Consider an arbitrary real quadratic form

Its coefficient matrix is ​​real symmetric. Therefore (see Chapter IX, § 13) it is orthogonally similar to some real diagonal matrix, i.e. there is a real orthogonal matrix such that

Here are the characteristic numbers of the matrix.

Because for orthogonal matrix, then from (41) it follows that the form under the orthogonal transformation of variables

or more detailed record

(42")

goes into shape

. (43)

Theorem 7. Real quadratic form can always be given using orthogonal transformation To canonical form(43); in this case, are the characteristic numbers of the matrix.

Reducing a quadratic form using an orthogonal transformation to the canonical form (43) is called reduction to the principal axes. This name is due to the fact that the equation of the central hypersurface of the second order,

, (44)

with an orthogonal transformation of variables (42) takes the canonical form

. (45)

If we consider them as coordinates in some orthonormal basis of -dimensional Euclidean space, then they will be coordinates in a new orthonormal basis of the same space, and the “rotation” of the axes is carried out by orthogonal transformation (42). The new coordinate axes are the axes of symmetry of the central surface (44) and are usually called the main axes of this surface.

From formula (43) it follows that the rank of the form equal to the number non-zero characteristic numbers of the matrix, and the signature is equal to the difference between the number of positive and the number of negative characteristic numbers of the matrix.

From here, in particular, the following proposal follows:

If, with a continuous change in the coefficients of a quadratic form, its rank remains unchanged, then with this change in the coefficients, its signature also remains unchanged.

In this case, we proceed from the fact that a continuous change in the coefficients entails a continuous change in the characteristic numbers. The signature can change only when some characteristic number changes sign. But then at some intermediate moment the characteristic number in question will go to zero, which entails a change in the rank of the form.. (48)

The theory of reducing a quadratic form to canonical form, set out in the previous paragraph, 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, in our theory the use of any non-degenerate linear transformations is allowed, while bringing a second-order curve to the canonical form is achieved by using linear transformations very special type(2), which are rotations of the plane. This geometric theory can, however, be generalized to the case of quadratic forms in unknowns with real coefficients by requiring that the transformation matrix be orthogonal. This transformation is called orthogonal, and the procedure itself reducing quadratic forms to the principal axes.

THEOREM. Each quadratic form can be reduced to canonical form by some orthogonal transformation.

PROOF. We will look at a matrix of quadratic form as a matrix of some linear operator in Euclidean space. If the matrix is ​​of quadratic form, then it is symmetric of order . If some orthonormal basis of dimensional Euclidean space, then the matrix defines a symmetric operator in this basis. By the main theorem about symmetric operators in Euclidean space, in a suitable orthonormal basis its matrix will be diagonal. Let the transition matrix from to , then .

But the matrix , as a transition matrix from one orthonormal basis to another, according to Theorem 2 §1.6 will be orthogonal, and therefore . That's why . Namely, this is how a matrix of a quadratic form is transformed, subjected to a linear transformation of unknowns with the matrix .

So, the transformation of unknowns having a matrix is ​​orthogonal, and the matrix, being diagonal, corresponds to the quadratic form canonical form. □

The fact that the matrix of a linear operator in a basis composed of eigenvectors, has a diagonal form (with eigenvalues ​​along the main diagonal), gives us a method for practically finding the canonical form of a quadratic form, as well as this orthogonal transformation itself.

Example 2. Find an orthogonal transformation that reduces the quadratic form

to the canonical view and write this canonical view.

Solution. The matrix of this form has the form

,

Let's find her characteristic polynomial:

.

Thus, the matrix has a double root and a simple root. Therefore, the canonical form of this quadratic form will be

.

Let us find an orthogonal transformation that implements this reduction. To do this, we find the eigenvectors corresponding to the found eigenvalues , i.e., we solve systems of linear homogeneous equations for each .

When we have

.

Where , i.e. there are 2 independent variables, and fundamental set the solutions will be:

Applying the orthogonalization process to them, we obtain.

The theory of reducing a quadratic form to canonical form, set out in the previous paragraph, 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 (2), which are rotations of the plane. This geometric theory can, however, be generalized to the case of quadratic forms in unknowns with real coefficients by requiring that the transformation matrix be orthogonal. This transformation is called orthogonal, and the procedure itself reducing quadratic forms to the principal axes.

THEOREM. Each quadratic form can be reduced to canonical form by some orthogonal transformation.

PROOF. We will look at a matrix of quadratic form as the matrix of some linear operator in Euclidean space. If the matrix is ​​of quadratic form, then it is symmetric of order . If some orthonormal basis of dimensional Euclidean space, then the matrix defines a symmetric operator in this basis. By the main theorem about symmetric operators in Euclidean space, in a suitable orthonormal basis its matrix will be diagonal. Let the transition matrix from to , then .

But the matrix , as a transition matrix from one orthonormal basis to another, according to Theorem 2 §1.6 will be orthogonal, and therefore . That's why . Namely, this is how a matrix of a quadratic form is transformed, subjected to a linear transformation of unknowns with the matrix .

So, a transformation of unknowns having a matrix is ​​orthogonal, and the matrix, being diagonal, corresponds to a quadratic form of the canonical form. □

The fact that the matrix of a linear operator in a basis composed of eigenvectors has a diagonal form (with eigenvalues ​​along the main diagonal) gives us a method for practically finding the canonical form of the quadratic form, as well as this orthogonal transformation itself.

Example 2. Find an orthogonal transformation that reduces the quadratic form

to the canonical view and write this canonical view.

Solution. The matrix of this form has the form

,

Let's find its characteristic polynomial:

.

Thus, the matrix has a double root and a simple root. Therefore, the canonical form of this quadratic form will be

.

Let us find an orthogonal transformation that implements this reduction. To do this, we find the eigenvectors corresponding to the found eigenvalues , i.e., we will solve systems of linear homogeneous equations for each .

When we have

.

Where , i.e. there are 2 independent variables, and the fundamental set of solutions will be:

Applying the orthogonalization process to them, we obtain:

When we have

.

This system is equivalent to the following:

,

whose solution will be

- Linear algebra

Reducing quadratic form to principal axes

Previously, we considered the problem of reducing a real

q(x)= \sum_(n=1)^(n) \sum_(j=1)^(n) a_(ij)x_ix_j=x^TAx

n variables to

\widetilde(q)(y)=\lambda_1y_1^2+ \lambda_2y_2^2+ \ldots+ \lambda_ny_n^2

using a non-degenerate linear change of variables x=Sy. To solve this problem we used .

Let's consider another approach to the solution. A linear non-degenerate change of variables x=Sy with an orthogonal matrix S~(S^(-1)=S^T) will be called an orthogonal change of variables (or orthogonal transformation of variables).

Let's formulate the problem reducing a quadratic form to the principal axes: it is required to find an orthogonal change of variables x=Sy (S^(-1)=S^T), bringing the quadratic form (9.23) to the canonical form (9.24).

To solve we use the following geometric meaning tasks. We will count the variables x_1,x_2,\ldots,x_n coordinates of the vector \boldsymbol(x) of the n-dimensional Euclidean space \mathbb(E) in an orthonormal basis (\boldsymbol(e))= (\boldsymbol(e)_1,\ldots,\boldsymbol(e)_n), and the matrix A of quadratic form (9.23) is the matrix of some linear transformation \mathcal(A)\colon \mathbb(E)\to \mathbb(E) on the same basis. Moreover, this transformation is self-adjoint, since its matrix is ​​symmetric: A^T=A. The quadratic form (9.23) can be represented as a scalar product

q(\boldsymbol(x))= \bigl\langle \mathcal(A)(\boldsymbol(x)), \boldsymbol(x)\bigr\rangle= \bigl\langle \boldsymbol(x), \mathcal(A )(\boldsymbol(x))\bigr\rangle.

An orthogonal change of variables x=Sy corresponds to the transition from one orthonormal basis to another. Indeed, let S be the transition matrix from an orthonormal basis (\boldsymbol(e)) to an orthonormal basis (\boldsymbol(s))= (\boldsymbol(s)_1,\ldots,\boldsymbol(s)_n), i.e. (\boldsymbol(s))= (\boldsymbol(e))S and S^(-1)=S^T . Then the x coordinates of the vector \boldsymbol(x) in the basis (\boldsymbol(e)) and the y coordinates of the same vector in the basis (\boldsymbol(s)) are related by formula (8.11): x=Sy .

Thus, the problem of reducing a quadratic form to the principal axes can be formulated as follows: it is required to find in the space \mathbb(E) a basis in which the self-adjoint transformation matrix \mathcal(A) has a diagonal form. According to Theorem 9.10, it is necessary to choose an orthonormal basis from the eigenvectors of the self-adjoint transformation. In this case, the transition matrix S to the canonical basis turns out to be orthogonal: S^T=S^(-1) .

Let us formulate this result for the quadratic form.

Theorem (9.12) on reducing the quadratic form to the principal axes

The real quadratic form (9.23) can be reduced to the canonical form (9.24) using an orthogonal transformation of the variables x=Sy, where - eigenvalues matrices A.

Consequence. The quadratic form (9.23) is positive definite (non-negative definite) if and only if all the eigenvalues ​​of its matrix are positive (non-negative).

Notes 9.10

1. With a linear non-degenerate replacement variable matrix quadratic form changes according to formula (6.10): A"=S^TAS. For an orthogonal matrix S, this formula takes the form A"=S^(-1)AS, which coincides with formula (9.4) for changing the linear transformation matrix when changing the basis.

2. To find the canonical form (9.24), it is enough to determine all the roots \lambda_1, \lambda_2,\ldots,\lambda_m(among which there may be equal ones) (equations) \det(A-\lambda E)=0, where E is the identity matrix.

3. The corollary of Theorem 9.12 can be used to analyze the sign of a quadratic form:

– if all eigenvalues ​​are positive (negative), then the quadratic form is positive (negative) definite;

– if all eigenvalues ​​are non-negative (non-positive), then the quadratic form is non-negative (non-positive) definite;

– if there are eigenvalues ​​of different signs, then the quadratic form is indefinite (alternating).

4. The results formulated in paragraph 3 of the comments can be used to verify sufficient and necessary conditions second order in the problem of searching for the unconditional extremum of functions. To do this, you need to find the eigenvalues \lambda_1, \lambda_2,\ldots,\lambda_m \dfrac(d^2f(x))(dx^Tdx) in each stationary points x^(\ast) functions f(x)=f(x_1,\ldots,x_n).

If all eigenvalues ​​are positive: \lambda_i>0,~ i=1,\ldots,n, then at point x^(\ast) local minimum;

– if all eigenvalues ​​are negative: \lambda_i<0,~ i=1,\ldots,n , then at the point x^(\ast) there is a local maximum;

– if all eigenvalues ​​are non-negative: \lambda_i\geqslant0,~ i=1,\ldots,n, then at the point x^(\ast) there may be a local minimum;

– if all eigenvalues ​​are non-positive: \lambda_i\leqslant0,~ i=1,\ldots,n, then at the point x^(\ast) there may be a local maximum;

– if the eigenvalues \lambda_i,~ i=1,\ldots,n, different signs, then there is no extremum at point x^(\ast);

– if all eigenvalues ​​are zero: \lambda_i=0,~ i=1,\ldots,n, then additional research is required.

5. The problem of reducing a quadratic form to the main axes is solved using an algorithm for reducing a self-adjoint transformation to a diagonal form. In this case, the diagonal form of the matrix of the quadratic form and the orthogonal matrix S of the change of variables x=Sy are found, bringing the quadratic form to the canonical form (to the main axes).

Example 9.7. Determine the sign of the quadratic form of three variables

q(x)= x_1^2+2x_1x_2+2x_1x_3+x_2^2+2x_2x_3+x_3^2

Solution. We compose a matrix of quadratic form: A=\begin(pmatrix) 1&1&1\\ 1&1&1\\ 1&1&1 \end(pmatrix). In Example 9.6, the eigenvalues ​​of this matrix were found: \lambda_(1,2)=0, \lambda_3=3. All eigenvalues ​​are non-negative, so the quadratic form is non-negative definite (see point 4 of Remarks 9.10).

An orthogonal matrix was found

S=\begin(pmatrix) \dfrac(\sqrt(2))(2)& \dfrac(\sqrt(6))(6)& \dfrac(\sqrt(3))(3)\\ 0&-\ dfrac(\sqrt(6))(3)& \dfrac(\sqrt(3))(3)\\ -\dfrac(\sqrt(2))(2)& \dfrac(\sqrt(6))( 6)& \dfrac(\sqrt(3))(3) \end(pmatrix)\!,

reducing matrix A to diagonal form \Lambda= \operatorname(diag) (0,0,3). We write down the required orthogonal change of variables x=Sy:

x_1= \frac(\sqrt(2))(2)\,y_1+ \frac(\sqrt(6))(6)\,y_2+ \frac(\sqrt(3))(3)\,y_3;\quad x_2= -\frac(\sqrt(6))(3)\,y_2+ \frac(\sqrt(3))(3)\,y_3;\quad x_3= -\frac(\sqrt(2))(2 )\,y_1+ \frac(\sqrt(6))(6)\,y_2+ \frac(\sqrt(3))(3)\,y_3.

and the quadratic form in canonical form: \widetilde(q)(y)= 3y_3^2.

Example 9.8. Find local extremum points of a function of two variables using matrices

f(x)=3x_1^5+x_1^4-5x_1^3-2x_1^2x_2+x_2^2.

Solution. In step 1, the gradient of the function was found, and from the necessary condition for a first-order extremum, three stationary points:

x^0= \begin(pmatrix)0&0 \end(pmatrix)^T,\qquad x^1=\begin(pmatrix) 1&1 \end(pmatrix)^T,\qquad x^2=\begin(pmatrix) - 1&1\end(pmatrix)^T.

\frac(df(x))(dx^Tdx)= \begin(pmatrix) 60x_1^3+12x_1^2-30x_1-4x_2&-4x_1\\-4x_1&2 \end(pmatrix)\!.

Let us find the eigenvalues ​​of the Hessian matrix at each stationary point:

\frac(df(x^0))(dx^Tdx)= \begin(pmatrix)0&0\\ 0&2 \end(pmatrix)\!;\quad \frac(df(x^1))(dx^Tdx ) = \begin(pmatrix)38&-4\\ -4&2 \end(pmatrix)\!,\quad \frac(df(x^2))(dx^Tdx)= \begin(pmatrix) -22&4\\4&2\ end(pmatrix)

At the point x^0=\begin(pmatrix)0\\0 \end(pmatrix) the Hessian matrix has the form \begin(pmatrix) 0&0\\ 0&2\end(pmatrix). From Eq. \begin(vmatrix) -\lambda&0\\ 0&2-\lambda\end(vmatrix)=0 we find \lambda_1=0, \lambda_2=2 . Since all eigenvalues ​​are non-negative, there may be a local minimum at the point x^0 and additional research is required for a final conclusion (see example 6.13).

At the point x^1=\begin(pmatrix)1\\1 \end(pmatrix) the Hessian matrix has the form \begin(pmatrix) 38&-4\\ -4&2 \end(pmatrix). From Eq. \begin(vmatrix) 38-\lambda&-4\\ -4&2-\lambda\end(vmatrix)=0, or \lambda^2-40 \lambda+60=0 we get \lambda_(1,2)= 20\pm2\sqrt(85). Since all eigenvalues ​​are positive, then at the point x^1 there is a local minimum of the function.

At the point x^2=\begin(pmatrix)-1\\1 \end(pmatrix) the Hessian matrix has the form \begin(pmatrix) -22&4\\ 4&2 \end(pmatrix). From Eq. \begin(vmatrix) -22-\lambda&4\\ 4&2-\lambda\end(vmatrix)=0, or \lambda^2+40 \lambda-60=0 we get \lambda_(1,2)=-10\pm4\sqrt(10). Since the eigenvalues ​​have different signs, there is no extremum at the point x^2.