Quadratic shapes

Quadratic shape f(x 1, x 2,...,x n) of n variables is a sum, each term of which is either the square of one of the variables, or the product of two different variables, taken with a certain coefficient: f(x 1, x 2, ...,x n) = (a ij = a ji).

The matrix A composed of these coefficients is called a matrix of quadratic form. It's always symmetrical matrix (i.e. a matrix symmetrical about the main diagonal, a ij = a ji).

In matrix notation, the quadratic form is f(X) = X T AX, where

Indeed

For example, let's write in matrix form quadratic form.

To do this, we find a matrix of quadratic form. Its diagonal elements are equal to the coefficients of the squared variables, and the remaining elements are equal to the halves of the corresponding coefficients of the quadratic form. That's why

Let the matrix-column of variables X be obtained by a non-degenerate linear transformation of the matrix-column Y, i.e. X = CY, where C is a non-singular matrix of nth order. Then the quadratic form
f(X) = X T AX = (CY) T A(CY) = (Y T C T)A(CY) = Y T (C T AC)Y.

Thus, with a non-degenerate linear transformation C, the matrix of quadratic form takes the form: A * = C T AC.

For example, let's find the quadratic form f(y 1, y 2), obtained from the quadratic form f(x 1, x 2) = 2x 1 2 + 4x 1 x 2 - 3x 2 2 by linear transformation.

The quadratic form is called canonical(has canonical view), if all its coefficients a ij = 0 for i ≠ j, i.e.
f(x 1, x 2,...,x n) = a 11 x 1 2 + a 22 x 2 2 + … + a nn x n 2 = .

Its matrix is diagonal.

Theorem(proof not given here). Any quadratic form can be reduced to canonical form using non-degenerate linear transformation.

For example, let us reduce the quadratic form to canonical form
f(x 1, x 2, x 3) = 2x 1 2 + 4x 1 x 2 - 3x 2 2 – x 2 x 3.

To do this, we first select perfect square with variable x 1:

f(x 1, x 2, x 3) = 2(x 1 2 + 2x 1 x 2 + x 2 2) - 2x 2 2 - 3x 2 2 – x 2 x 3 = 2(x 1 + x 2) 2 - 5x 2 2 – x 2 x 3.

Now we select a complete square with the variable x 2:

f(x 1, x 2, x 3) = 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 =
= 2(x 1 + x 2) 2 – 5(x 2 – (1/10)x 3) 2 - (1/20)x 3 2.

Then the non-degenerate linear transformation y 1 = x 1 + x 2, y 2 = x 2 – (1/10)x 3 and y 3 = x 3 brings this quadratic form to the canonical form f(y 1, y 2, y 3) = 2y 1 2 - 5y 2 2 - (1/20)y 3 2 .

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

Let us verify this by bringing the same quadratic form to canonical form in a different way. Let's start the transformation with the variable x 2:
f(x 1, x 2, x 3) = 2x 1 2 + 4x 1 x 2 - 3x 2 2 – x 2 x 3 = -3x 2 2 – x 2 x 3 + 4x 1 x 2 + 2x 1 2 = - 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 =
= -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 = f (y 1 , y 2 , y 3) = -3y 1 2 -
-3y 2 2 + 2y 3 2, where y 1 = - (2/3)x 1 + x 2 – (1/6) x 3, y 2 = (2/3)x 1 + (1/6) x 3 and y 3 = x 1 . Here there is a positive coefficient of 2 at y 3 and two negative coefficients (-3) at y 1 and y 2 (and using another method we got a positive coefficient of 2 at y 1 and two negative coefficients - (-5) at y 2 and (-1 /20) at y 3).

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

The quadratic form f(X) is called positively (negative) certain, if for all values of the variables that are not simultaneously equal to zero, it is positive, i.e. f(X) > 0 (negative, i.e.
f(X)< 0).

For example, the quadratic form f 1 (X) = x 1 2 + x 2 2 is positive definite, because is a sum of squares, and the quadratic form f 2 (X) = -x 1 2 + 2x 1 x 2 - x 2 2 is negative definite, because represents it can be represented as f 2 (X) = -(x 1 - x 2) 2.

In most practical situations, it is somewhat more difficult to establish the definite sign of a quadratic form, so for this we use one of the following theorems (we will formulate them without proof).

Theorem. A quadratic form is positive (negative) definite if and only if all eigenvalues its matrices are positive (negative).

Theorem (Sylvester criterion). A quadratic form is positive definite if and only if all the leading minors of the matrix of this form are positive.

Main (corner) minor The kth order matrix A of the nth order is called the determinant of the matrix, composed of the first k rows and columns of the matrix A ().

Note that for negative definite quadratic forms the signs of the principal minors alternate, and the first-order minor must be negative.

For example, let us examine the quadratic form f(x 1, x 2) = 2x 1 2 + 4x 1 x 2 + 3x 2 2 for sign definiteness.

= (2 - l)*
*(3 - l) – 4 = (6 - 2l - 3l + l 2) – 4 = l 2 - 5l + 2 = 0; D = 25 – 8 = 17;
. Therefore, the quadratic form is positive definite.

Method 2. Principal minor of the first order of matrix A D 1 = a 11 = 2 > 0. Principal minor of the second order D 2 = = 6 – 4 = 2 > 0. Therefore, according to Sylvester’s criterion, the quadratic form is positive definite.

Let us examine another quadratic form for sign definiteness, f(x 1, x 2) = -2x 1 2 + 4x 1 x 2 - 3x 2 2.

Method 1. Let's construct a matrix of quadratic form A = . Characteristic equation will look like = (-2 - l)*
*(-3 - l) – 4 = (6 + 2l + 3l + l 2) – 4 = l 2 + 5l + 2 = 0; D = 25 – 8 = 17;
. Therefore, the quadratic form is negative definite.

In this paragraph we will focus on a special, but important class positive quadratic forms.

Definition 3. A real quadratic form is called non-negative (non-positive) if, for any real values of the variables

. (35)

In this case, the symmetric matrix of coefficients is called positive semidefinite (negative semidefinite).

Definition 4. A real quadratic form is called positive definite (negative definite) if, for any real values of the variables that are not simultaneously zero,

. (36)

In this case, the matrix is also called positive definite (negative definite).

The class of positive definite (negative definite) forms is part of the class of non-negative (resp. non-positive) forms.

Let a non-negative form be given. Let's imagine it as a sum of independent squares:

. (37)

In this representation, all squares must be positive:

. (38)

Indeed, if there were any , then it would be possible to select values such that

But then, with these values of the variables, the form would have a negative value, which is impossible by condition. Obviously, conversely, from (37) and (38) it follows that the form is positive.

Thus, the non-negative quadratic form is characterized by the equalities.

Let now be a positive definite form. Then it is a non-negative form. Therefore, it can be represented in the form (37), where all are positive. From the positive definiteness of the form it follows that . Indeed, in the case it is possible to select values that are not simultaneously equal to zero, at which all would turn to zero. But then, by virtue of (37), at , which contradicts condition (36).

It is easy to see that conversely, if in (37) and are all positive, then it is a positive definite form.

In other words, a non-negative form is positive definite if and only if it is not singular.

The following theorem gives a criterion for the positive definiteness of a form in the form of inequalities that the form coefficients must satisfy. In this case, the notation already encountered in previous paragraphs for successive principal minors of the matrix is used:

Theorem 3. In order for a quadratic form to be positive definite, it is necessary and sufficient that the inequalities be satisfied

Proof. The sufficiency of conditions (39) follows directly from Jacobi formula (28). The necessity of conditions (39) is established as follows. From the positive definiteness of the form follows the positive definiteness of “truncated” forms

But then all these forms must be non-singular, i.e.

Now we have the opportunity to use the Jacobi formula (28) (at ). Since on the right side of this formula all squares must be positive, then

This implies inequalities (39). The theorem has been proven.

Since any principal minor of a matrix, with proper renumbering of variables, can be placed in the upper left corner, then we have

Consequence. In positive definite quadratic form, all major minors of the coefficient matrix are positive:

Comment. From the non-negativity of successive principal minors

the non-negativity of the form does not follow. Indeed, the form

in which , satisfies the conditions , but is not non-negative.

However, the following holds

Theorem 4. In order for a quadratic form to be non-negative, it is necessary and sufficient that all the major minors of its coefficient matrix are non-negative:

Proof. Let's introduce auxiliary form was non-positive, it is necessary and sufficient for inequalities to occur

Positive definite quadratic forms

Definition. Quadratic form from n unknowns are called positive definite, if its rank is equal to the positive inertia index and equal to the number of unknowns.

Theorem. A quadratic form is positive definite if and only if, on any nonzero set of variable values, it takes positive values.

Proof. Let the quadratic form be a non-degenerate linear transformation of the unknowns

brought back to normal

For any non-zero set of variable values, at least one of the numbers different from zero, i.e. . The necessity of the theorem is proven.

Suppose that the quadratic form takes positive values on any non-zero set of variables, but its positive inertia index is a non-degenerate linear transformation of the unknowns

Let's bring it to normal form. Without loss of generality, we can assume that in this normal form the square of the last variable is either absent or included with a minus sign, i.e. , where or . Let us assume that is a non-zero set of variable values obtained as a result of solving the system linear equations

In this system, the number of equations is equal to the number of variables and the determinant of the system is nonzero. According to Cramer's theorem, the system has the only solution, and it is non-zero. For this set. Contradiction with the condition. We come to a contradiction with the assumption, which proves the sufficiency of the theorem.

Using this criterion, it is impossible to determine from the coefficients whether the quadratic form is positive definite. The answer to this question is given by another theorem, for the formulation of which we introduce another concept. Principal diagonal minors of a matrix– these are minors located in its upper left corner:

, , , … , .

Theorem.A quadratic form is positive definite if and only if all its principal diagonal minors are positive.

Proof we will carry out the method of complete mathematical induction by number n quadratic variables f.

Induction hypothesis. Let us assume that for quadratic forms with fewer variables n the statement is true.

Consider the quadratic form of n variables. Let's put all the terms containing . The remaining terms form a quadratic form of the variables. According to the induction hypothesis, the statement is true for her.

Let us assume that the quadratic form is positive definite. Then the quadratic form is positive definite. If we assume that this is not the case, then there is a non-zero set of variable values , for which and, accordingly, , and this contradicts the fact that the quadratic form is positive definite. By the induction hypothesis, all principal diagonal minors of a quadratic form are positive, i.e. all first principal minors of quadratic form f are positive. Last principal minor of quadratic form – this is the determinant of its matrix. This determinant is positive, since its sign coincides with the sign of its matrix normal looking, i.e. with the sign of the determinant of the identity matrix.

Let all the principal diagonal minors of the quadratic form be positive. Then all the principal diagonal minors of the quadratic form are positive from the equality . By the induction hypothesis, the quadratic form is positive definite, so there is a non-degenerate linear transformation of the variables that reduces the form to the form of the sum of squares of the new variables. This linear transformation can be extended to a non-degenerate linear transformation of all variables by setting . This transformation reduces the quadratic form to the form

The matrix A composed of these coefficients is called a matrix of quadratic form. It's always symmetrical matrix (i.e. a matrix symmetrical about the main diagonal, a ij =a ji).

In matrix notation, the quadratic form is f(X) = X T AX, where

Indeed

For example, let's write the quadratic form in matrix form.

Let the matrix-column of variables X be obtained by a non-degenerate linear transformation of the matrix-column Y, i.e. X = CY, where C is a non-singular matrix of nth order. Then the quadratic form f(X) = X T AX = (CY) T A(CY) = (Y T C T)A(CY) =Y T (C T AC)Y.

Thus, with a non-degenerate linear transformation C, the matrix of quadratic form takes the form: A * =C T AC.

For example, let's find the quadratic form f(y 1, y 2), obtained from the quadratic form f(x 1, x 2) = 2x 1 2 + 4x 1 x 2 - 3x 2 2 by linear transformation.

The quadratic form is called canonical(has canonical view), if all its coefficientsa ij = 0 for i≠j, i.e. f(x 1, x 2,...,x n) = a 11 x 1 2 + a 22 x 2 2 + … + a nn x n 2 = .

Its matrix is diagonal.

Theorem(proof not given here). Any quadratic form can be reduced to canonical form using a non-degenerate linear transformation.

For example, let’s bring to canonical form the quadratic form f(x 1, x 2, x 3) = 2x 1 2 + 4x 1 x 2 - 3x 2 2 – x 2 x 3.

To do this, first select a complete square with the variable x 1:

f(x 1, x 2, x 3) = 2(x 1 2 + 2x 1 x 2 + x 2 2) - 2x 2 2 - 3x 2 2 – x 2 x 3 = 2(x 1 + x 2) 2 - 5x 2 2 – x 2 x 3.

Now we select a complete square with the variable x 2:

f(x 1, x 2, x 3) = 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 = = 2(x 1 + x 2) 2 – 5(x 2 – (1/10)x 3) 2 - (1/20)x 3 2.

Then the non-degenerate linear transformation y 1 = x 1 + x 2,y 2 = x 2 – (1/10)x 3 and y 3 = x 3 brings this quadratic form to the canonical formf(y 1,y 2,y 3) = 2y 1 2 - 5y 2 2 - (1/20)y 3 2 .

Note that the canonical form of a quadratic form is determined ambiguously (the same quadratic form can be reduced to canonical form in different ways 1). However, canonical forms obtained by various methods have a number of common properties. In particular, the number of terms with positive (negative) coefficients of a quadratic form does not depend on the method of reducing the form to this form (for example, in the example considered there will always be two negative and one positive coefficient). This property is called law of inertia of quadratic forms.

Let us verify this by bringing the same quadratic form to canonical form in a different way. Let's start the transformation with the variable x 2:f(x 1, x 2, x 3) = 2x 1 2 + 4x 1 x 2 - 3x 2 2 – x 2 x 3 = -3x 2 2 – x 2 x 3 + 4x 1 x 2 + 2x 1 2 = -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 = = -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 =f(y 1 ,y 2 ,y 3) = -3y 1 2 - -3y 2 2 + 2y 3 2 , where y 1 = - (2/3)x 1 + x 2 – (1/6) x 3 ,y 2 = (2/3)x 1 + (1/6) x 3 and y 3 = x 1 . Here there is a positive coefficient of 2 for y 3 and two negative coefficients (-3) for y 1 and y 2 (and using another method, we got a positive coefficient of 2 for y 1 and two negative ones - (-5) for y 2 and (-1/20) for y 3 ).

It should also be noted that the rank of a matrix of quadratic form, called rank of quadratic form, is equal to the number of nonzero coefficients of the canonical form and does not change under linear transformations.

The quadratic form f(X) is called positively(negative)certain, if for all values of the variables that are not simultaneously equal to zero, it is positive, i.e. f(X) > 0 (negative, i.e. f(X)< 0).

Theorem. A quadratic form is positive (negative) definite if and only if all eigenvalues of its matrix are positive (negative).

Theorem (Sylvester criterion). A quadratic form is positive definite if and only if all the leading minors of the matrix of this form are positive.

Main (corner) minor The k-th order matrices of the An-th order are called the determinant of the matrix, composed of the first k rows and columns of the matrix A ().

Note that for negative definite quadratic forms the signs of the principal minors alternate, and the first-order minor must be negative.

For example, let us examine the quadratic form f(x 1, x 2) = 2x 1 2 + 4x 1 x 2 + 3x 2 2 for sign definiteness.

= (2 -)* *(3 -) – 4 = (6 - 2- 3+ 2) – 4 = 2 - 5+ 2 = 0;D= 25 – 8 = 17; . Therefore, the quadratic form is positive definite.

Method 2. Principal minor of the first order of the matrix A  1 =a 11 = 2 > 0. Principal minor of the second order  2 = = 6 – 4 = 2 > 0. Therefore, according to Sylvester’s criterion, the quadratic form is positive definite.

Let us examine another quadratic form for sign definiteness, f(x 1, x 2) = -2x 1 2 + 4x 1 x 2 - 3x 2 2.

Method 1. Let's construct a matrix of quadratic form A = . The characteristic equation will have the form = (-2 -)* *(-3 -) – 4 = (6 + 2+ 3+ 2) – 4 = 2 + 5+ 2 = 0;D= 25 – 8 = 17 ; . Therefore, the quadratic form is negative definite.

Method 2. Principal minor of the first order of the matrix A  1 =a 11 = = -2< 0. Главный минор второго порядка 2 = = 6 – 4 = 2 >0. Consequently, according to Sylvester’s criterion, the quadratic form is negative definite (the signs of the principal minors alternate, starting with the minus).

And as another example, we examine the sign-determined quadratic form f(x 1, x 2) = 2x 1 2 + 4x 1 x 2 - 3x 2 2.

Method 1. Let's construct a matrix of quadratic form A = . The characteristic equation will have the form = (2 -)* *(-3 -) – 4 = (-6 - 2+ 3+ 2) – 4 = 2 +- 10 = 0;D= 1 + 40 = 41; . One of these numbers is negative and the other is positive. The signs of the eigenvalues are different. Consequently, the quadratic form can be neither negative nor positive definite, i.e. this quadratic form is not sign-definite (it can take values of any sign).

Method 2. Principal minor of the first order of matrix A  1 =a 11 = 2 > 0. Principal minor of the second order 2 = = -6 – 4 = -10< 0. Следовательно, по критерию Сильвестра квадратичная форма не является знакоопределенной (знаки главных миноров разные, при этом первый из них – положителен).

1The considered method of reducing a quadratic form to canonical form is convenient to use when non-zero coefficients are encountered with the squares of variables. If they are not there, it is still possible to carry out the conversion, but you have to use some other techniques. For example, let f(x 1, x 2) = 2x 1 x 2 = x 1 2 + 2x 1 x 2 + x 2 2 - x 1 2 - x 2 2 =

= (x 1 + x 2) 2 - x 1 2 - x 2 2 = (x 1 + x 2) 2 – (x 1 2 - 2x 1 x 2 + x 2 2) - 2x 1 x 2 = (x 1 + x 2) 2 – - (x 1 - x 2) 2 - 2x 1 x 2 ; 4x 1 x 2 = (x 1 + x 2) 2 – (x 1 - x 2) 2 ;f(x 1, x 2) = 2x 1 x 2 = (1/2)* *(x 1 + x 2 ) 2 – (1/2)*(x 1 - x 2) 2 =f(y 1 ,y 2) = (1/2)y 1 2 – (1/2)y 2 2, where y 1 = x 1 + x 2, аy 2 = x 1 – x 2.

Square shapes.
Sign definiteness of forms. Sylvester criterion

The adjective “quadratic” immediately suggests that something here is connected with a square (the second degree), and very soon we will find out this “something” and what the shape is. It turned out to be a tongue twister :)

Welcome to my new lesson, and as an immediate warm-up we'll look at the striped shape linear. Linear form variables called homogeneous 1st degree polynomial:

- some specific numbers * (we assume that at least one of them is non-zero), a are variables that can take arbitrary values.

* Within the framework of this topic we will only consider real numbers .

We have already encountered the term “homogeneous” in the lesson about homogeneous systems of linear equations, and in in this case it implies that the polynomial does not have a plus constant.

For example: – linear form of two variables

Now the shape is quadratic. Quadratic shape variables called homogeneous polynomial of 2nd degree, each term of which contains either the square of the variable or doubles product of variables. So, for example, the quadratic form of two variables has the following form:

Attention! This standard notation, and you don’t need to change anything in it! Despite the “scary” appearance, everything is simple here - double subscripts of constants signal which variables are included in which term:
– this term contains the product and (square);
- here is the work;
- and here is the work.

– I immediately anticipate a gross mistake when they lose the “minus” of a coefficient, not understanding that it refers to a term:

Sometimes there is a “school” design option in the spirit, but only sometimes. By the way, note that the constants don’t tell us anything at all here, and therefore it’s more difficult to remember the “easy notation”. Especially when there are more variables.

And quadratic form of three variables already contains six members:

...why are “two” factors placed in “mixed” terms? This is convenient, and it will soon become clear why.

However general formula Let’s write it down, it’s convenient to arrange it as a “sheet”:

– we carefully study each line – there’s nothing wrong with that!

The quadratic form contains terms with the squares of the variables and terms with their paired products (cm. combinatorial combination formula) . Nothing more - no “lonely X” and no added constant (then you will get not a quadratic form, but heterogeneous polynomial of 2nd degree).

Matrix notation of quadratic form

Depending on the values, the form in question can take on both positive and negative values, and the same applies to any linear form - if at least one of its coefficients is different from zero, then it can be either positive or negative (depending on the values).

This form is called alternating sign. And if with linear form everything is transparent, then with the quadratic form things are much more interesting:

It is absolutely clear that this form can take on the meaning of any sign, thus the quadratic form can also be alternating.

Or maybe not:

– always, unless simultaneously equal to zero.

– for anyone vector except zero.

And in general, if for anyone non-zero vector , , then the quadratic form is called positive definite; if so then negative definite.

And everything would be fine, but the definiteness of the quadratic form is visible only in simple examples, and this visibility is lost even with a slight complication:
– ?

One might assume that the form is positive definite, but is this really so? Suddenly there are values at which it less than zero?

There is a theorem: if ALL eigenvalues matrices of quadratic form are positive * , then it is positive definite. If all are negative, then negative.

* It has been proven in theory that all eigenvalues of a real symmetric matrix valid

Let's write the matrix of the above form:
and from Eq. let's find her eigenvalues:

Let's solve the good old quadratic equation:

, which means the form is defined positively, i.e. for any non-zero values it greater than zero.

The considered method seems to work, but there is one big BUT. Already for a three-by-three matrix, looking for proper numbers is a long and unpleasant task; with a high probability you will get a polynomial of the 3rd degree with irrational roots.

What should I do? There is an easier way!

Sylvester criterion

No, not Sylvester Stallone :) First, let me remind you what it is corner minors matrices. This qualifiers which “grow” from her left top corner:

and the last one is exactly equal to the determinant of the matrix.

Now, actually, criterion:

1) Quadratic form is defined positively if and only if ALL its angular minors are greater than zero: .

2) Quadratic form is defined negative if and only if its angular minors alternate in sign, with the 1st minor being less than zero: , , if – even or , if – odd.

If at least one corner minor opposite sign, then the form alternating sign. If the angular minors are of “that” sign, but there are zero ones among them, then this is special case, which I will discuss a little later, after we click on more common examples.

Let's analyze the angular minors of the matrix :

And this immediately tells us that the form is not negatively defined.

Conclusion: all corner minors are greater than zero, which means the form is defined positively.

There is a difference with the method eigenvalues? ;)

Let us write the form matrix from Example 1:

the first is its angular minor, and the second , from which it follows that the shape is alternating in sign, i.e. depending on the values, it can take both positive and negative values. However, this is already obvious.

Let's take the form and its matrix from Example 2:

There’s no way to figure this out without insight. But with Sylvester’s criterion we don’t care:
, therefore, the form is definitely not negative.

, and definitely not positive (since all angular minors must be positive).

Conclusion: the shape is alternating.

Warm-up examples for independent decision:

Example 4

Investigate quadratic forms for sign definiteness

In these examples everything is smooth (see the end of the lesson), but in fact, to complete such a task Sylvester's criterion may not be sufficient.

The point is that there are “edge” cases, namely: if for any non-zero vector, then the shape is determined non-negative, if – then negative. These forms have non-zero vectors for which .

Here you can quote the following “accordion”:

Highlighting perfect square, we see right away non-negativity form: , and it is equal to zero for any vector with equal coordinates, for example: .

"Mirror" example negative a certain form:

and an even more trivial example:
– here the form is equal to zero for any vector , where is an arbitrary number.

How to identify non-negative or non-positive forms?

For this we need the concept major minors matrices. A major minor is a minor composed of elements that stand at the intersection of rows and columns with the same numbers. Thus, the matrix has two main minors of the 1st order:
(the element is located at the intersection of the 1st row and 1st column);
(the element is at the intersection of the 2nd row and 2nd column),

and one major minor of 2nd order:
– composed of elements of the 1st, 2nd row and 1st, 2nd column.

The matrix is “three by three” There are seven main minors, and here you’ll have to flex your biceps:
– three minors of the 1st order,
three 2nd order minors:
– composed of elements of the 1st, 2nd row and 1st, 2nd column;
– composed of elements of the 1st, 3rd row and 1st, 3rd column;
– composed of elements of the 2nd, 3rd row and 2nd, 3rd column,
and one 3rd order minor:
– composed of elements of the 1st, 2nd, 3rd row and 1st, 2nd and 3rd column.
Exercise for understanding: write down all the major minors of the matrix .
We check at the end of the lesson and continue.

Schwarzenegger criterion:

1) Non-zero* quadratic form defined non-negative if and only if ALL of its major minors non-negative(greater than or equal to zero).

* The zero (degenerate) quadratic form has all coefficients equal to zero.

2) Non-zero quadratic form with matrix is defined negative if and only if:
– major minors of the 1st order non-positive(less than or equal to zero);
– major minors of the 2nd order non-negative;
– major minors of the 3rd order non-positive(alternation began);
…
– major minor of the th order non-positive, if – odd or non-negative, if – even.

If at least one minor is of the opposite sign, then the form is sign-alternating.

Let's see how the criterion works in the above examples:

Let's create a shape matrix, and first of all Let's calculate the angular minors - what if it is defined positively or negatively?

The obtained values do not satisfy the Sylvester criterion, but the second minor not negative, and this makes it necessary to check the 2nd criterion (in the case of the 2nd criterion will not be fulfilled automatically, i.e. the conclusion is immediately drawn that the shape is alternating in sign).

Main minors of the 1st order:
– positive,
major minor of 2nd order:
– not negative.

Thus, ALL major minors are not negative, which means the form non-negative.

Let's write the shape matrix , for which the Sylvester criterion is obviously not satisfied. But we also did not receive opposite signs (since both angular minors are equal to zero). Therefore, we check the fulfillment of the non-negativity/non-positivity criterion. Main minors of the 1st order:
– not positive,
major minor of 2nd order:
– not negative.

Thus, according to Schwarzenegger’s criterion (point 2), the form is non-positively defined.

Now let’s take a closer look at a more interesting problem:

Example 5

Examine the quadratic form for sign definiteness

This form adorns the order "alpha", which can be equal to anyone real number. But it will only be more fun we decide.

First, let’s write down the form matrix; many people have probably already gotten used to doing this orally: on main diagonal We put the coefficients for the squares, and in the symmetrical places we put half the coefficients of the corresponding “mixed” products:

Let's calculate the angular minors:

I will expand the third determinant on the 3rd line: