Find your own vector matrix example. Matrixes and vectors

With a matrix A, if there is such a number L, that ah \u003d lx.

In this case, the number l is called own meaning Operator (matrix a) corresponding to the vector x.

In other words, eigenvector - This is a vector that under the action of a linear operator goes into a collinear vector, i.e. Just multiplied by some number. In contrast, incompatible vectors are converted more difficult.

We write the definition of your own vector in the form of a system of equations:

We transfer all the components to the left:

The last system can be written in matrix form as follows:

(A - L) x \u003d about

The resulting system always has a zero solution X \u003d O. Such systems in which all free members are zero, called uniform. If the matrix of such a system is square, and its determinant is not equal to zero., then according to the crawler formulas, we always get only decision - zero. It can be proved that the system has non-zero solutions if and only if the determinant of this matrix is \u200b\u200bzero, i.e.

| A - L | \u003d. = 0

This equation with an unknown L is called characteristic equation (characteristic polynomial) Matrix A (linear operator).

It is possible to prove that the characteristic polynomial of the linear operator does not depend on the choice of the base.

For example, we will find eigenvalues \u200b\u200band eigenvectors of the linear operator specified by the matrix A \u003d.

To do this, make a characteristic equation | A - L | \u003d. \u003d (1 - L) 2 - 36 \u003d 1 - 2L + L 2 - 36 \u003d L 2 - 2L - 35 \u003d 0; D \u003d 4 + 140 \u003d 144; eigenvalues \u200b\u200bL 1 \u003d (2 - 12) / 2 \u003d -5; L 2 \u003d (2 + 12) / 2 \u003d 7.

To find your own vectors, solve two systems of equations

(A + 5e) x \u003d o

(A - 7E) x \u003d o

For the first of them, the extended matrix takes the form

,

where x 2 \u003d s, x 1 + (2/3) c \u003d 0; x 1 \u003d - (2/3) s, i.e. X (1) \u003d (- (2/3) s; c).

For the second of them, the extended matrix will take the form

,

where x 2 \u003d C 1, x 1 - (2/3) C 1 \u003d 0; x 1 \u003d (2/3) C 1, i.e. X (2) \u003d ((2/3) C 1; C 1).

Thus, their own vectors of this linear operator are all vectors of the form (- (2/3) C; C) with its own value (-5) and all vectors of the form ((2/3) C 1; C 1) with its own value 7 .

It can be proved that the matrix of the operator A in the base consisting of its own vectors is diagonal and has the form:

,

where L i is the eigenvalues \u200b\u200bof this matrix.

It is true and the opposite: if the matrix and in some base is diagonal, then all the vectors of this base will be our own vectors of this matrix.

It can also be proved that if the linear operator has N in pairwise different values, then the corresponding eigen vectors are linearly independent, and the matrix of this operator in the corresponding basis has a diagonal view.

Let us explain it on the previous example. Take arbitrary nonzero values \u200b\u200bC and C 1, but such that the vectors x (1) and x (2) are linearly independent, i.e. For the basis of the basis. For example, let C \u003d C 1 \u003d 3, then x (1) \u003d (-2; 3), x (2) \u003d (2; 3).

Make sure in linear independence of these vectors:

12 ≠ 0. In this new basis, the matrix and will take the form a * \u003d.

To make sure that we use the formula A * \u003d C -1 speakers. First we find with -1.

C -1 \u003d. ;

Quadratic forms

Quadratic form F (x 1, x 2, x n) from n variables call the amount, each member of which is either a square of one of the variables or the product of two different variables, taken with some coefficient: F (x 1, x 2, x n) \u003d (A ij \u003d a ji).

The matrix A composed of these coefficients is called matrixquadratic shape. It is always symmetric Matrix (i.e. matrix, symmetrical relative to the main diagonal, A ij \u003d a ji).

In the matrix record, the quadratic form has the form f (x) \u003d x t Ax, where

Indeed

For example, write a quadratic form in matrix form.

To do this, find the quadratic form matrix. Its diagonal elements are equal to coefficients at squares of variables, and the remaining elements are half the corresponding coefficients of the quadratic form. therefore

Let the matrix column of the variables x are obtained by a non-degenerate linear conversion of the matrix-column Y, i.e. X \u003d CY, where C is a non-degenerate matrix of the N-th order. Then the quadratic form f (x) \u003d x t ah \u003d (CY) T A (CY) \u003d (y t C T) A (CY) \u003d Y T (C T AC) Y.

