But with a 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) \u003d x 1 y 1 + x 2 y 2.

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

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

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

When passing to a new basis from the variables x 1 and x 2, we pass 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 the equations of curves and surfaces of the second order to the canonical form.

Example. Convert the quadratic form to canonical form

F (x 1, x 2) \u003d 27.

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

We compose the 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 the canonical form:

17x2 + 12xy + 8y2 - 20 = 0.

Coefficients a 11 \u003d 17, a 12 \u003d 6, and 22 \u003d 8. A \u003d

Let's make the characteristic equation:

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

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

l 2 - 25l + 100 = 0

l 1 \u003d 5, l 2 \u003d 20.

Total: - the canonical equation of the ellipse.

Solution: We compose the characteristic equation of the quadratic form: when

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

Find the coordinates of the eigenvectors:

Eigenvectors:

The canonical equation of the line in the new coordinate system will look like:

Example. Using the theory of quadratic forms, bring the second-order line equation to the canonical form. Schematically represent the graph.

Solution: Let's make a characteristic quadratic form equation: when

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

Find the coordinates of the eigenvectors:

setting m 1 = 1, we get n 1 =

setting m 2 = 1, we get n 2 =

Eigenvectors:

We find the coordinates of the unit vectors of the new basis.

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

The canonical equation of the line in the new coordinate system will look like:

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

Double-click the icon to launch the program:

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

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

This method consists in successive selection of full squares in quadratic form.

Let a quadratic form be given

Recall that, due to the symmetry of the matrix

,

Two cases are possible:

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

2. All coefficients

but there is a non-zero coefficient (for definiteness, let it be).

In the first case we transform the quadratic form as follows:

,

and all other terms are denoted by.

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

She is treated in the same way, and so on.

notice, that

Second case change of variables

comes down to the first.

Example 1: Convert a quadratic form to a canonical form using a non-degenerate linear transformation.

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

.

(Because .)

or

(3)

or


(4)

and from unknown
form will take the form. Next, we set

or

and from unknown
form takes the canonical form

Let us solve 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
ratios

We met with LU - decomposition in the workshop 2_1

Recall the statements from the 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 notebook using the F9 button.

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

Ax=X."*A*X % get a 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 permuting the rows of the matrix A

% When converting a matrix to a stepped form

% without row permutations, we get the matrix M1 and U3

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

% with such a 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

% of M1 is easy to get L1 by changing signs

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

0.5000 1.0000 0

0.5000 0 1.0000

% L1 such that

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

% Elements on the main diagonal U -

% are coefficients at squares y i ^2

% in converted quadratic form

% in our case, there is only one coefficient

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

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

% columns of the matrix L1 is the expansion of Y in X

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

% on the second we see y2=x2; on the third y3=x3.

% if transpose L1,

% i.e. 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

% Notice 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 quadratic form representation 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)

% Find the transformation,

% quadratic Ax=X."*A*X

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

Ay =4*y1^2 - y2*y3

x1 - x2/2 + x3/2

% matrix of the second transformation,

% which is much simpler.

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

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

R=Q1*Q2 % non-degenerate linear transformation

% reducing the operator matrix to the canonical form.

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

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

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

Let us formulate an algorithm for reducing the quad ratic form to the canonical form by an orthogonal transformation:

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 us write out the matrix of the quadratic form A in basis ( e}, A(e) =
. When changing the basis, the matrix of the quadratic form changes according to the formula A(f) = C t A(e)C, where C is the transition matrix from the basis ( e) to the basis ( f), a C t is the transposed matrix C.

Definition11.12. The kind of 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 the 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 are vector coordinates x in basis ( f). Expression (3) is called canonical view quadratic form. Coefficients  1 , λ 2 , …, λ n called canonical; the basis in which the 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 to r, where r ≤ n. The matrix of a quadratic form in the canonical form has a diagonal form. A(f) =
, because its rank is r, then among the coefficients  i should be r, not equal to zero. This implies that the number of non-zero 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, where the old variables are expressed in terms of the 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 + a n 2 y 2 + … + α nn y n .

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

Theorem 11.2 (basic theorem on quadratic forms). Any quadratic form A(x, x) specified in n-dimensional vector space V, with the help of a non-degenerate linear transformation of coordinates can be reduced to the canonical form.

Proof. (Lagrange method) The idea of ​​this method is to successively complement the square trinomial in each variable to a full 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 the other variable is different from zero, then by renumbering the variables it is possible to achieve that a 11 ≠ 0. The renumbering of variables is a non-degenerate linear transformation. If all the coefficients of the squares of the 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) there will be squares of two variables at once.

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

Let's transform the allocated sum to the form:

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

while the coefficients a ij change to . Consider a 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 the canonical form is resolved.

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

Comment 1. Necessary transformation of initial 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 a 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 quadratic form.

