1 + . . . + β n e n = 0

or e k = (- β 1) e 1 + . . . + (- β k - 1) e k - 1 + (- β k + 1) e k + 1 + . . . + (- β n) e n

It follows that one of the vectors of the system is expressed through all other vectors of the system. Which is what needed to be proven (etc.).

Adequacy

Let one of the vectors be linearly expressed through all other vectors of the system:

e k = γ 1 e 1 + . . . + γ k - 1 e k - 1 + γ k + 1 e k + 1 + . . . + γ n e n

We move the vector e k to the right side of this equality:

0 = γ 1 e 1 + . . . + γ k - 1 e k - 1 - e k + γ k + 1 e k + 1 + . . . + γ n e n

Since the coefficient of the vector e k is equal to - 1 ≠ 0, we get a non-trivial representation of zero by a system of vectors e 1, e 2, . . . , e n , and this, in turn, means that this system of vectors is linearly dependent. Which is what needed to be proven (etc.).

Consequence:

• A system of vectors is linearly independent when none of its vectors can be expressed in terms of all other vectors of the system.
• A system of vectors that contains a zero vector or two equal vectors is linearly dependent.

Properties of linearly dependent vectors

1. For 2- and 3-dimensional vectors, the following condition is met: two linearly dependent vectors are collinear. Two collinear vectors are linearly dependent.
2. For 3-dimensional vectors, the following condition is satisfied: three linearly dependent vectors are coplanar. (3 coplanar vectors are linearly dependent).
3. For n-dimensional vectors, the following condition is satisfied: n + 1 vectors are always linearly dependent.

Examples of solving problems involving linear dependence or linear independence of vectors

Let's check the vectors a = 3, 4, 5, b = - 3, 0, 5, c = 4, 4, 4, d = 3, 4, 0 for linear independence.

Solution. Vectors are linearly dependent because the dimension of vectors is less than the number of vectors.

Example 4

Let's check the vectors a = 1, 1, 1, b = 1, 2, 0, c = 0, - 1, 1 for linear independence.

Solution. We find the values ​​of the coefficients at which the linear combination will be equal to the zero vector:

x 1 a + x 2 b + x 3 c 1 = 0

We write the vector equation in linear form:

x 1 + x 2 = 0 x 1 + 2 x 2 - x 3 = 0 x 1 + x 3 = 0

We solve this system using the Gaussian method:

1 1 0 | 0 1 2 - 1 | 0 1 0 1 | 0 ~

From the 2nd line we subtract the 1st, from the 3rd - the 1st:

~ 1 1 0 | 0 1 - 1 2 - 1 - 1 - 0 | 0 - 0 1 - 1 0 - 1 1 - 0 | 0 - 0 ~ 1 1 0 | 0 0 1 - 1 | 0 0 - 1 1 | 0 ~

From the 1st line we subtract the 2nd, to the 3rd we add the 2nd:

~ 1 - 0 1 - 1 0 - (- 1) | 0 - 0 0 1 - 1 | 0 0 + 0 - 1 + 1 1 + (- 1) | 0 + 0 ~ 0 1 0 | 1 0 1 - 1 | 0 0 0 0 | 0

From the solution it follows that the system has many solutions. This means that there is a non-zero combination of values ​​of such numbers x 1, x 2, x 3 for which the linear combination of a, b, c equals the zero vector. Therefore, the vectors a, b, c are linearly dependent. ​​​​​​​

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Let L is an arbitrary linear space, a i Î L,- its elements (vectors).

Definition 3.3.1. Expression , Where , - arbitrary real numbers, called a linear combination vectors a 1 , a 2 ,…, a n.

If the vector R = , then they say that R decomposed into vectors a 1 , a 2 ,…, a n.

Definition 3.3.2. A linear combination of vectors is called non-trivial, if among the numbers there is at least one non-zero. Otherwise, the linear combination is called trivial.

Definition 3.3.3 . Vectors a 1 , a 2 ,…, a n are called linearly dependent if there exists a nontrivial linear combination of them such that

= 0 .

Definition 3.3.4. Vectors a 1 ,a 2 ,…, a n are called linearly independent if the equality = 0 is possible only in the case when all the numbers l 1, l 2,…, l n are simultaneously equal to zero.

Note that any non-zero element a 1 can be considered as a linearly independent system, since the equality l a 1 = 0 possible only if l= 0.

Theorem 3.3.1. A necessary and sufficient condition for the linear dependence a 1 , a 2 ,…, a n is the possibility of decomposing at least one of these elements into the rest.

Proof. Necessity. Let the elements a 1 , a 2 ,…, a n linearly dependent. It means that = 0 , and at least one of the numbers l 1, l 2,…, l n different from zero. Let for certainty l 1 ¹ 0. Then

i.e. element a 1 is decomposed into elements a 2 , a 3 , …, a n.

Adequacy. Let element a 1 be decomposed into elements a 2 , a 3 , …, a n, i.e. a 1 = . Then = 0 , therefore, there is a non-trivial linear combination of vectors a 1 , a 2 ,…, a n, equal 0 , so they are linearly dependent .

Theorem 3.3.2. If at least one of the elements a 1 , a 2 ,…, a n zero, then these vectors are linearly dependent.

Proof . Let a n= 0 , then = 0 , which means the linear dependence of these elements.

Theorem 3.3.3. If among n vectors any p (p< n) векторов линейно зависимы, то и все n элементов линейно зависимы.

Proof. Let, for definiteness, the elements a 1 , a 2 ,…, a p linearly dependent. This means that there is a non-trivial linear combination such that = 0 . The specified equality will be preserved if we add the element to both its parts. Then + = 0 , and at least one of the numbers l 1, l 2,…, lp different from zero. Therefore, vectors a 1 , a 2 ,…, a n are linearly dependent.

Corollary 3.3.1. If n elements are linearly independent, then any k of them are linearly independent (k< n).

Theorem 3.3.4. If the vectors a 1 , a 2 ,…, a n- 1 are linearly independent, and the elements a 1 , a 2 ,…, a n- 1,a n are linearly dependent, then the vector a n can be expanded into vectors a 1 , a 2 ,…, a n- 1 .

Proof. Since by condition a 1 , a 2 ,…, a n- 1,a n are linearly dependent, then there is a nontrivial linear combination of them = 0 , and (otherwise, the vectors a 1 , a 2 ,…, a n- 1). But then the vector

Q.E.D.

In other words, the linear dependence of a group of vectors means that there is a vector among them that can be represented by a linear combination of other vectors in this group.

Let's say. Then

Therefore the vector x linearly dependent of the vectors of this group.

Vectors x, y, ..., z are called linear independent vectors, if it follows from equality (0) that

α=β= ...= γ=0.

That is, groups of vectors are linearly independent if no vector can be represented by a linear combination of other vectors in this group.

Determination of linear dependence of vectors

Let m string vectors of order n be given:

Having made a Gaussian exception, we reduce matrix (2) to upper triangular form. The elements of the last column change only when the rows are rearranged. After m elimination steps we get:

Where i 1 , i 2 , ..., i m - row indices obtained by possible permutation of rows. Considering the resulting rows from the row indices, we exclude those that correspond to the zero row vector. The remaining lines form linearly independent vectors. Note that when composing matrix (2), by changing the sequence of row vectors, you can obtain another group of linearly independent vectors. But the subspace that both these groups of vectors form coincides.