Theorem on the matrix of a quadratic form. Positive definite quadratic forms

Definition.Quadratic Form, corresponding to the symmetric bilinear form on the linear space V , is called a function of one vector argument .

Let a quadratic form , be a symmetric bilinear form corresponding to it. Then

whence it follows that the symmetric bilinear form corresponding to a quadratic form is also uniquely determined. So, between symmetric bilinear and quadratic forms on a linear space V a one-to-one correspondence is established, so quadratic forms can be studied using symmetric bilinear forms.

Consider n-dimensional linear space. Matrix of quadratic form in a given basis of a linear space is called a matrix of the corresponding symmetric bilinear form in the same basis. A quadratic matrix is always symmetric.

Denote the matrix of the quadratic form in some space basis . If, as usual, we denote X the coordinate column of the vector in the same basis, then from equality 5.5 we obtain the matrix form of the quadratic form:

Theorem 5.4. Let two bases be given in a linear space

(5.10)

, (5.11)

and let and be quadratic matrices in bases (5.10) and (5.11), respectively. Then where T is the transition matrix from (5.10) to (5.11).

The proof follows from Theorem 5.2 and the definition of a matrix of a quadratic form.

Due to the fact that the transition matrix T is non-degenerate, then the rank of the matrix of the quadratic form does not change when passing to a new basis. Therefore, we can formulate the following definition.

Definition. rank of a quadratic form defined on a linear space is called the rank of its matrix in some, and hence in any basis of the space (denoted by ).

Now we write the quadratic form in coordinate form. To do this, we expand the vector in terms of the basis (5.10): . If is a matrix of a quadratic form in the same basis, then, in accordance with equality (5.4), we have

– (5.12)

coordinate form of a quadratic form. Let us write (5.12) in detail for n= 3, given that

So, if a basis is given in, then the quadratic form in coordinate notation looks like a homogeneous polynomial of the second degree in n variables – vector coordinates in the given basis. This polynomial is called view quadratic form in a given basis. But in applications, such polynomials often arise independently, without visible connection with linear spaces (for example, the second differentials of functions), so we formulate one more definition of a quadratic form.

Definition. quadratic form from n variables is a homogeneous second-degree polynomial in these variables, i.e., a function of the form (5.12). A matrix of a quadratic form (5.12) is a symmetric matrix.

Example compiling a matrix of a quadratic form. Let be

It can be seen from (5.12) and (5.13) that the coefficient of at coincides with , i.e., the diagonal elements of the matrix of the quadratic form are the coefficients of the squares. In the same way, we see that is half the coefficient of the product. Thus, the quadratic form matrix (5.14) looks like this:

We now choose in space again two bases (5.10) and (5.11) and denote, as usual, are the coordinate columns of the vector in bases (5.10) and (5.11), respectively. When passing from the basis (5.10) to the basis (5.11), the coordinates of the vector change according to the law:

where is the transition matrix from (5.10) to (5.11). Note that the matrix is nondegenerate. We write equality (5.15) in coordinate form:

or in detail:

(5.17)

With the help of equality (5.17) (or (5.16), which is the same), we pass from variables to variables .

Definition. Linear non-degenerate transformation of variables is a transformation of variables defined by a system of equalities (5.16) or (5.17), or a single matrix equality (5.15), provided that is a nonsingular matrix. Matrix T is called the matrix of this transformation of variables.

If in (5.12) instead of variables we substitute their expressions in terms of variables according to formulas (5.17), open brackets and give similar ones, then we get another homogeneous polynomial of the second degree:

In this case, the linear non-degenerate transformation of variables (5.17) is said to take the quadratic form to the quadratic form . The values of the variables and related by relation (5.15) (or relations (5.16) or (5.17)) will be called relevant for a given linear nondegenerate transformation of variables.

Definition. The set of variables is called non-trivial , if the value of at least one of the variables in it is nonzero. Otherwise, the set of variables is called trivial .

Lemma 5.2. Under a linear nondegenerate transformation of variables, a trivial set of variables corresponds to a trivial set.

It obviously follows from equality (5.15): if , then and . On the other hand, using the nonsingularity of the matrix T, again from (5.15) we obtain , whence it is clear that for , also .◄

Consequence. Under a linear nondegenerate transformation of variables, a nontrivial set of variables corresponds to a nontrivial set.

Theorem 5.5. If the linear non-degenerate transformation (5.15) takes the quadratic form with matrix BUT into a quadratic form with matrix BUT", then (another formulation of Theorem 5.4).

Consequence. Under a linear non-degenerate transformation of variables, the determinant of a matrix of a quadratic form does not change sign.

Comment. Unlike the transition matrix and the matrix of a linear operator, the matrix of a linear nondegenerate transformation of variables is written not by columns, but by rows.

Let two linear nondegenerate transformations of variables be given:

Let's apply them in sequence:

Composition of linear non-degenerate transformations of variables(5.18) and (5.19) is their sequential application, i.e., transformation of variables From (5.20) it is clear that the composition of two linear non-degenerate transformations of variables is also a linear non-degenerate transformation of variables.

Definition. The quadratic forms are called equivalent , if there exists a linear non-degenerate transformation of variables that transforms one of them into another.

Quadratic forms

quadratic form f(x 1, x 2,..., x n) of n variables is called the 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 the quadratic form matrix. It's always symmetrical matrix (i.e., a matrix symmetric about the main diagonal, a ij = a ji).

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

Indeed

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

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

Let the matrix-column of variables X be obtained by a nondegenerate linear transformation of the matrix-column Y, i.e. X = CY, where C is a non-degenerate matrix of order n. Then the quadratic form
f(X) \u003d X T AX \u003d (CY) T A (CY) \u003d (Y T C T) A (CY) \u003d Y T (C T AC) Y.

Thus, under a non-degenerate linear transformation C, the matrix of the 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 a linear transformation.

The quadratic form is called canonical(It 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(the proof is not given here). Any quadratic form can be reduced to a canonical form using a non-degenerate linear transformation.

For example, let us reduce to the canonical form the quadratic form
f (x 1, x 2, x 3) \u003d 2x 1 2 + 4x 1 x 2 - 3x 2 2 - x 2 x 3.

To do this, first select the full square for the variable x 1:

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

Now we select the full square for the variable x 2:

f (x 1, x 2, x 3) \u003d 2 (x 1 + x 2) 2 - 5 (x 2 2 - 2 * x 2 * (1/10) x 3 + (1/100) x 3 2) - (5/100) x 3 2 =
\u003d 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 \u003d x 1 + x 2, y 2 \u003d x 2 - (1/10) x 3 and y 3 \u003d 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 defined ambiguously (the same quadratic form can be reduced to the canonical form in different ways). 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 how the form is reduced to this form (for example, in the considered example there will always be two negative and one positive coefficient). This property is called the law of inertia of quadratic forms.

Let us verify this by reducing the same quadratic form to the canonical form in a different way. Let's start the transformation with the variable x 2:
f (x 1, x 2, x 3) \u003d 2x 1 2 + 4x 1 x 2 - 3x 2 2 - x 2 x 3 \u003d -3x 2 2 - x 2 x 3 + 4x 1 x 2 + 2x 1 2 \u003d - 3(x 2 2 -
- 2 * x 2 ((1/6) x 3 + (2/3) x 1) + ((1/6) x 3 + (2/3) x 1) 2) - 3 ((1/6) x 3 + (2/3) x 1) 2 + 2x 1 2 =
\u003d -3 (x 2 - (1/6) x 3 - (2/3) x 1) 2 - 3 ((1/6) x 3 + (2/3) x 1) 2 + 2x 1 2 \u003d f (y 1, y 2, y 3) = -3y 1 2 -
-3y 2 2 + 2y 3 2, where y 1 \u003d - (2/3) x 1 + x 2 - (1/6) x 3, y 2 \u003d (2/3) x 1 + (1/6) x 3 and y 3 = x 1 . Here, a positive coefficient 2 at y 3 and two negative coefficients (-3) at y 1 and y 2 (and using another method, we got a positive coefficient 2 at y 1 and two negative coefficients - (-5) at y 2 and (-1 /20) for y 3).

It should also be noted that the rank of a matrix of a quadratic form, called the rank of the quadratic form, is equal to the number of non-zero 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).

For example, the quadratic form f 1 (X) \u003d x 1 2 + x 2 2 is positive definite, because is the sum of squares, and the quadratic form f 2 (X) \u003d -x 1 2 + 2x 1 x 2 - x 2 2 is negative definite, because represents it can be represented as f 2 (X) \u003d - (x 1 - x 2) 2.

In most practical situations, it is somewhat more difficult to establish the sign-definiteness of a quadratic form, so one of the following theorems is used for this (we formulate them without proofs).

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

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

Major (corner) minor The k-th order of the matrix A of the n-th 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, we 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 \u003d (6 - 2l - 3l + l 2) - 4 \u003d l 2 - 5l + 2 \u003d 0; D \u003d 25 - 8 \u003d 17;
. Therefore, the quadratic form is positive definite.

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

We examine another quadratic form for sign-definiteness, f (x 1, x 2) \u003d -2x 1 2 + 4x 1 x 2 - 3x 2 2.

Method 1. Let's construct a matrix of quadratic form А = . The characteristic equation will have the form = (-2 - l)*
*(-3 - l) - 4 \u003d (6 + 2l + 3l + l 2) - 4 \u003d l 2 + 5l + 2 \u003d 0; D \u003d 25 - 8 \u003d 17;
. Therefore, the quadratic form is negative definite.

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

In matrix notation, the quadratic form has the form 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 nondegenerate linear transformation of the matrix-column Y, i.e. X = CY, where C is a non-degenerate matrix of order n. 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 the 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 a linear transformation.

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

Its matrix is diagonal.

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

For example, let's bring to the 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 the full square for the variable x 1:

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

Now we select the full square for the variable x 2:

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

Then the non-degenerate linear transformation y 1 \u003d x 1 + x 2, y 2 \u003d x 2 - (1/10) x 3 and y 3 \u003d x 3 brings this quadratic form to the canonical form f (y 1, y 2, y 3) \u003d 2y 1 2 - 5y 2 2 - (1/20)y 3 2 .

Note that the canonical form of a quadratic form is defined ambiguously (the same quadratic form can be reduced to canonical form in different ways1). 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 how the form is reduced to this form (for example, in the considered example there will always be two negative and one positive coefficient). This property is called the law of inertia of quadratic forms.

Let us verify this by reducing the same quadratic form to the canonical form in a different way. Let's start the transformation with the variable x 2: f (x 1, x 2, x 3) \u003d 2x 1 2 + 4x 1 x 2 - 3x 2 2 - x 2 x 3 \u003d -3x 2 2 - x 2 x 3 + 4x 1 x 2 + 2x 1 2 \u003d -3 (x 2 2 - - 2 * x 2 ((1/6) x 3 + (2/3) x 1) + ((1/6) x 3 + (2/3) x 1) 2) - 3 ((1/6) x 3 + (2/3) x 1) 2 + 2x 1 2 = = -3 (x 2 - (1/6) x 3 - (2/3) x 1) 2 - 3 ((1/6) x 3 + (2/3) x 1) 2 + 2x 1 2 \u003d f (y 1, y 2, y 3) \u003d -3y 1 2 - -3y 2 2 + 2y 3 2, where y 1 = - (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 a positive coefficient 2 at y 3 and two negative coefficients (-3) at y 1 and y 2 ).

The quadratic form f(X) is called positively(negative)certain, if for all values of 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's criterion). A quadratic form is positive definite if and only if all principal minors of the matrix of this form are positive.

Major (corner) minor The k-th order of the An-th order matrix 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, we 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. The main minor of the first order of the matrix A  1 = a 11 = 2 > 0. The main minor of the second order  2 = = 6 - 4 = 2 > 0. Therefore, according to the Sylvester criterion, the quadratic form is positive definite.

We examine another quadratic form for sign-definiteness, f (x 1, x 2) \u003d -2x 1 2 + 4x 1 x 2 - 3x 2 2.

Method 1. Let's construct a matrix of quadratic form А = . 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. The main minor of the first order of the matrix A  1 \u003d a 11 \u003d \u003d -2< 0. Главный минор второго порядка 2 = = 6 – 4 = 2 >0. Therefore, according to the Sylvester criterion, the quadratic form is negative definite (the signs of the principal minors alternate, starting from minus).

And as another example, we examine the quadratic form f (x 1, x 2) \u003d 2x 1 2 + 4x 1 x 2 - 3x 2 2 for sign-definiteness.

Method 1. Let's construct a matrix of quadratic form А = . 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. Therefore, a quadratic form cannot be either negative or positive definite, i.e. this quadratic form is not sign-definite (it can take values of any sign).

Method 2. The main minor of the first order of the matrix A  1 = a 11 = 2 > 0. The main 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 the quadratic form to the canonical form is convenient to use when non-zero coefficients occur under the squares of the variables. If they are not there, it is still possible to carry out the conversion, but you have to use some other tricks. 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 =

\u003d (x 1 + x 2) 2 - x 1 2 - x 2 2 \u003d (x 1 + x 2) 2 - (x 1 2 - 2x 1 x 2 + x 2 2) - 2x 1 x 2 \u003d (x 1 + x 2) 2 - - (x 1 - x 2) 2 - 2x 1 x 2; 4x 1 x 2 \u003d (x 1 + x 2) 2 - (x 1 - x 2) 2; f (x 1, x 2) \u003d 2x 1 x 2 \u003d (1/2) * * (x 1 + x 2 ) 2 - (1/2) * (x 1 - x 2) 2 \u003d f (y 1, y 2) \u003d (1/2) y 1 2 - (1/2) y 2 2, where y 1 \u003d x 1 + x 2, ay 2 \u003d x 1 - x 2.

Square shapes.
Significance of forms. Sylvester's criterion

The adjective "square" immediately suggests that something here is connected with a square (second degree), and very soon we will know this "something" and what a form is. Turned out straight away :)

Welcome to my new lesson, and as an immediate warm-up, we will 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 different from zero), and are variables that can take arbitrary values.

* In 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 this case it implies that the polynomial does not have an added constant .

For example: – linear form of two variables

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

Attention! This is a standard entry, and you do not need to change anything in it! Despite the “terrible” look, everything is simple here - double subscripts of constants signal which variables are included in one or another 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 the coefficient, not realizing that it refers to the term:

Sometimes there is a "school" version of the design in the spirit, but then only sometimes. By the way, note that the constants here do not tell us anything at all, and therefore it is more difficult to remember the “easy notation”. Especially when there are more variables.

And the quadratic form of three variables already contains six terms:

... why are "two" multipliers put in the "mixed" terms? This is convenient, and it will soon become clear why.

However, we will write down the general formula, it is convenient to arrange it with a “sheet”:

- carefully study each line - there's nothing wrong with that!

The quadratic form contains terms with squared variables and terms with their pair products (cm. combinatorial formula of combinations ) . Nothing else - no "lonely x" and no added constant (then you get not a quadratic form, but heterogeneous 2nd degree polynomial).

Matrix notation of a quadratic form

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

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

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

It may not be:

– always, unless both are equal to zero.

- for anyone vector except for zero.

And generally speaking, if for any non-zero vector , , then the quadratic form is called positive definite; if - 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 already with a slight complication:
– ?

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

On this account, there theorem: if all eigenvalues matrices of quadratic form are positive * , then it is positively defined. If all are negative, then it is negative.

* It is proved in theory that all eigenvalues of a real symmetric matrix valid

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

We solve the good old quadratic equation :

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

The considered method seems to be working, but there is one big BUT. Already for the “three by three” matrix, looking for eigenvalues is a long and unpleasant task; with a high probability you get a polynomial of the 3rd degree with irrational roots.

How to be? There is an easier way!

Sylvester's criterion

No, not Sylvester Stallone :) First, let me remind you what angular minors matrices. This is determinants which "grow" from its upper left corner:

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

Now, in fact, criterion:

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

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

If at least one angular minor has the opposite sign, then the form sign-alternating. If the angular minors are of “that” sign, but there are zeros among them, then this is a special case, which I will analyze a little later, after we have gone over the more common examples.

Let us analyze the angular minors of the matrix :

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

Conclusion: all angle minors are greater than zero, so the shape positively defined.

Is there a difference with the eigenvalue method? ;)

We write the shape matrix from Example 1:

its first angular minor, and the second , whence it follows that the form is sign-alternating, i.e. depending on the values , can take both positive and negative values. However, this is so obvious.

Take the form and its matrix from Example 2:

here at all without insight not to understand. But with the Sylvester criterion, we don’t care:
, hence the form is definitely not negative.

, and definitely not positive. (because all angle minors must be positive).

Conclusion: the shape is alternating.

Warm-up examples for self-solving:

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 "boundary" cases, namely: if for any non-zero vector , then the shape is defined non-negative, if - then non-positive. These forms have non-zero vectors for which .

Here you can bring such a "button accordion":

Highlighting full square , we immediately see non-negativity form: , moreover, it is equal to zero for any vector with equal coordinates, for example: .

"Mirror" example non-positive 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 reveal the non-negativity or non-positiveness of a form?

For this we need the concept major minors matrices. The main minor is a minor composed of elements that are at the intersection of rows and columns with the same numbers. So, the matrix has two principal minors of the 1st order:
(the element is 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 2nd order minor:
- composed of elements of the 1st, 2nd row and 1st, 2nd column.

Matrix "three by three" There are seven main minors, and here you already have to wave your biceps:
- three minors of the 1st order,
three minors of the 2nd order:
- 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 columns.
Exercise for understanding: write down all the main 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 principal minors non-negative(greater than or equal to zero).

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

2) Nonzero quadratic form with matrix defined non-positive if and only if its:
– principal minors of the 1st order non-positive(less than or equal to zero);
are principal minors of the 2nd order non-negative;
– principal minors of the 3rd order non-positive(alternation has begun);
…
– major minor of the th order non-positive, if is odd or non-negative, if is 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 make a shape matrix, and primarily let's calculate the angular minors - what if it is positively or negatively defined?

The obtained values do not satisfy the Sylvester criterion, however, the second minor not negative, and this makes it necessary to check the 2nd criterion (in the case of the 2nd criterion, it will not be fulfilled automatically, i.e., a conclusion is immediately made about the sign alternation of the form).

Major minors of the 1st order:
- are positive
2nd order major minor:
- not negative.

Thus, ALL major minors are non-negative, so the form non-negative.

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

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

Now, fully armed, we will analyze a more entertaining problem:

Example 5

Examine the quadratic form for sign-definiteness

This form is decorated with the order "alpha", which can be equal to any real number. But it'll only be more fun decide.

First, let's write down the form matrix, probably, many have already adapted to do it orally: on main diagonal we put the coefficients at the squares, and at the symmetrical places - the half coefficients of the corresponding "mixed" products:

Let's calculate the angular minors:

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

Portal for the student. Self-training

Quadratic forms

Square shapes. Significance of forms. Sylvester's criterion

Matrix notation of a quadratic form

Sylvester's criterion

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Square shapes.
Significance of forms. Sylvester's criterion