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

Solution. Insofar as 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, you can select the square of one unknown, let it be y one . 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's 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].

Get A(x, x) = 2– 2–– 8z 2 z 3 . We single out 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 Matrix Transformation 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].

Received: A(x, x) = 2– 2+ 6canonical form of the quadratic form, while [ x] = A[y], [y] = B[z], [z] = C[t], hence [ 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 ,

Introduction

quadratic form canonical form equation

Initially, the theory of quadratic forms was used to study curves and surfaces given by second-order equations containing two or three variables. Later, this theory found other applications. In particular, in the mathematical modeling of economic processes, objective functions may contain quadratic terms. Numerous applications of quadratic forms required the construction of a general theory, when the number of variables is equal to any, and the coefficients of a quadratic form are not always real numbers.

The theory of quadratic forms was first developed by the French mathematician Lagrange, who owns many ideas in this theory, in particular, he introduced the important concept of reduced form, with the help of which he proved the finiteness of the number of classes of binary quadratic forms of a given discriminant. Then this theory was significantly expanded by Gauss, who introduced many new concepts, on the basis of which he was able to obtain proofs of difficult and deep theorems in number theory that eluded his predecessors in this field.

The aim of the work is to study the types of quadratic forms and ways to reduce quadratic forms to the canonical form.

In this work, the following tasks are set: to select the necessary literature, to consider definitions and main theorems, to solve a number of problems on this topic.

Reduction of a quadratic form to a canonical form

The origins of the theory of quadratic forms lie in analytic geometry, namely in the theory of curves (and surfaces) of the second order. It is known that the equation of the central curve of the second order on the plane, after transferring the origin rectangular coordinates to the center of this curve, has the form

that in the new coordinates the equation of our curve will have the "canonical" form

in this equation, the coefficient in the product of the unknowns is, therefore, zero. The transformation of coordinates (2) can obviously be interpreted as a linear transformation of the unknowns, moreover, non-degenerate, since the determinant of its coefficients is equal to one. This transformation is applied to the left side of equation (1), and therefore it can be said that the left side of equation (1) is transformed by a non-degenerate linear transformation (2) into the left side of equation (3).

Numerous applications required the construction of a similar theory for the case when the number of unknowns instead of two is equal to any, and the coefficients are either real or any complex numbers.

Generalizing the expression on the left side of equation (1), we arrive at the following concept.

A quadratic form in unknowns is a sum in which each term is either the square of one of these unknowns or the product of two different unknowns. A quadratic form is called real or complex depending on whether its coefficients are real or can be any complex numbers.

Assuming that the reduction of similar terms has already been done in the quadratic form, we introduce the following notation for the coefficients of this form: we denote the coefficient of at by, and the coefficient of the product of for - by (compare with (1)!).

Since, however, the coefficient of this product could also be denoted by, i.e. the notation introduced by us implies the validity of the equality

The term can now be written in the form

and the entire quadratic form - as a sum of all possible terms, where and independently of each other take values ​​from 1 to:

in particular, for , the term

From the coefficients one can obviously compose a square matrix of order; it is called the matrix of the quadratic form, and its rank is called the rank of this quadratic form.

If, in particular, i.e. matrix is ​​non-degenerate, then the quadratic form is also called non-degenerate. In view of equality (4), the elements of the matrix A, which are symmetric with respect to the main diagonal, are equal to each other, i.e. matrix A is symmetric. Conversely, for any symmetric matrix A of order, one can indicate a well-defined quadratic form (5) in unknowns, which has elements of the matrix A by its coefficients.

The quadratic form (5) can be written in a different form using rectangular matrix multiplication. Let us first agree on the following notation: if a square or generally rectangular matrix A is given, then the matrix obtained from the matrix A by transposition will be denoted by. If the matrices A and B are such that their product is defined, then the equality takes place:

those. the matrix obtained by transposing the product is equal to the product of the matrices obtained by transposing the factors, moreover, taken in reverse order.

Indeed, if the product AB is defined, then the product will be defined, as it is easy to check, and the product: the number of columns of the matrix is ​​equal to the number of rows of the matrix. The element of the matrix, which is in its th row and m column, in the matrix AB is located in the th row and m column. It is therefore equal to the sum of the products of the corresponding elements of the th row of matrix A and the th column of matrix B, i.e. is equal to the sum products of the corresponding elements of the th column of the matrix and the th row of the matrix. This proves equality (6).

Note that the matrix A will be symmetric if and only if it coincides with its transposed one, i.e. if

We now denote by a column composed of unknowns.

is a matrix with rows and one column. Transposing this matrix, we get the matrix

Made up of one line.

The quadratic form (5) with matrix can now be written as the following product:

Indeed, the product will be a matrix consisting of one column:

Multiplying this matrix from the left by a matrix, we get a "matrix" consisting of one row and one column, namely the right side of equality (5).

What happens to a quadratic form if the unknowns included in it are subjected to a linear transformation

Hence by (6)

Substituting (9) and (10) into the record (7) of the form, we get:

Matrix B will be symmetric, because in view of equality (6), which is obviously valid for any number of factors, and equality equivalent to the symmetry of the matrix, we have:

Thus, the following theorem has been proved:

A quadratic form in unknowns with a matrix, after performing a linear transformation of the unknowns with a matrix, turns into a quadratic form in new unknowns, and the product is the matrix of this form.

Let us now assume that we perform a non-degenerate linear transformation, i.e. , and therefore and are nondegenerate matrices. The product is obtained in this case by multiplying the matrix by non-degenerate matrices, and therefore, the rank of this product is equal to the rank of the matrix. Thus, the rank of a quadratic form does not change when performing a non-degenerate linear transformation.

Let us now consider, by analogy with the geometric problem indicated at the beginning of the section of reducing the equation of the central curve of the second order to the canonical form (3), the question of reducing an arbitrary quadratic form by some non-degenerate linear transformation to the form of the sum of squares of unknowns, i.e. to such a form when all the coefficients in the products of various unknowns are equal to zero; this special kind of quadratic form is called canonical. Assume first that the quadratic form in the unknowns has already been reduced by a non-degenerate linear transformation to the canonical form

where are the new unknowns. Some of the coefficients may Of course, be zeros. Let us prove that the number of nonzero coefficients in (11) is necessarily equal to the rank of the form.

Indeed, since we arrived at (11) using a non-degenerate transformation, the quadratic form on the right side of equality (11) must also be of rank.

However, the matrix of this quadratic form has a diagonal form

and requiring this matrix to have a rank is tantamount to assuming that its main diagonal contains exactly non-zero entries.

Let us proceed to the proof of the following main theorem on quadratic forms.

Any quadratic form can be reduced to a canonical form by some non-degenerate linear transformation. If a real quadratic form is considered, then all coefficients of the specified linear transformation can be considered real.

This theorem is true for the case of quadratic forms in one unknown, since any such form has a form that is canonical. We can therefore carry out the proof by induction on the number of unknowns, i.e. prove the theorem for quadratic forms in n unknowns, assuming it has already been proven for forms with fewer unknowns.

Given a quadratic form

from n unknowns. We will try to find such a non-degenerate linear transformation that would single out one of the unknowns from the square, i.e. would lead to the form of the sum of this square and some quadratic form from the remaining unknowns. This goal is easily achieved if among the coefficients of the forms on the main diagonal in the matrix there are non-zero, i.e. if the square of at least one of the unknowns enters (12) with a difference from zero coefficients

Let, for example, . Then, as is easy to check, the expression, which is a quadratic form, contains the same terms with an unknown as our form, and therefore the difference

will be a quadratic form containing only unknowns, but not. From here

If we introduce the notation

then we get

where is now the quadratic form in the unknowns. Expression (14) is the desired expression for the form, since it is obtained from (12) by a non-degenerate linear transformation, namely, by the inverse transformation of linear transformation (13), which has its own determinant and is therefore not degenerate.

If there are equalities, then you first need to perform an auxiliary linear transformation, which leads to the appearance of squares of unknowns in our form. Since among the coefficients in the notation (12) of this form there must be non-zero ones, otherwise there would be nothing to prove, then let, for example, i.e. is the sum of a term and terms, each of which includes at least one of the unknowns.

Let's do a linear transformation

It will be non-degenerate, since it has a determinant

As a result of this transformation, our form member will take the form

those. in the form, with non-zero coefficients, the squares of two unknowns at once will appear, and they cannot cancel with any of the other terms, since each of these last includes at least one of the unknowns; now we are in the conditions of the case already considered above, those. by another non-degenerate linear transformation, we can bring the form to the form (14).

To complete the proof, it remains to note that the quadratic form depends on fewer than the number of unknowns, and therefore, by the inductive assumption, it is reduced to the canonical form by some non-degenerate transformation of the unknowns. This transformation, considered as a (non-degenerate, as it is easy to see) transformation of all unknowns, under which it remains unchanged, consequently reduces (14) to the canonical form. Thus, the quadratic form by two or three non-degenerate linear transformations, which can be replaced by one non-degenerate transformation - their product, is reduced to the form of the sum of squares of unknowns with some coefficients. The number of these squares is equal, as we know, to the rank of the form. If, moreover, the quadratic form is real, then the coefficients both in the canonical form of the form and in the linear transformation leading to this form will be real; indeed, both the linear transformation inverse (13) and the linear transformation (15) have real coefficients.