Thus, with a nondegenerate linear transformation with a quadratic form matrix takes the form: A * \u003d C T AC.

For example, we will find the quadratic form f (y 1, y 2), obtained from a quadratic form F (x 1, x 2) \u003d 2x 1 2 + 4x 1 x 2 - 3x 2 2 linear conversion.

The quadratic form is called canonical (It has canonical species) if all of its coefficients are ij \u003d 0 at i ≠ j, i.e.
f (x 1, x 2, x n) \u003d a 11 x 1 2 + a 22 x 2 2 + a nn x n 2 \u003d.

Its matrix is \u200b\u200bdiagonal.

Theorem (The proof here is not given). Any quadratic form can be given to canonical form with a nondegenerate linear transformation.

For example, we give the canonical species of the quadratic form
f (x 1, x 2, x 3) \u003d 2x 1 2 + 4x 1 x 2 - 3x 2 2 - x 2 x 3.

To do this, first allocate a full square with a variable x 1:

f (x 1, x 2, x 3) \u003d 2 (x 1 2 + 2x 1 x 2 + x 2 2) - 2x 2 2 - 3x 2 2 - x 2 x 3 \u003d 2 (x 1 + x 2) 2 - 5x 2 2 - x 2 x 3.

Now allocate a full square with a variable x 2:

f (x 1, x 2, x 3) \u003d 2 (x 1 + x 2) 2 - 5 (x 2 2 + 2 * x 2 * (1/10) x 3 + (1/100) x 3 2) + (5/100) x 3 2 \u003d
\u003d 2 (x 1 + x 2) 2 - 5 (x 2 - (1/10) x 3) 2 + (1/20) x 3 2.

Then the nondegenerate linear transformation y 1 \u003d x 1 + x 2, y 2 \u003d x 2 + (1/10) x 3 and y 3 \u003d x 3 gives this quadratic shape to the canonical form F (y 1, y 2, y 3) \u003d 2Y 1 2 - 5Y 2 2 + (1/20) Y 3 2.

Note that the canonical type of quadratic form is determined ambiguously (one and the same quadratic form can be given to canonical different ways). However, received different ways Canonical forms have near common properties. In particular, the number of components with positive (negative) coefficients of a quadratic form does not depend on the method of bringing the form to this type (for example, in the considered example there will always be two negative and one positive coefficient). This property is called the law of inertia of quadratic forms.

I will be convinced of this, in a different way, leading the same quadratic form to canonical form. Let's start the transformation from the variable x 2:

f (x 1, x 2, x 3) \u003d 2x 1 2 + 4x 1 x 2 - 3x 2 2 - x 2 x 3 \u003d -3x 2 2 - x 2 x 3 + 4x 1 x 2 + 2x 1 2 \u003d - 3 (x 2 2 +
+ 2 * x 2 ((1/6) x 3 - (2/3) x 1) + ((1/6) x 3 - (2/3) x 1) 2) + 3 ((1/6) x 3 - (2/3) x 1) 2 + 2x 1 2 \u003d
\u003d -3 (x 2 + (1/6) x 3 - (2/3) x 1) 2 + 3 ((1/6) x 3 + (2/3) x 1) 2 + 2x 1 2 \u003d F (Y 1, Y 2, Y 3) \u003d -3y 1 2 -
+ 3Y 2 2 + 2Y 3 2, where y 1 \u003d - (2/3) x 1 + x 2 + (1/6) x 3, y 2 \u003d (2/3) x 1 + (1/6) x 3 and y 3 \u003d x 1. Here negative coefficient -3 for y 1 and two positive coefficients 3 and 2 at y 2 and y 3 (and when using another method, we obtained a negative coefficient (-5) at y 2 and two positive: 2 at y 1 and 1/20 at y 3 ).

It should also be noted that the rank of the matrix of the quadratic form, called rank quadratic form, equal to the number Non-zero canonical form coefficients and does not change with linear transformations.

The quadratic form F (X) is called positively (negative) definedif with all values \u200b\u200bof variables that are not equal at the same time zero, it is positive, i.e. f (x)\u003e 0 (negative, i.e.
F (X)< 0).

For example, a quadratic shape F 1 (x) \u003d x 1 2 + x 2 2 is positively defined, because represents the sum of the squares, and the quadratic shape F 2 (x) \u003d -x 1 2 + 2x 1 x 2 - x 2 2 is negatively defined, because It represents it can be represented as F 2 (x) \u003d - (x 1 - x 2) 2.

In most practical situations Set the value of the quadratic form is somewhat more complicated, so this uses one of the following theorems (we formulate them without evidence).

Theorem. The quadratic form is positive (negatively) determined if and only if all eigenvalues \u200b\u200bof its matrix are positive (negative).

Theorem(Sylvester Criterion). The quadratic form is positively defined then and only if all major minorities of the matrix of this form are positive.

The main (angular) minor K-th order of the matrix A N-th order is called the matrix determinant, compiled from the first k rows and columns of the matrix A ().

Note that for negatively defined quadratic forms, the signs of the main minorities alternate, and the minor of the first order should be negative.

For example, we investigate the quadratic shape F (x 1, x 2) \u003d 2x 1 2 + 4x 1 x 2 + 3 2 2.

\u003d (2 - L) *
* (3 - L) - 4 \u003d (6 - 2L - 3L + L 2) - 4 \u003d L 2 - 5L + 2 \u003d 0; D \u003d 25 - 8 \u003d 17;
. Consequently, the quadratic form is positively defined.

Method 2. Chief Minor of the first order of the matrix A D 1 \u003d A 11 \u003d 2\u003e 0. The main minor of the second order d 2 \u003d 6 - 4 \u003d 2\u003e 0. Therefore, according to the criterion of the Sylvester, the quadratic form is positively defined.

We investigate on the meaningfulness of another quadratic shape, F (x 1, x 2) \u003d -2x 1 2 + 4x 1 x 2 - 3x 2 2.

Method 1. We construct the quadratic matrix A \u003d. The characteristic equation will be viewed \u003d (-2 - L) *
* (- 3 - L) - 4 \u003d (6 + 2L + 3L + L 2) - 4 \u003d L 2 + 5L + 2 \u003d 0; D \u003d 25 - 8 \u003d 17;
. Consequently, the quadratic form is negatively defined.

Method 2. Chief Minor of the first order of the matrix A D 1 \u003d A 11 \u003d
= -2 < 0. Главный минор второго порядка D 2 = = 6 - 4 = 2 > 0. Consequently, according to the criterion of the Sylvester, the quadratic form is negatively defined (the signs of the main miners alternate, starting with a minus).

And as another example, we investigate the quadratic shape f (x 1, x 2) \u003d 2x 1 2 + 4x 1 x 2 - 3x 2 2.

Method 1. We construct the quadratic matrix A \u003d. The characteristic equation will be viewed \u003d (2 - L) *
* (- 3 - L) - 4 \u003d (-6 - 2L + 3L + L 2) - 4 \u003d L 2 + L - 10 \u003d 0; D \u003d 1 + 40 \u003d 41;
.

One of these numbers is negative, and the other is positive. Signs of their own values \u200b\u200bare different. Consequently, the quadratic form cannot be negative, nor positively defined, i.e. This quadratic form is not distinct (can take values \u200b\u200bof any sign).

Method 2. Chief Minor of the first order of the matrix A D 1 \u003d A 11 \u003d 2\u003e 0. The main minor of the second order d 2 \u003d -6 - 4 \u003d -10< 0. Следовательно, по критерию Сильвестра квадратичная форма не является знакоопределенной (знаки главных миноров разные, при этом первый из них - положителен).

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Any fractal is based on a specific rule that is consistently applied to an unlimited number of times. Everyone is called iteration.

The iterative algorithm for constructing the menger sponge is quite simple: the source cube with a side 1 is divided by planes parallel to its faces, on 27 equal cubes. One central cube and 6 adjacent cubes are removed from it. A set consisting of 20 remaining smaller cubes is obtained. By doing the same with each of these cubes, we obtain a set, consisting already from 400 smaller cubes. Continuing this process infinitely, we get a sponge of Menger.

Definition 9.3. Vector h. called own vector Matrians BUTif there is such a number λ, What equality is performed: BUT h.= λ h., That is the result of applying to h. Linear conversion specified by the matrix BUTis the multiplication of this vector to the number λ . Name λ called own number Matrians BUT.

Substituting in formula (9.3) x` J \u003d λx j, We obtain a system of equations for determining the coordinates of its own vector:

. (9.5)

This linear homogeneous system will have a nontrivial solution only if its main determinant is 0 (crawler rule). After writing this condition in the form:

we obtain the equation to determine own numbers λ called characteristic equation. Briefly can be submitted:

| A - λE. | = 0, (9.6)

since its left part is the determinant of the matrix A-λE.. Polynomial relative λ | A - λE.| called characteristic polynomial Matrix A.

Characteristic polynomial properties:

1) The characteristic polynomial linear transformation does not depend on the choice of the base. Evidence. (see (9.4)), but hence, . Thus, it does not depend on the choice of the base. So, and | A-λE.| Does not change when moving to a new basis.

2) if the matrix BUT Linear transformation is symmetric (those. and ij \u003d a ji), all the roots of the characteristic equation (9.6) are valid numbers.

Properties of own numbers and eigenvectors:

1) If you select the basis from your own vectors x 1, x 2, x 3 corresponding to their own λ 1, λ 2, λ 3 Matrians BUTIn this basis, the linear transformation A has a diagonal form matrix:

(9.7) Proof of this property follows from the definition of eigenvectors.

2) if your own conversion values BUT Different, then the corresponding eigen vectors are linearly independent.

3) if the characteristic polynomial matrix BUT has three different roots, then in some base matrix BUT has a diagonal view.

We will find eigenvalues \u200b\u200band eigenvectric vectors of the matrix to be the characteristic equation: (1- λ )(5 - λ )(1 - λ ) + 6 - 9(5 - λ ) - (1 - λ ) - (1 - λ ) = 0, λ ³ - 7. λ ² + 36 \u003d 0, λ 1 = -2, λ 2 = 3, λ 3 = 6.

We will find the coordinates of their own vectors corresponding to each found value. λ. From (9.5) it follows that if h. (1) ={x 1, x 2, x 3) - own vector corresponding to λ 1 \u003d -2, then

- joint, but uncertain system. Its solution can be written as h. (1) ={a.,0,-a.), where a is any number. In particular, if you need to | x. (1) |=1, h. (1) =

Substituting into the system (9.5) λ 2 \u003d 3, we obtain the system to determine the coordinates of the second own vector - x. (2) ={y 1, Y 2, Y 3}:

From! h. (2) ={b, -b, b) or, provided | x. (2) |=1, x. (2) =

For λ 3 \u003d 6 Find your own vector x. (3) ={z 1, z 2, z 3}:

, x. (3) ={c.,2C, C.) or in the normalized version

x (3) = It can be noted that h. (1) h. (2) = aB - AB= 0, x. (1) x. (3) = aC - AC.= 0, x. (2) x. (3) = bC.- 2bC + BC.\u003d 0. Thus, the eigenvectors of this matrix are pairwise orthogonal.

Lecture 10.

Quadratic forms and their connection with symmetric matrices. Properties of own vectors and own numbers of a symmetric matrix. Bringing a quadratic form to canonical form.

Definition 10.1.Quadratic form valid variables x 1, x 2, ..., x n The polynomial of the second degree is called relative to these variables, which does not contain a free member and members of the first degree.

Examples of quadratic forms:

(n. = 2),

(n. = 3). (10.1)

We will remind this in the last lecture the definition of a symmetric matrix:

Definition 10.2. Square matrix is \u200b\u200bcalled symmetricIf, that is, if the elements of the matrix are equal to the main diagonal.

Properties of own numbers and eigenvets of a symmetric matrix:

1) All eigenvalues \u200b\u200bof the symmetric matrix are valid.

Proof (for n. = 2).

Let the matrix BUT It has the form: . Make a characteristic equation:

(10.2) We will find a discriminant:

Consequently, the equation has only valid roots.

2) Own vectors of the symmetric matrix are orthogonal.

Proof (for n.= 2).

Coordinates of own vectors and must satisfy the equations.

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