Find the decomposition of the vector in the basis. Linear dependence and linear independence of vectors
In vector calculus and its applications great importance has a decomposition task consisting in representing a given vector as a sum of several vectors called components of a given
vector. This problem, which in general has an infinite number of solutions, becomes completely defined if we specify some elements of the component vectors.
2. Examples of decomposition.
Let us consider several very common cases of decomposition.
1. Decompose a given vector c into two component vectors of which one, for example a, is given in magnitude and direction.
The problem comes down to determining the difference between two vectors. Indeed, if the vectors are components of the vector c, then the equality must be satisfied
From here the second component vector is determined
2. Decompose the given vector c into two components, one of which must lie in a given plane and the second must lie on a given straight line a.
To determine the component vectors, we move the vector c so that its beginning coincides with the point of intersection of the given straight line with the plane (point O - see Fig. 18). From the end of vector c (point C) we draw a straight line to
intersection with the plane (B is the point of intersection), and then from point C we draw a straight line parallel
The vectors and will be the desired ones, i.e. Naturally, the indicated expansion is possible if the straight line a and the plane are not parallel.
3. Given three coplanar vectors a, b and c, and the vectors are not collinear. It is required to decompose the vector c into vectors
Let us bring all three given vectors to one point O. Then, due to their coplanarity, they will be located in the same plane. On given vector with how on the diagonal we will construct a parallelogram, the sides of which are parallel to the lines of action of the vectors (Fig. 19). This construction is always possible (unless the vectors are collinear) and unique. From Fig. 19 it is clear that
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 cover two sections at once. higher mathematics, and we'll see how they get along 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, basis of vectors 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
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 forefinger 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, , 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 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'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 rectangular system coordinates can be completely determined through 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, and 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 little creative example for independent decision:
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 this 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:. 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 figures good and different:
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 be valid for space. I tried to minimize the theory notes because lion's share information has already been chewed up. 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, 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 ourselves 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, 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.
The basis of space they call such a system of vectors in which all other vectors in space can be represented as a linear combination of vectors included in the basis.
In practice, this is all implemented quite simply. The basis, as a rule, is checked on a plane or in space, and for this you need to find the determinant of a second, third order matrix composed of vector coordinates. Below are schematically written conditions under which vectors form a basis
To expand vector b into basis vectors
e,e...,e[n] it is necessary to find the coefficients x, ..., x[n] for which the linear combination of vectors e,e...,e[n] is equal to the vector b:
x1*e+ ... + x[n]*e[n] = b.
To do this, the vector equation should be converted to the system linear equations and find solutions. This is also quite simple to implement.
The found coefficients x, ..., x[n] are called coordinates of vector b in the basis e,e...,e[n].
Let's move on to the practical side of the topic.
Decomposition of a vector into basis vectors
Task 1. Check whether vectors a1, a2 form a basis on the plane
1) a1 (3; 5), a2 (4; 2)
Solution: We compose a determinant from the coordinates of the vectors and calculate it
Determinant is not zero, hence the vectors are linearly independent, which means they form a basis.
2) a1 (2;-3), a2 (5;-1)
Solution: We calculate the determinant made up of vectors 
The determinant is equal to 13 (not equal to zero) - from this it follows that the vectors a1, a2 are a basis on the plane.
---=================---
Let's look at typical examples from the MAUP program in the discipline "Higher Mathematics".
Task 2. Show that the vectors a1, a2, a3 form the basis of a three-dimensional vector space, and expand the vector b according to this basis (use Cramer’s method when solving a system of linear algebraic equations).
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 matrix A
built on non-zero vectors. The matrix contains one zero element, so it is more appropriate to calculate the determinant as a schedule in the first column or third row. 
As a result of the calculations, we found that the determinant is different from zero, therefore vectors a1, a2, a3 are linearly independent.
By definition, vectors form a basis in R3. Let's write down the schedule of vector b based on 
Vectors are equal when their corresponding coordinates are equal.
Therefore, from the vector equation we obtain a system of linear equations 
Let's solve SLAE Cramer's method. To do this, we write the system of equations in the form
The main determinant of a SLAE is always equal to the determinant composed of basis vectors
Therefore, in practice it is not counted twice. To find auxiliary determinants, we put a column of free terms in place of each column of the main determinant. Determinants are calculated using the triangle rule 
Let's substitute the found determinants into Cramer's formula 
So, the expansion of the vector b in terms of the basis has the form b=-4a1+3a2-a3. The coordinates of 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: We check the vectors for a basis - we compose 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 vector b through this basis. To do this, we write the vector equation 
and transform to a system of linear equations 
Let's write it down matrix equation
Next, for Cramer’s formulas we find auxiliary determinants 
We apply Cramer's formulas 
So a given vector b has a schedule through two basis vectors b=-2a1+5a3, and its coordinates in the basis are equal to b(-2,0, 5).