The proof of the main theorem is complete. The method used in this proof can be applied in specific examples to actually reduce a quadratic form to canonical form. It is only necessary instead of induction, which we used in the proof, to consistently extract the squares of the unknowns using the above method.

Example 1. Canonicalize a quadratic form

In view of the absence of unknown squares in this form, we first perform a non-degenerate linear transformation

with matrix

after which we get:

Now the coefficients at are nonzero, and therefore we can extract the square of one unknown from our form. Assuming

those. doing a linear transformation for which the inverse would have a matrix

we will bring to mind

So far, only the square of the unknown has stood out, since the form still contains the product of two other unknowns. Using the inequality zero of the coefficient at, we once again apply the above method. Making a linear transformation

for which the inverse has the matrix

we will finally bring the form to the canonical form

A linear transformation that reduces (16) immediately to the form (17) will have as its matrix the product

One can also check by direct substitution that the non-degenerate (since the determinant is equal) linear transformation

turns (16) into (17).

The theory of reduction of a quadratic form to a canonical form is constructed by analogy with the geometric theory of central curves of the second order, but cannot be considered a generalization of this latter theory. Indeed, in our theory, any non-degenerate linear transformations are allowed, while the reduction of the second-order curve to the canonical form is achieved by applying linear transformations of a very special form,

which are rotations of the plane. This geometric theory can, however, be generalized to the case of quadratic forms in unknowns with real coefficients. An exposition of this generalization, called the reduction of quadratic forms to the principal axes, will be given below.

Reduction of quadratic forms

Consider the simplest and most often used in practice method of reducing a quadratic form to a canonical form, called Lagrange method. It is based on the selection of a complete square in quadratic form.

Theorem 10.1(Lagrange's theorem). Any quadratic form (10.1):

using a non-singular linear transformation (10.4) can be reduced to the canonical form (10.6):

□ Let us prove the theorem in a constructive way, using the Lagrange method of selecting perfect squares. The problem is to find a non-singular matrix such that the linear transformation (10.4) results in the quadratic form (10.6) of the canonical form. This matrix will be obtained gradually as a product of a finite number of matrices of a special type.

Item 1 (preparatory).

1.1. We single out among the variables one that enters the quadratic form in the square and in the first degree simultaneously (we call it leading variable). Let's move on to point 2.

1.2. If there are no leading variables in the quadratic form (for all : ), then we choose a pair of variables whose product enters the form with a nonzero coefficient and proceed to step 3.

1.3. If there are no products of oppositely named variables in a quadratic form, then the given quadratic form is already represented in the canonical form (10.6). The proof of the theorem is complete.

Point 2 (highlighting the full square).

2.1. Based on the leading variable, we select the full square. Without loss of generality, we assume that the leading variable is the variable . Grouping the terms containing , we obtain

Singling out the full square with respect to the variable in , we obtain

Thus, as a result of selecting the full square for a variable, we obtain the sum of the square of the linear form

which includes the leading variable , and a quadratic form in variables , in which the leading variable is no longer included. Let's make a change of variables (introduce new variables)

we get the matrix

() non-singular linear transformation , as a result of which the quadratic form (10.1) takes the following form

We will proceed with the quadratic form in the same way as in paragraph 1.

2.1. If the leading variable is the variable , then there are two ways to do it: either select the full square for this variable, or execute renaming (renumbering) variables:

with a non-singular transformation matrix:

Point 3 (creation of a leading variable). The selected pair of variables will be replaced by the sum and difference of two new variables, and the rest of the old variables will be replaced by the corresponding new variables. If, for example, in paragraph 1 the term

then the corresponding change of variables has the form

and in quadratic form (10.1) the leading variable will be obtained.

For example, in case of variable substitution:

the matrix of this non-singular linear transformation has the form

As a result of the above algorithm (successive application of points 1, 2, 3), the quadratic form (10.1) will be reduced to the canonical form (10.6).

Note that as a result of the transformations performed on the quadratic form (selecting the full square, renaming and creating the leading variable), we used elementary non-singular matrices of three types (they are transition matrices from basis to basis). The desired matrix of a nonsingular linear transformation (10.4), in which the form (10.1) has the canonical form (10.6), is obtained by multiplying a finite number of elementary nonsingular matrices of three types. ■

Example 10.2. Bring a quadratic form

to the canonical form by the Lagrange method. Specify the corresponding non-singular linear transformation. Run a check.

Solution. We choose the leading variable (coefficient ). Grouping the terms containing , and selecting a full square on it, we get

where indicated

Let's make a change of variables (introduce new variables)

Expressing the old variables in terms of the new ones:

we get the matrix

Let us calculate the matrix of a non-singular linear transformation (10.4). Considering equality

we get that the matrix has the form

Let's check the performed calculations. The matrices of the original quadratic form and the canonical form have the form

Let us verify the validity of equality (10.5).