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 the matrix 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, with the help of a non-degenerate linear transformation of coordinates 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 ,

Introduction

quadratic form canonical form equation

Initially, the theory of quadratic forms was used to study curves and surfaces defined by second-order equations containing two or three variables. Later, this theory found other applications. In particular, when mathematical modeling 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 the quadratic form are not always real numbers.

The theory of quadratic forms was first developed by the French mathematician Lagrange, who owned many ideas in this theory; in particular, he introduced the important concept of a 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 of number theory that eluded his predecessors in this field.

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

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

Reducing a quadratic form to canonical form

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

that in the new coordinates the equation of our curve will have a “canonical” form

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

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 of 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 quadratic form, we introduce the following notation for the coefficients of this form: the coefficient for is denoted by, and the coefficient of the product for is denoted by (compare with (1)!).

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

The term can now be written in the form

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

in particular, when we get the term

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

If, in particular, i.e. If the matrix is ​​non-degenerate, then the quadratic form is called non-degenerate. In view of equality (4), the elements of matrix A, symmetrical 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 specify a well-defined quadratic form (5) of the unknowns, which has elements of matrix A with its coefficients.

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

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

In fact, if the product AB is defined, then the product will also be defined, as is easy to check: the number of columns of the matrix is ​​equal to the number of rows of the matrix. The matrix element located in its th row and th column is located in the AB matrix in the th row and th 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. 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 matrix A then and only then will be symmetric if it coincides with its transpose, i.e. If

Let us now denote by a column composed of unknowns.

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

Composed of one line.

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 on the left by a matrix, we get a “matrix” consisting of one row and one column, namely the right side of equality (5).

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

From here by (6)

Substituting (9) and (10) into entry (7) of the form, we obtain:

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

Thus, the following theorem is proven:

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

Let us now assume that we are performing a non-degenerate linear transformation, i.e. , and therefore and are non-singular matrices. The product is obtained in this case by multiplying the matrix by non-singular matrices and therefore, the rank of this product is equal to the rank of the matrix. Thus, the rank of the 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 a second-order central curve to the canonical form (3), the question of reducing an arbitrary quadratic form by some non-degenerate linear transformation to the form of a sum of squares of unknowns, i.e. to such a form when all coefficients in the products of various unknowns are equal to zero; this special kind of quadratic form is called canonical. Let us first assume 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 odds 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.

In fact, 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 that this matrix have rank is equivalent to requiring that its main diagonal contains exactly zero elements.

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

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

This theorem is true for the case of quadratic forms in one unknown, since every 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, considering it already proven for forms with a smaller number of unknowns.

Empty given quadratic form

from n unknowns. We will try to find a non-degenerate linear transformation that would separate the square of one of the unknowns, i.e. would lead to the form of the sum of this square and some quadratic form of the remaining unknowns. This goal is easily achieved if among the coefficients in the form matrix on the main diagonal there are non-zero coefficients, i.e. if (12) contains the square of at least one of the unknowns 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 the 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 will now be a quadratic form about the unknowns. Expression (14) is the desired expression for the form, since it is obtained from (12) by a non-degenerate linear transformation, namely the transformation inverse to the linear transformation (13), which has as its determinant and is therefore not degenerate.

If there are equalities, then we first need to perform an auxiliary linear transformation, leading to the appearance of squares of unknowns in our form. Since among the coefficients in the entry (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 us now perform a linear transformation

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

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

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

To complete the proof, it remains to note that the quadratic form depends on less than the number of unknowns and therefore, by the induction hypothesis, is reduced to a canonical form by some non-degenerate transformation of the unknowns. This transformation, considered as a (non-degenerate, as is easy to see) transformation of all unknowns, which remains unchanged, leads, therefore, to (14) in 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 a 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; in fact, 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 its canonical form. It is only necessary, instead of induction, which we used in the proof, to consistently isolate the squares of the unknowns using the method outlined above.

Example 1. Reduce a quadratic form to canonical form

Due to the absence of squared unknowns in this form, we first perform a non-degenerate linear transformation

with matrix

after which we get:

Now the coefficients for are different from zero, and therefore from our form we can isolate the square of one unknown. Believing

those. performing a linear transformation for which the inverse will have a matrix

we will bring to mind

So far, only the square of the unknown has been isolated, since the form still contains the product of two other unknowns. Using the inequality of the coefficient at to zero, we will once again apply the method outlined above. Performing a linear transformation

for which the inverse has the matrix

we will finally bring the form to the canonical form

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

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

turns (16) into (17).

The theory of reducing a quadratic form to 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. In fact, our theory allows the use of any non-degenerate linear transformations, while bringing a second-order curve to its canonical form is achieved by using linear transformations of a very special type,

being the rotation 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.

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 seen, geometric meaning numerical value of the quadratic form Ф 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 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 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.

Reduction of quadratic forms

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

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

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

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

Point 1 (preparatory).

1.1. Let us select among the variables one that is included in the quadratic form squared and to the first power at the same time (let’s 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 select a pair of variables whose product is included in the form with a non-zero coefficient and move on to step 3.

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

Point 2 (selecting a complete square).

2.1. Using the leading variable, we select a complete square. Without loss of generality, assume that the leading variable is . Grouping the terms containing , we get

Isolating a complete square with respect to the variable in , we obtain

Thus, as a result of isolating the complete square with a variable, we obtain the sum of the square of the linear form

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

we get a matrix

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

We will do the same with the quadratic form as in point 1.

2.1. If the leading variable is the variable , then you can do it in two ways: either select a complete square for this variable, or perform renaming (renumbering) variables:

with a non-singular transformation matrix:

Point 3 (creating a leading variable). We replace the selected pair of variables with the sum and difference of two new variables, and replace the remaining old variables with the corresponding new variables. If, for example, in paragraph 1 the term was highlighted

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 replacement:

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

As a result of the above algorithm (sequential 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 a complete square, renaming and creating a leading variable), we used elementary non-singular matrices of three types (they are matrices of transition from basis to basis). The required matrix of the non-singular linear transformation (10.4), under which the form (10.1) has the canonical form (10.6), is obtained by multiplying a finite number of elementary non-singular matrices of three types. ■

Example 10.2. Give quadratic form

to canonical form by the Lagrange method. Indicate the corresponding non-singular linear transformation. Perform check.

Solution. Let's choose the leading variable (coefficient). Grouping the terms containing , and selecting a complete square from it, we obtain

where indicated

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

Expressing the old variables in terms of the new ones:

we get a matrix

Let us calculate the matrix of the non-singular linear transformation (10.4). Given the equalities

we find that the matrix has the form

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

Let us verify the validity of equality (10.5).