Reducing a quadratic form to a diagonal form. Theorem on the possibility of reducing a quadratic form to canonical form

Definition 10.4.Canonical view quadratic form (10.1) is called the following form: . (10.4)

Let us show that in a basis of eigenvectors, the quadratic form (10.1) takes on a canonical form. Let

- normalized eigenvectors corresponding to eigenvalues λ 1 ,λ 2 ,λ 3 matrices (10.3) in an orthonormal basis. Then the transition matrix from the old basis to the new one will be the matrix

. In the new basis the matrix A will take the diagonal form (9.7) (by the property of eigenvectors). Thus, transforming the coordinates using the formulas:

,

in the new basis we obtain the canonical form of a quadratic form with coefficients equal to the eigenvalues λ 1, λ 2, λ 3:

Remark 1. From a geometric point of view, the considered coordinate transformation is a rotation of the coordinate system, combining the old coordinate axes with the new ones.

Remark 2. If any eigenvalues ​​of the matrix (10.3) coincide, we can add a unit vector orthogonal to each of them to the corresponding orthonormal eigenvectors, and thus construct a basis in which the quadratic form takes the canonical form.

Let us bring the quadratic form to canonical form

x² + 5 y² + z² + 2 xy + 6xz + 2yz.

Its matrix has the form In the example discussed in Lecture 9, the eigenvalues ​​and orthonormal eigenvectors of this matrix are found:

Let's create a transition matrix to the basis from these vectors:

(the order of the vectors is changed so that they form a right-handed triple). Let's transform the coordinates using the formulas:

.

So, the quadratic form is reduced to canonical form with coefficients equal to the eigenvalues ​​of the matrix of the quadratic form.

Lecture 11.

Second order curves. Ellipse, hyperbola and parabola, their properties and canonical equations. Reducing a second order equation to canonical form.

Definition 11.1.Second order curves on a plane are called the lines of intersection of a circular cone with planes that do not pass through its vertex.

If such a plane intersects all the generatrices of one cavity of the cone, then in the section it turns out ellipse, at the intersection of the generatrices of both cavities – hyperbola, and if the cutting plane is parallel to any generatrix, then the section of the cone is parabola.

Comment. All second-order curves are specified by second-degree equations in two variables.

Ellipse.

Definition 11.2.Ellipse is the set of points in the plane for which the sum of the distances to two fixed points is F 1 and F tricks, is a constant value.

Comment. When the points coincide F 1 and F 2 the ellipse turns into a circle.

Let us derive the equation of the ellipse by choosing the Cartesian system

y M(x,y) coordinates so that the axis Oh coincided with a straight line F 1 F 2, beginning

r 1 r 2 coordinates – with the middle of the segment F 1 F 2. Let the length of this

segment is equal to 2 With, then in the chosen coordinate system

F 1 O F 2 x F 1 (-c, 0), F 2 (c, 0). Let the point M(x, y) lies on the ellipse, and

the sum of the distances from it to F 1 and F 2 equals 2 A.

Then r 1 + r 2 = 2a, But ,

therefore, introducing the notation b² = a²- c² and after carrying out simple algebraic transformations, we obtain canonical ellipse equation: (11.1)

Definition 11.3.Eccentricity of an ellipse is called the magnitude e=s/a (11.2)

Definition 11.4.Headmistress D i ellipse corresponding to the focus F i F i relative to the axis OU perpendicular to the axis Oh on distance a/e from the origin.

Comment. With a different choice of coordinate system, the ellipse can be specified not by the canonical equation (11.1), but by a second-degree equation of a different type.

Ellipse properties:

1) An ellipse has two mutually perpendicular axes of symmetry (the main axes of the ellipse) and a center of symmetry (the center of the ellipse). If an ellipse is given by a canonical equation, then its main axes are the coordinate axes, and its center is the origin. Since the lengths of the segments formed by the intersection of the ellipse with the main axes are equal to 2 A and 2 b (2a>2b), then the main axis passing through the foci is called the major axis of the ellipse, and the second main axis is called the minor axis.

2) The entire ellipse is contained within the rectangle

3) Ellipse eccentricity e< 1.

Really,

4) The directrixes of the ellipse are located outside the ellipse (since the distance from the center of the ellipse to the directrix is a/e, A e<1, следовательно, a/e>a, and the entire ellipse lies in a rectangle)

5) Distance ratio r i from ellipse point to focus F i to the distance d i from this point to the directrix corresponding to the focus is equal to the eccentricity of the ellipse.

Proof.

Distances from point M(x, y) up to the foci of the ellipse can be represented as follows:

Let's create the directrix equations:

(D 1), (D 2). Then From here r i / d i = e, which was what needed to be proven.

Hyperbola.

Definition 11.5.Hyperbole is the set of points in the plane for which the modulus of the difference in distances to two fixed points is F 1 and F 2 of this plane, called tricks, is a constant value.

Let us derive the canonical equation of a hyperbola by analogy with the derivation of the equation of an ellipse, using the same notation.

|r 1 - r 2 | = 2a, from where If we denote b² = c² - a², from here you can get

- canonical hyperbola equation. (11.3)

Definition 11.6.Eccentricity a hyperbola is called a quantity e = c/a.

Definition 11.7.Headmistress D i hyperbola corresponding to the focus F i, is called a straight line located in the same half-plane with F i relative to the axis OU perpendicular to the axis Oh on distance a/e from the origin.

Properties of a hyperbola:

1) A hyperbola has two axes of symmetry (the main axes of the hyperbola) and a center of symmetry (the center of the hyperbola). In this case, one of these axes intersects with the hyperbola at two points, called the vertices of the hyperbola. It is called the real axis of the hyperbola (axis Oh for the canonical choice of the coordinate system). The other axis has no common points with the hyperbola and is called its imaginary axis (in canonical coordinates - the axis OU). On both sides of it are the right and left branches of the hyperbola. The foci of a hyperbola are located on its real axis.

2) The branches of the hyperbola have two asymptotes, determined by the equations

3) Along with hyperbola (11.3), we can consider the so-called conjugate hyperbola, defined by the canonical equation

for which the real and imaginary axis are swapped while maintaining the same asymptotes.

4) Eccentricity of the hyperbola e> 1.

5) Distance ratio r i from hyperbola point to focus F i to the distance d i from this point to the directrix corresponding to the focus is equal to the eccentricity of the hyperbola.

The proof can be carried out in the same way as for the ellipse.

Parabola.

Definition 11.8.Parabola is the set of points on the plane for which the distance to some fixed point is F this plane is equal to the distance to some fixed straight line. Dot F called focus parabolas, and the straight line is its headmistress.

To derive the parabola equation, we choose the Cartesian

coordinate system so that its origin is the middle

D M(x,y) perpendicular FD, omitted from focus on the directive

r su, and the coordinate axes were located parallel and

perpendicular to the director. Let the length of the segment FD

D O F x is equal to R. Then from the equality r = d follows that

because the

Using algebraic transformations, this equation can be reduced to the form: y² = 2 px, (11.4)

called canonical parabola equation. Magnitude R called parameter parabolas.

Properties of a parabola:

1) A parabola has an axis of symmetry (parabola axis). The point where the parabola intersects the axis is called the vertex of the parabola. If a parabola is given by a canonical equation, then its axis is the axis Oh, and the vertex is the origin of coordinates.

2) The entire parabola is located in the right half-plane of the plane Ooh.

Comment. Using the properties of the directrixes of an ellipse and a hyperbola and the definition of a parabola, we can prove the following statement:

The set of points on the plane for which the relation e the distance to some fixed point to the distance to some straight line is a constant value, it is an ellipse (with e<1), гиперболу (при e>1) or parabola (with e=1).

Related information.

And with the matrix.

This symmetric transformation can be written as:

y 1 = a 11 x 1 + a 12 x 2

y 2 = a 12 x 1 + a 22 x 2

where y 1 and y 2 are the coordinates of the vector in the basis.

Obviously, the quadratic form can be written as:

F(x 1, x 2) = x 1 y 1 + x 2 y 2.

As you can see, the geometric meaning of the numerical value of the quadratic form Ф at a point with coordinates x 1 and x 2 is a scalar product.

If we take another orthonormal basis on the plane, then the quadratic form Ф will look different in it, although its numerical value at each geometric point will not change. If we find a basis in which the quadratic form does not contain coordinates to the first power, but only coordinates in the square, then the quadratic form can be reduced to canonical form.

If we take the set of eigenvectors of a linear transformation as a basis, then in this basis the linear transformation matrix has the form:

When moving to a new basis from the variables x 1 and x 2, we move to the variables and. Then:

The expression is called canonical view quadratic form. Similarly, a quadratic form with a large number of variables can be reduced to canonical form.

The theory of quadratic forms is used to reduce equations of curves and second-order surfaces to canonical form.

Example. Reduce the quadratic form to canonical form

F(x 1, x 2) = 27.

Odds: a 11 = 27, a 12 = 5, and 22 = 3.

Let's create a characteristic equation: ;

(27 - l)(3 - l) - 25 = 0

l 2 - 30l + 56 = 0

l 1 = 2; l 2 = 28;

Example. Bring the second order equation to canonical form:

17x 2 + 12xy + 8y 2 - 20 = 0.

Coefficients a 11 = 17, a 12 = 6, and 22 = 8. A =

Let's create a characteristic equation:

(17 - l)(8 - l) - 36 = 0

136 - 8l - 17l + l 2 - 36 = 0

l 2 - 25l + 100 = 0

l 1 = 5, l 2 = 20.

Total: - canonical equation of an ellipse.

Solution: Let's create a characteristic equation of quadratic form: when

Solving this equation, we get l 1 = 2, l 2 = 6.

Let's find the coordinates of the eigenvectors:

Eigenvectors:

The canonical equation of a line in the new coordinate system will have the form:

Example. Using the theory of quadratic forms, bring the equation of a second-order line to canonical form. Draw a schematic diagram of the graph.

Solution: Let's create a characteristic equation of quadratic form: when

Solving this equation, we get l 1 = 1, l 2 = 11.

Let's find the coordinates of the eigenvectors:

putting m 1 = 1, we get n 1 =

putting m 2 = 1, we get n 2 =

Eigenvectors:

Find the coordinates of the unit vectors of the new basis.

We have the following equation of the line in the new coordinate system:

The canonical equation of a line in the new coordinate system will have the form:

When using the computer version “ Higher mathematics course” it is possible to run a program that solves the above examples for any initial conditions.

To start the program, double-click on the icon:

In the program window that opens, enter the coefficients of the quadratic form and press Enter.

Note: To run the program, the Maple program (Ó Waterloo Maple Inc.) of any version, starting with MapleV Release 4, must be installed on your computer.

Given a quadratic form (2) A(x, x) = , where x = (x 1 , x 2 , …, x n). Consider a quadratic form in space R 3, that is x = (x 1 , x 2 , x 3), A(x, x) =
+
+
+
+
+
+ +
+
+
=
+
+
+ 2
+ 2
+ + 2
(we used the condition of shape symmetry, namely A 12 = A 21 , A 13 = A 31 , A 23 = A 32). Let's write out a matrix of quadratic form A in basis ( e}, A(e) =
. When the basis changes, the matrix of quadratic form changes according to the formula A(f) = C t A(e)C, Where C– transition matrix from the basis ( e) to basis ( f), A C t– transposed matrix C.

Definition11.12. The form of a quadratic form with a diagonal matrix is ​​called canonical.

