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 can 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)
it is sometimes called the triple dot product of vectors, apparently because 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 more) 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 vector systems.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 a 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.
Introduced by us linear operations on vectors make it possible to create various expressions for vector quantities and transform them using the properties set for these operations.
Based on a given set of vectors a 1, ..., a n, you can create an expression of the form
where a 1, ..., and n are arbitrary real numbers. This expression is called linear combination of vectors a 1, ..., a n. The numbers α i, i = 1, n, represent linear combination coefficients. A set of vectors is also called system of vectors.
In connection with the introduced concept of a linear combination of vectors, the problem arises of describing a set of vectors that can be written as a linear combination of a given system of vectors a 1, ..., a n. In addition, there are natural questions about the conditions under which there is a representation of a vector in the form of a linear combination, and about the uniqueness of such a representation.
Definition 2.1. Vectors a 1, ..., and n are called linearly dependent, if there is a set of coefficients α 1 , ... , α n such that
α 1 a 1 + ... + α n а n = 0 (2.2)
and at least one of these coefficients is non-zero. If the specified set of coefficients does not exist, then the vectors are called linearly independent.
If α 1 = ... = α n = 0, then, obviously, α 1 a 1 + ... + α n a n = 0. With this in mind, we can say this: vectors a 1, ..., and n are linearly independent if it follows from equality (2.2) that all coefficients α 1 , ... , α n are equal to zero.
The following theorem explains why the new concept is called the term "dependence" (or "independence"), and provides a simple criterion for linear dependence.
Theorem 2.1. In order for the vectors a 1, ..., and n, n > 1, to be linearly dependent, it is necessary and sufficient that one of them is a linear combination of the others.
◄ Necessity. Let us assume that the vectors a 1, ..., and n are linearly dependent. According to Definition 2.1 of linear dependence, in equality (2.2) on the left there is at least one non-zero coefficient, for example α 1. Leaving the first term on the left side of the equality, we move the rest to the right side, changing their signs, as usual. Dividing the resulting equality by α 1, we get
a 1 =-α 2 /α 1 ⋅ a 2 - ... - α n /α 1 ⋅ a n
those. representation of vector a 1 as a linear combination of the remaining vectors a 2, ..., a n.
Adequacy. Let, for example, the first vector a 1 can be represented as a linear combination of the remaining vectors: a 1 = β 2 a 2 + ... + β n a n. Transferring all terms from the right side to the left, we obtain a 1 - β 2 a 2 - ... - β n a n = 0, i.e. a linear combination of vectors a 1, ..., a n with coefficients α 1 = 1, α 2 = - β 2, ..., α n = - β n, equal to zero vector. In this linear combination, not all coefficients are zero. According to Definition 2.1, the vectors a 1, ..., and n are linearly dependent.
The definition and criterion for linear dependence are formulated to imply the presence of two or more vectors. However, we can also talk about a linear dependence of one vector. To realize this possibility, instead of “vectors are linearly dependent,” you need to say “the system of vectors is linearly dependent.” It is easy to see that the expression “a system of one vector is linearly dependent” means that this single vector is zero (in a linear combination there is only one coefficient, and it should not be equal to zero).
The concept of linear dependence has a simple geometric interpretation. The following three statements clarify this interpretation.
Theorem 2.2. Two vectors are linearly dependent if and only if they collinear.
◄ If vectors a and b are linearly dependent, then one of them, for example a, is expressed through the other, i.e. a = λb for some real number λ. According to definition 1.7 works vectors per number, vectors a and b are collinear.
Let now vectors a and b be collinear. If they are both zero, then it is obvious that they are linearly dependent, since any linear combination of them is equal to the zero vector. Let one of these vectors not be equal to 0, for example vector b. Let us denote by λ the ratio of vector lengths: λ = |a|/|b|. Collinear vectors can be unidirectional or oppositely directed. In the latter case, we change the sign of λ. Then, checking Definition 1.7, we are convinced that a = λb. According to Theorem 2.1, vectors a and b are linearly dependent.
Remark 2.1. In the case of two vectors, taking into account the criterion of linear dependence, the proven theorem can be reformulated as follows: two vectors are collinear if and only if one of them is represented as the product of the other by a number. This is a convenient criterion for the collinearity of two vectors.
Theorem 2.3. Three vectors are linearly dependent if and only if they coplanar.
◄ If three vectors a, b, c are linearly dependent, then, according to Theorem 2.1, one of them, for example a, is a linear combination of the others: a = βb + γс. Let us combine the origins of vectors b and c at point A. Then the vectors βb, γс will have a common origin at point A and along according to the parallelogram rule, their sum is those. vector a will be a vector with origin A and the end, which is the vertex of a parallelogram built on component vectors. Thus, all vectors lie in the same plane, i.e., coplanar.
Let vectors a, b, c be coplanar. If one of these vectors is zero, then it will obviously be a linear combination of the others. It is enough to take all coefficients of a linear combination equal to zero. Therefore, we can assume that all three vectors are not zero. Compatible started of these vectors at a common point O. Let their ends be points A, B, C, respectively (Fig. 2.1). Through point C we draw lines parallel to lines passing through pairs of points O, A and O, B. Designating the points of intersection as A" and B", we obtain a parallelogram OA"CB", therefore, OC" = OA" + OB". Vector OA" and a non-zero vector a = OA are collinear, and therefore the first of them can be obtained by multiplying the second by a real number α:OA" = αOA. Similarly, OB" = βOB, β ∈ R. As a result, we obtain that OC" = α OA + βOB, i.e. vector c is a linear combination of vectors a and b. According to Theorem 2.1, vectors a, b, c are linearly dependent.
Theorem 2.4. Any four vectors are linearly dependent.
◄ We carry out the proof according to the same scheme as in Theorem 2.3. Consider arbitrary four vectors a, b, c and d. If one of the four vectors is zero, or among them there are two collinear vectors, or three of the four vectors are coplanar, then these four vectors are linearly dependent. For example, if vectors a and b are collinear, then we can make their linear combination αa + βb = 0 with non-zero coefficients, and then add the remaining two vectors to this combination, taking zeros as coefficients. We obtain a linear combination of four vectors equal to 0, in which there are non-zero coefficients.
Thus, we can assume that among the selected four vectors, no vectors are zero, no two are collinear, and no three are coplanar. Let us choose point O as their common beginning. Then the ends of the vectors a, b, c, d will be some points A, B, C, D (Fig. 2.2). Through point D we draw three planes parallel to the planes OBC, OCA, OAB, and let A", B", C" be the points of intersection of these planes with the straight lines OA, OB, OS, respectively. We obtain a parallelepiped OA" C "B" C" B"DA", and vectors a, b, c lie on its edges emerging from vertex O. Since the quadrilateral OC"DC" is a parallelogram, then OD = OC" + OC". In turn, the segment OC" is a diagonal parallelogram OA"C"B", so OC" = OA" + OB" and OD = OA" + OB" + OC" .
It remains to note that the pairs of vectors OA ≠ 0 and OA" , OB ≠ 0 and OB" , OC ≠ 0 and OC" are collinear, and, therefore, it is possible to select the coefficients α, β, γ so that OA" = αOA , OB" = βOB and OC" = γOC. We finally get OD = αOA + βOB + γOC. Consequently, the OD vector is expressed through the other three vectors, and all four vectors, according to Theorem 2.1, are linearly dependent.
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 will turn out to be linearly dependent n- 1). But then the vector
,
Q.E.D.
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 a given system of vectors is linearly dependent or linearly independent:
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.
Tasks. 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.
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 at 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 .