# Advice 1: How to prove that the vectors form a basis

Basis in n-dimensional space is called the system of n vectors, when all the other vectors in the space can be represented as a combination of vectors included in the basis. In three-dimensional space, any basis consists of three vectors. But not any three form a basis, therefore, there is the problem of system test vectors on the opportunity to build their base. You will need
• - the ability to calculate the determinant of a matrix
Instruction
1
Suppose that in a linear n-dimensional space there exists a system of vectors e1, E2, E3, ... , EN. Their coordinates are: e1 = (e11; e21; e31; ... ; en1), E2 = (E12; E22; е32; ... ; EP2), ... , EN = (e1n; e2n; e3n; ... ; enn). To find out whether they form a basis in this space, make a matrix with columns e1, E2, E3, ... , EN. Find its determinant and compare it with zero. If the determinant of the matrix of these vectors is not zero, then these vectors form a basis in this n-dimensional linear space.
2
For example, suppose you are given three vectors in three dimensions a1, a2 and a3. Their coordinates: A1 = (3; 1; 4), A2 = (-4; 2; 3) and A3 = (2; -1; -2). Need to find out whether these form a vector basis in three-dimensional space. Make a matrix from the vectors, as shown in the figure.
3
Calculate the determinant of the resulting matrix. The figure shows a simple method of calculating the determinant of the matrix 3 by 3. Elements are connected by a line should be multiplied. The works outlined in red are included in the total amount with the sign "+", and the United blue line - with the sign "-". det A = 3*2*(-2) + 1*2*3 + 4*(-4)*(-1) - 2*2*4 - 1*(-4)*(-2) - 3*3*(-1) = -12 + 6 + 16 - 16 - 8 + 9 = -5 -5≠0, therefore, A1, A2 and A3 form a basis.

# Advice 2 : How to find the basis of the system of vector-columns

Before considering this question it is worth Recalling that any ordered system of n linearly independent vectors of the space R^n is called a basis of this space. In this form the system of vectors are considered linearly independent if any linear combination of the zero is only possible at the expense of equality to zero of all coefficients of this combination. You will need
• paper;
• - handle.
Instruction
1
Using only basic definitions to check linear independence of a system of vector-columns, respectively, and give a conclusion on the existence of the basis, is very difficult. Therefore, in this case, you may be able to use some special signs.
2
It is known that the vectors are linearly independent if derived from the determinant is not zero.Therefore, it is possible enough to explain the fact that the system of vectors forms a basis. So, in order to prove that the vectors form a basis, should be the coordinates of their determinant and make sure it is not zero.Further, in order to reduce and simplify records, the representation of the vector-column matrix column will replace the transpose by the matrix-line.
3
Example 1. Whether form a basis in R^3 vector-columns (1, 3, 5)^T, (2, 6, 4)^T, (3, 9, 0)^T. the Solution. Make up a determinant |A|, whose rows are the elements of the specified column (see Fig.1).Expanding this determinant by the rule of triangles, you get: |A| = 0+90+36-90-36-0=0. Therefore, these vectors cannot form a basis.
4
Example. 2. The system of vectors consists of (10, 3, 6)^T, (1, 3, 4)^T, (3, 9, 2)^T. if they Can form a basis?Solution. By analogy with the first example, make a key (see Fig.2): |A| =60+54+36-54-360-6=270, ie is not zero. Therefore, this system is a vector-column suitable for use as the basis in R^3.
5
Now obviously becomes clear, that for finding of the basis system of the vector-column is sufficient to take the determinant of any suitable dimension different from zero. Elements its columns form a base system. Moreover, always it is desirable to have a simple basis. Since the determinant of the identity matrix is always nonzero (for any dimension), as the basis you can always choose a system (1, 0, 0,...,0)^T, (0, 1, 0,...,0)^T, (0, 0, 1,...,0)^T,..., (0, 0, 0,...,1)^T.

# Advice 3 : How to sort vector

Any vector can be decomposed into the sum of vectors, and such possibilities are endless. The task to decompose the vector can be given as in the geometric form, and the form of formulas, this will depend on the solution of the problem. You will need
• - initial vector;
• - vector for which you want to decompose.
Instruction
1
If you want to decompose a vector in the drawing, select the direction for parts. For convenience of calculations most often used decomposition of the vectora, parallel to the coordinate axes, but you can choose absolutely any convenient direction.
2
Draw one of the components of the vectors, in which case it must proceed from the same point as the original (the length you choose). Connect the ends of the source and the received vectorand another vectorom. Please note: two of the received vectorand the result should lead you to the same point as the source (if you move the arrows).
3
Move the resulting vectorand in the place where they are convenient to use, while maintaining the direction and length. Regardless of where the vectorand will be, in total they will be equal to the original. Please note that if you place the resulting vectorand so that they proceeded from the same point as the source, and the dotted line connecting their ends will produce a parallelogram, and the original vector will coincide with one of the diagonals.
4
If you need to decompose the vector {x1,x2,X3} with respect to the basis, that is, given the vectorof am {P1, P2, P3}, {q1,q2,q3}, {r1,r2,r3}, we proceed as follows. Substitute the coordinate values into the formula x=ar+βq+γr.
5
As a result, you will obtain a system of three equations P1A+q1β+r1γ=x1, p2α+q2β+r2γ=x2, p3α+q3β+r3γ=X3. Solve this system using method of summations or matrices, find the coefficients α, β, γ. If the task is given in the plane, the solution is more simple, because instead of three variables and equations you will receive only two (they will be in the form P1A+q1β=x1, p2α+q2β=x2). Write your answer in the form x=AP+βq+γr.
6
If you get an infinite number of solutions, make a conclusion that the vectors p, q, r lie in the same plane with the vectorω of x and decompose it in the specified way definitely not.
7
If solutions, the system has to write the answer to the problem: the vectors p, q, r lie in one plane, and the vector x in another, so it is impossible to decompose the given image.

# Advice 4 : How to find the coordinates of the vector in the basis

A pair of points called ordered, if it is known which of the points is first and which is second. The segment ends with ordered is called a directed segment or vector. Basis in a vector space is an ordered linearly independent system of vectors in that any vector space is decomposed by it. The coefficients in this decomposition are the coordinates of the vector in this basis. Instruction
1
Let a system of vectors a1,a2,...,ak. It is linearly independent when the zero vector is routed through it the only way. In other words, only the trivial combination of these vectors will result in zero vector. Trivial decomposition implies the equality to zero of all coefficients.
2
The system, consisting of a single nonzero vector is always linearly independent. The system of the two vectors are linearly independent if they are not collinear. To a system of three vectors were linearly independent, it is necessary that they were non-complanar. Of four or more vectors, it is impossible to compose a linearly independent system.
3
Thus, the null space basis is not. In one-dimensional space a basis can be any nonzero vector. In space dimension two the basis can be any ordered pair of non-collinear vectors. Finally ordered three non-complanar vectors will form a basis for three-dimensional space.
4
The vector can be decomposed on the basis of, for example, p= λ1•a1+ λ2•a2+...+ λk•ak. The coefficients of the decomposition λ1,..., λk are the coordinates of the vector in this basis. Sometimes they are also called components of the vector. Since a basis is a linearly independent system, the expansion coefficients are determined uniquely and the only way.
5
Suppose we have a basis consisting of one vector e. Any vector in this basis will be only one coordinate: p=a•e. If p is collinear basis vectors, the number a show the ratio of the lengths of the vectors p and e. If the opposite number a is even and negative. In the case of arbitrary direction of the vector p relative to the vector e in component a will include the cosine of the angle between them.
6
In the basis of higher-order decay to represent a more complex equation. However, it is possible to consistently decompose a given vector on the basis vectors, similar to one-dimensional.
7
To find the coordinates of the vector in the basis, position in the drawing vector near the base. If necessary, draw the projection of the vector on the coordinate axis. Compare the length of the vector basis, shall describe the angles between them and basic vectors. Use the trigonometric functions: sine, cosine, tangent. Decompose a vector in terms of the basis, and the coefficients of the decomposition are its coordinates.
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