So let A(f) =
, Then A"(x, x) =
+
+
, Where x" 1 , x" 2 , x" 3 – vector coordinates x in a new basis ( f}.

Definition11.13. Let in n V such a basis is chosen f = {f 1 , f 2 , …, f n), in which the quadratic form has the form

A(x, x) =
+
+ … +
, (3)

Where y 1 , y 2 , …, y n– vector coordinates x in basis ( f). Expression (3) is called canonical view quadratic form. Coefficients  1, λ 2, …, λ n are called canonical; a basis in which a quadratic form has a canonical form is called canonical basis.

Comment. If the quadratic form A(x, x) is reduced to canonical form, then, generally speaking, not all coefficients  i are different from zero. The rank of a quadratic form is equal to the rank of its matrix in any basis.

Let the rank of the quadratic form A(x, x) is equal r, Where r ≤ n. A matrix of quadratic form in canonical form has a diagonal form. A(f) =
, since its rank is equal r, then among the coefficients  i there must be r, not equal to zero. It follows that the number of nonzero canonical coefficients is equal to the rank of the quadratic form.

Comment. A linear transformation of coordinates is a transition from variables x 1 , x 2 , …, x n to variables y 1 , y 2 , …, y n, in which old variables are expressed through new variables with some numerical coefficients.

x 1 = α 11 y 1 + α 12 y 2 + … + α 1 n y n ,

x 2 = α 2 1 y 1 + α 2 2 y 2 + … + α 2 n y n ,

………………………………

x 1 = α n 1 y 1 + α n 2 y 2 + … + α nn y n .

Since each basis transformation corresponds to a non-degenerate linear coordinate transformation, the question of reducing a quadratic form to a canonical form can be solved by choosing the corresponding non-degenerate coordinate transformation.

Theorem 11.2 (main theorem about quadratic forms). Any quadratic form A(x, x), specified in n-dimensional vector space V, using a non-degenerate linear coordinate transformation can be reduced to canonical form.

Proof. (Lagrange method) The idea of ​​this method is to sequentially complement the quadratic trinomial for each variable to a complete square. We will assume that A(x, x) ≠ 0 and in the basis e = {e 1 , e 2 , …, e n) has the form (2):

A(x, x) =
.

If A(x, x) = 0, then ( a ij) = 0, that is, the form is already canonical. Formula A(x, x) can be transformed so that the coefficient a 11 ≠ 0. If a 11 = 0, then the coefficient of the square of another variable is different from zero, then by renumbering the variables it is possible to ensure that a 11 ≠ 0. Renumbering of variables is a non-degenerate linear transformation. If all the coefficients of the squared variables are equal to zero, then the necessary transformations are obtained as follows. Let, for example, a 12 ≠ 0 (A(x, x) ≠ 0, so at least one coefficient a ij≠ 0). Consider the transformation

x 1 = y 1 – y 2 ,

x 2 = y 1 + y 2 ,

x i = y i, at i = 3, 4, …, n.

This transformation is non-degenerate, since the determinant of its matrix is ​​non-zero
= = 2 ≠ 0.

Then 2 a 12 x 1 x 2 = 2 a 12 (y 1 – y 2)(y 1 + y 2) = 2
– 2
, that is, in the form A(x, x) squares of two variables will appear at once.

A(x, x) =
+ 2
+ 2
+
. (4)

Let's convert the allocated amount to the form:

A(x, x) = a 11
, (5)

while the coefficients a ij change to . Consider the non-degenerate transformation

y 1 = x 1 + + … + ,

y 2 = x 2 ,

y n = x n .

Then we get

A(x, x) =
. (6).

If the quadratic form
= 0, then the question of casting A(x, x) to canonical form is resolved.

If this form is not equal to zero, then we repeat the reasoning, considering coordinate transformations y 2 , …, y n and without changing the coordinate y 1 . It is obvious that these transformations will be non-degenerate. In a finite number of steps, the quadratic form A(x, x) will be reduced to canonical form (3).

Comment 1. The required transformation of the original coordinates x 1 , x 2 , …, x n can be obtained by multiplying the non-degenerate transformations found in the process of reasoning: [ x] = A[y], [y] = B[z], [z] = C[t], Then [ x] = AB[z] = ABC[t], that is [ x] = M[t], Where M = ABC.

Comment 2. Let A(x, x) = A(x, x) =
+
+ …+
, where  i ≠ 0, i = 1, 2, …, r, and  1 > 0, λ 2 > 0, …, λ q > 0, λ q +1 < 0, …, λ r < 0.

Consider the non-degenerate transformation

y 1 = z 1 , y 2 = z 2 , …, y q = z q , y q +1 =
z q +1 , …, y r = z r , y r +1 = z r +1 , …, y n = z n. As a result A(x, x) will take the form: A(x, x) = + + … + – – … – which is called normal form of quadratic form.

Example11.1. Reduce the quadratic form to canonical form A(x, x) = 2x 1 x 2 – 6x 2 x 3 + 2x 3 x 1 .

Solution. Because the a 11 = 0, use the transformation

x 1 = y 1 – y 2 ,

x 2 = y 1 + y 2 ,

x 3 = y 3 .

This transformation has a matrix A =
, that is [ x] = A[y] we get A(x, x) = 2(y 1 – y 2)(y 1 + y 2) – 6(y 1 + y 2)y 3 + 2y 3 (y 1 – y 2) =

2– 2– 6y 1 y 3 – 6y 2 y 3 + 2y 3 y 1 – 2y 3 y 2 = 2– 2– 4y 1 y 3 – 8y 3 y 2 .

Since the coefficient at is not equal to zero, we can select the square of one unknown, let it be y 1 . Let us select all terms containing y 1 .

A(x, x) = 2(– 2y 1 y 3) – 2– 8y 3 y 2 = 2(– 2y 1 y 3 + ) – 2– 2– 8y 3 y 2 = 2(y 1 – y 3) 2 – 2– 2– 8y 3 y 2 .

Let us perform a transformation whose matrix is ​​equal to B.

z 1 = y 1 – y 3 ,  y 1 = z 1 + z 3 ,

z 2 = y 2 ,  y 2 = z 2 ,

z 3 = y 3 ;  y 3 = z 3 .

B =
, [y] = B[z].

We get A(x, x) = 2– 2–– 8z 2 z 3. Let us select the terms containing z 2. We have A(x, x) = 2– 2(+ 4z 2 z 3) – 2= 2– 2(+ 4z 2 z 3 + 4) + + 8 – 2 = 2– 2(z 2 + 2z 3) 2 + 6.

Performing a transformation with a matrix C:

t 1 = z 1 ,  z 1 = t 1 ,

t 2 = z 2 + 2z 3 ,  z 2 = t 2 – 2t 3 ,

t 3 = z 3 ;  z 3 = t 3 .

C =
, [z] = C[t].

Got: A(x, x) = 2– 2+ 6canonical form of a quadratic form, with [ x] = A[y], [y] = B[z], [z] = C[t], from here [ x] = ABC[t];

ABC =


=
. The conversion formulas are as follows

x 1 = t 1 – t 2 + t 3 ,

x 2 = t 1 + t 2 – t 3 ,

This method consists of sequentially selecting complete squares in quadratic form.

Let the quadratic form be given

Recall that, due to the symmetry of the matrix

,

There are two possible cases:

1. At least one of the coefficients of squares is different from zero. Without loss of generality, we will assume (this can always be achieved by appropriate renumbering of variables);

2. All coefficients

but there is a coefficient different from zero (for definiteness, let it be).

In the first case transform the quadratic form as follows:

,

and all other terms are denoted by.

is a quadratic form of (n-1) variables.

They treat her in the same way and so on.

notice, that

Second case substitution of variables

comes down to the first.

Example 1: Reduce the quadratic form to canonical form through a non-degenerate linear transformation.

Solution. Let's collect all the terms containing the unknown , and add them to a complete square

.

(Because .)

or

(3)

or


(4)

and from unknown
form will take the form. Next we assume

or

and from unknown
form will take the canonical form

Let us resolve equalities (3) with respect to
:

or

Sequential execution of linear transformations
And
, Where

,

has a matrix

Linear transformation of unknowns
gives a quadratic form to the canonical form (4). Variables
associated with new variables
relations

We got acquainted with LU decomposition in workshop 2_1

Let's remember the statements from workshop 2_1

Statements(see L.5, p. 176)

This script is designed to understand the role of LU in the Lagrange method; you need to work with it in the EDITOR notepad using the F9 button.

And in the tasks attached below, it is better to create your own M-functions that help calculate and understand linear algebra problems (within the framework of this work)

Ax=X."*A*X % we get the quadratic form

Ax=simple(Ax) % simplify it

4*x1^2 - 4*x1*x2 + 4*x1*x3 + x2^2 - 3*x2*x3 + x3^2

% find the LU decomposition without rearranging the rows of the matrix A

% When converting a matrix to echelon form

%without row permutations, we get a matrix of M1 and U3

% U is obtained from A U3=M1*A,

% with this matrix of elementary transformations

0.5000 1.0000 0

0.5000 0 1.0000

%we get U3=M1*A, where

4.0000 -2.0000 2.0000

% from M1 it is easy to obtain L1 by changing the signs

% in the first column in all rows except the first.

0.5000 1.0000 0

0.5000 0 1.0000

% L1 is such that

A_=L1*U % this is the LU decomposition we need

% Elements on the main diagonal U -

% are coefficients of squares y i ^2

% in converted quadratic form

% in our case, there is only one coefficient

% means that in the new coordinates there will only be 4y 1 2 squared,

% for the remaining 0y 2 2 and 0y 3 2 coefficients are equal to zero

% columns of matrix L1 are the decomposition of Y by X

% in the first column we see y1=x1-0.5x2+0.5x3

% for the second we see y2=x2; according to the third y3=x3.

% if L1 is transposed,

% that is T=L1."

% T - transition matrix from (X) to (Y): Y=TX

0.5000 1.0000 0

1.0000 -0.5000 0.5000

% A2 – matrix of transformed quadratic form

% Note U=A2*L1." and A=L1* A2*L1."

4.0000 -2.0000 2.0000

1.0000 -0.5000 0.5000

% So, we got the decomposition A_=L1* A2*L1." or A_=T."* A2*T

% showing change of variables

% y1=x1-0.5x2+0.5x3

% and representation of quadratic form in new coordinates

A_=T."*A2*T % T=L1." transition matrix from (X) to (Y): Y=TX

isequal(A,A_) % must match the original A

4.0000 -2.0000 2.0000

2.0000 1.0000 -1.5000

2.0000 -1.5000 1.0000

Q1=inv(T) % find the transition matrix from (Y) to (X)

% Let's find the transformation,

% quadratic Ax=X."*A*X

% to the new type Ay=(Q1Y)."*A*Q1Y=Y." (Q1."*A*Q1)*Y=Y." (U)*Y

Ay =4*y1^2 - y2*y3

x1 - x2/2 + x3/2

% second transformation matrix,

% which is much simpler to compose.

4*z1^2 - z2^2 + z3^2

% R=Q1*Q2, X=R*Z

R=Q1*Q2 % non-degenerate linear transformation

% bringing the operator matrix to canonical form.

det(R) % determinant is not equal to zero - the transformation is non-degenerate

4*z1^2 - z2^2 + z3^2 ok

4*z1^2 - z2^2 + z3^2

Let us formulate an algorithm for reducing quads ratical form to canonical form by orthogonal transformation: