Find the eigenvalue of a linear operator. Eigenvectors of a linear operator

With matrix A, if there is a number l such that AX = lX.

In this case, the number l is called eigenvalue operator (matrix A) corresponding to vector X.

In other words, an eigenvector is a vector that, under the action of a linear operator, transforms into a collinear vector, i.e. just multiply by some number. In contrast, improper vectors are more complex to transform.

Let's write down the definition of an eigenvector in the form of a system of equations:

Let's move all the terms to the left side:

The latter system can be written in matrix form as follows:

(A - lE)X = O

The resulting system always has a zero solution X = O. Such systems in which all free terms are equal to zero are called homogeneous. If the matrix of such a system is square and its determinant is not equal to zero, then according to Cramer’s formulas we will always get a unique solution - zero. It can be proven that a system has non-zero solutions if and only if the determinant of this matrix is equal to zero, i.e.

|A - lE| = = 0

This equation with unknown l is called characteristic equation (characteristic polynomial) matrix A (linear operator).

It can be proven that the characteristic polynomial of a linear operator does not depend on the choice of basis.

For example, let's find the eigenvalues and eigenvectors of the linear operator defined by the matrix A = .

To do this, let's create a characteristic equation |A - lE| = = (1 - l) 2 - 36 = 1 - 2l + l 2 - 36 = l 2 - 2l - 35 = 0; D = 4 + 140 = 144; eigenvalues l 1 = (2 - 12)/2 = -5; l 2 = (2 + 12)/2 = 7.

To find eigenvectors, we solve two systems of equations

(A + 5E)X = O

(A - 7E)X = O

For the first of them, the expanded matrix takes the form

whence x 2 = c, x 1 + (2/3)c = 0; x 1 = -(2/3)s, i.e. X (1) = (-(2/3)s; s).

For the second of them, the expanded matrix takes the form

from where x 2 = c 1, x 1 - (2/3)c 1 = 0; x 1 = (2/3)s 1, i.e. X (2) = ((2/3)s 1; s 1).

Thus, the eigenvectors of this linear operator are all vectors of the form (-(2/3)с; с) with eigenvalue (-5) and all vectors of the form ((2/3)с 1 ; с 1) with eigenvalue 7 .

It can be proven that the matrix of the operator A in the basis consisting of its eigenvectors is diagonal and has the form:

where l i are the eigenvalues of this matrix.

The converse is also true: if matrix A in some basis is diagonal, then all vectors of this basis will be eigenvectors of this matrix.

It can also be proven that if a linear operator has n pairwise distinct eigenvalues, then the corresponding eigenvectors are linearly independent, and the matrix of this operator in the corresponding basis has a diagonal form.

Let's illustrate this with the previous example. Let's take arbitrary non-zero values c and c 1, but such that the vectors X (1) and X (2) are linearly independent, i.e. would form a basis. For example, let c = c 1 = 3, then X (1) = (-2; 3), X (2) = (2; 3).

Let's make sure linear independence these vectors:

12 ≠ 0. In this new basis, matrix A will take the form A * = .

To verify this, let's use the formula A * = C -1 AC. First, let's find C -1.

C -1 = ;

Quadratic shapes

Quadratic shape f(x 1, x 2, x n) of n variables is called 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 matrix 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.

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(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(proof not given here). Any quadratic form can be reduced to canonical form using a 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, 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 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 different ways). However, the received different 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 the law of inertia of quadratic forms.

Let us verify this by bringing the same quadratic form to its 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 negative coefficient -3 at y 1 and two positive coefficients 3 and 2 at y 2 and y 3 (and using another method we got a negative coefficient (-5) at y 2 and two positive ones: 2 at y 1 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 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 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. Major minor first order 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.

We 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 - 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.

Method 2. Principal minor of the first order of matrix A D 1 = a 11 =
= -2 < 0. Главный минор второго порядка D 2 = = 6 - 4 = 2 >0. Consequently, according to Sylvester’s criterion, the quadratic form is negative definite (the signs of the main 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 - l)*
*(-3 - l) - 4 = (-6 - 2l + 3l + l 2) - 4 = l 2 + l - 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 negatively nor positively 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 D 1 = a 11 = 2 > 0. Principal minor of the second order D 2 = = -6 - 4 = -10< 0. Следовательно, по критерию Сильвестра квадратичная форма не является знакоопределенной (знаки главных миноров разные, при этом первый из них - положителен).

Vector X ≠ 0 is called eigenvector linear operator with matrix A, if there is a number such that AX =X.

In this case, the number is called eigenvalue operator (matrix A) corresponding to the vector x.

Let's write down the definition of an eigenvector in the form of a system of equations:

Let's move all the terms to the left side:

The latter system can be written in matrix form as follows:

(A - E)X = O

The resulting system always has a zero solution X = O. Such systems in which all free terms are equal to zero are called homogeneous. If the matrix of such a system is square and its determinant is not equal to zero, then using Cramer’s formulas we will always get a unique solution – zero. It can be proven that a system has non-zero solutions if and only if the determinant of this matrix is equal to zero, i.e.

|A - E| = = 0

This equation with an unknown is called characteristic equation(characteristic polynomial) matrix A (linear operator).

It can be proven that the characteristic polynomial of a linear operator does not depend on the choice of basis.

For example, let's find the eigenvalues and eigenvectors of the linear operator defined by the matrix A = .

To do this, let's create a characteristic equation |A - E| = = (1 -) 2 – 36 = 1 – 2+ 2 - 36 = 2 – 2- 35; D = 4 + 140 = 144; eigenvalues 1 = (2 - 12)/2 = -5; 2 = (2 + 12)/2 = 7.

To find eigenvectors, we solve two systems of equations

(A + 5E)X = O

(A - 7E)X = O

For the first of them, the expanded matrix takes the form

whence x 2 = c, x 1 + (2/3)c = 0; x 1 = -(2/3)s, i.e. X (1) = (-(2/3)s; s).

For the second of them, the expanded matrix takes the form

from where x 2 = c 1, x 1 - (2/3)c 1 = 0; x 1 = (2/3)s 1, i.e. X (2) = ((2/3)s 1; s 1).

Thus, the eigenvectors of this linear operator are all vectors of the form (-(2/3)с; с) with eigenvalue (-5) and all vectors of the form ((2/3)с 1 ; с 1) with eigenvalue 7 .

It can be proven that the matrix of the operator A in the basis consisting of its eigenvectors is diagonal and has the form:

where  i are the eigenvalues of this matrix.

The converse is also true: if matrix A in some basis is diagonal, then all vectors of this basis will be eigenvectors of this matrix.

The simplest linear operator is multiplication of a vector by a number \(\lambda\). This operator simply stretches all vectors by \(\lambda \) times. Its matrix form in any basis is \(diag(\lambda ,\lambda ,...,\lambda)\). For definiteness, we fix the basis \(\(e\)\) in the vector space \(\mathit(L)\) and consider a linear operator with a diagonal matrix form in this basis, \(\alpha = diag(\lambda _1,\lambda _2,...,\lambda _n)\). This operator, according to the definition of the matrix form, stretches \(e_k\) by \(\lambda _k\) times, i.e. \(Ae_k=\lambda _ke_k\) for all \(k=1,2,...,n\). It is convenient to work with diagonal matrices; functional calculus is simple to construct for them: for any function \(f(x)\) we can put \(f(diag(\lambda _1,\lambda _2,...,\lambda _n))= diag(f(\lambda _1),f(\lambda _2),...,f(\lambda _n))\). Thus, a natural question arises: let there be a linear operator \(A\), is it possible to choose such a basis in the vector space so that the matrix form of the operator \(A\) is diagonal in this basis? This question leads to the definition of eigenvalues and eigenvectors.

Definition. Let for the linear operator \(A\) there exist a nonzero vector \(u\) and a number \(\lambda \) such that \[ Au=\lambda \cdot u. \quad \quad(59) \] Then the vector \(u\) is called eigenvector operator \(A\), and the number \(\lambda \) - the corresponding eigenvalue operator \(A\). The set of all eigenvalues is called spectrum of the linear operator \(A\).

A natural problem arises: find for a given linear operator its eigenvalues and the corresponding eigenvectors. This problem is called the spectrum problem of a linear operator.

Eigenvalue equation

For definiteness, we fix the basis in vector space, i.e. We will assume that it is given once and for all. Then, as discussed above, the consideration of linear operators can be reduced to the consideration of matrices - matrix forms of linear operators. We rewrite equation (59) in the form \[ (\alpha -\lambda E)u=0. \] Here \(E\) is the identity matrix, and \(\alpha\) is the matrix form of our linear operator \(A\). This relation can be interpreted as a system \(n\) linear equations for \(n\) unknowns - the coordinates of the vector \(u\). Moreover, this is a homogeneous system of equations, and we should find it non-trivial solution. Previously, a condition for the existence of such a solution was given - for this it is necessary and sufficient that the rank of the system be less than the number of unknowns. This implies the equation for the eigenvalues: \[ det(\alpha -\lambda E)=0. \quad \quad(60) \]

Definition. Equation (60) is called characteristic equation for the linear operator \(A\).

Let us describe the properties of this equation and its solutions. If we write it out explicitly, we obtain an equation of the form \[ (-1)^n\lambda ^n+...+det(A)=0. \quad \quad(61) \] On the left side there is a polynomial in the variable \(\lambda \). Such equations are called algebraic of degree \(n\). Let's give necessary information about these equations.

Help about algebraic equations.

Theorem. Let all eigenvalues of the linear operator \(A\) be prime. Then the set of eigenvectors corresponding to these eigenvalues forms the basis of the vector space.

From the conditions of the theorem it follows that all eigenvalues of the operator \(A\) are different. Suppose that the set of eigenvectors is linearly dependent, so that there are constants \(c_1,c_2,...,c_n\), not all of which are zero, satisfying the condition: \[ \sum_(k=1)^nc_ku_k=0. \quad \quad(62) \]

Among such formulas, let us consider one that includes the minimum number of terms, and act on it with the operator \(A\). Due to its linearity, we obtain: \[ A\left (\sum_(k=1)^nc_ku_k \right)=\sum_(k=1)^nc_kAu_k=\sum_(k=1)^nc_k\lambda _ku_k=0. \quad \quad(63) \]

Let, for definiteness, \(c_1 \neq 0\). Multiplying (62) by \(\lambda _1\) and subtracting from (63), we obtain a relation of the form (62), but containing one less term. The contradiction proves the theorem.

So, under the conditions of the theorem, a basis appears associated with a given linear operator - the basis of its eigenvectors. Let us consider the matrix form of the operator in such a basis. As mentioned above, the \(k\)th column of this matrix is the decomposition of the vector \(Au_k\) with respect to the basis. However, by definition, \(Au_k=\lambda _ku_k\), so this expansion (what is written on the right side) contains only one term and the constructed matrix turns out to be diagonal. As a result, we find that under the conditions of the theorem, the matrix form of the operator in the basis of its eigenvectors is equal to \(diag(\lambda _1,\lambda _2,...,\lambda _n)\). Therefore, if it is necessary to develop functional calculus for a linear operator, it is reasonable to work in the basis of its eigenvectors.

If among the eigenvalues of a linear operator there are multiples, the description of the situation becomes more complicated and may include so-called Jordan cells. We refer the reader to more advanced tutorials for relevant situations.

Definition 5.3. Nonzero vector x in linear space L is called eigenvector of the linear operator A: L → L, if for some real number A the relation Ax = λx holds. In this case, the number λ is called eigenvalue (eigenvalue) of the linear operator A.

Example 5.3. The linear space K n [x] of polynomials of degree not higher than n contains polynomials of degree zero, i.e. permanent functions. Since dc/dx = 0 = 0 c, polynomials of degree zero p(x) = c ≠ 0 are the eigenvectors of the linear differentiation operator, and the number λ = 0 is the eigenvalue of this operator. #

The set of all eigenvalues of a linear operator is called spectrum of the linear operator . Each eigenvector is associated with its own eigenvalue. Indeed, if a vector x simultaneously satisfies two equalities Ax = λx and Ax = μx, then λx = μx, whence (λ - μ)x = 0. If λ - μ ≠ 0, multiply the equality by the number (λ - μ) -1 and as a result we get that x = 0. But this contradicts the definition of an eigenvector, since an eigenvector is always non-zero.

Each eigenvalue has its own eigenvectors, and there are infinitely many of them. Indeed, if x is an eigenvector of a linear operator A with an eigenvalue λ, i.e. Ах = λx, then for any non-zero real number α we have αx ≠ 0 and А(αх) = α(Ах) = αλx = λ(αx). This means that the vector αx is also an eigenvector for the linear operator.

Remark 5.1. They often talk about eigenvalues (numbers), spectrum and eigenvectors of a square matrix . This means the following. Matrix A of order n is matrix some linear operator in a fixed basis, operating in n-dimensional linear space. For example, if we stop at standard basis in linear arithmetic space R n , then the matrix A defines a linear operator A, mapping a vector x ∈ R n with a coordinate column x to a vector with a coordinate column Ax. Matrix A is precisely matrix A. It is natural to identify an operator with its matrix in the same way as an arithmetic vector is identified with a column of its coordinates. This identification, which is often used and not always specified, makes it possible to transfer “operator” terms to matrices.

The spectrum of a linear operator is closely related to its characteristic equation.

Theorem 5.3. In order to real numberλ was an eigenvalue of a linear operator; it is necessary and sufficient that it be the root of the characteristic equation of this operator.

◄ Necessity. Let the number λ be an eigenvalue of the linear operator A: L → L. This means that there is a vector x ≠ 0 for which

Ax = λx. (5.2)

Note that in L there is identity operator I: Ix = x for any vector x. Using this operator, we transform equality (5.2): Ах = λIx, or

(A - λI)х = 0. (5.3)

Let us write the vector equality (5.3) in some basis b. The matrix of the linear operator A - λI will be the matrix A - λE, where A is the matrix of the linear operator A in basis b, and E is the identity matrix, and let x be the column of coordinates of the eigenvector x. Then x ≠ 0, and the vector equality (5.3) is equivalent to the matrix

(A - λE)x = 0, (5.4)

which is a matrix form of recording a homogeneous system of linear algebraic equations(SLAU) with square matrix A - λE of order n. This system has a nonzero solution, which is the x-coordinate column of the eigenvector x. Therefore, the matrix A - λE of system (5.4) has a zero determinant, i.e. det(A - λE) = 0. This means that λ is the root of the characteristic equation of the linear operator A.

Adequacy. It is easy to verify that the above reasoning can be carried out in reverse order. If λ is the root of the characteristic equation, then in a given basis b the equality det (A - λE) = 0 holds. Consequently, the matrix of the homogeneous SLAE (5.4), written in matrix form, is degenerate, and the system has a non-zero solution x. This non-zero solution is a set of coordinates in the basis b of some non-zero vector x for which the vector equality (5.3) or its equivalent equality (5.2) holds. We come to the conclusion that the number λ is an eigenvalue of the linear operator A.

Each eigenvalue λ of the matrix (linear operator) is associated with its multiplicity, putting it equal to the multiplicity of the root λ of the characteristic equation of this matrix (of this linear operator).

The set of all eigenvectors corresponding to a given eigenvalue of a linear operator is not linear subspace, since this set does not contain zero vector, which, by definition, cannot be proper. But this formal and easily removable obstacle is the only one. Let us denote by £(A, λ) the set of all eigenvectors of the linear operator A in the linear space L corresponding to the eigenvalue λ, with the zero vector added to this set.

Theorem 5.4. The set £(A,λ) is a linear subspace in L.

◄ Let's choose arbitrary two vector x,y∈ £(A, λ) and prove that for any real α and β the vector αx + βу also belongs to £(A, λ). To do this, we calculate the image of this vector under the action of the linear operator A:

А(αх + βу) = А((αx) + А(βу) = αАх + βАу = α(λх) + β(λу) = λ(αx) + λ(βу) = λ(αx + βу).

Thus, for the vector z = αх + βу the relation Az = λz holds. If z is a zero vector, then it belongs to £(A,λ). If it is non-zero, then, according to the proven relation, it is an eigenvalue with an eigenvalue λ and again belongs to the set £(A, λ).

The linear subspace £(A,λ) is sometimes called eigensubspace of the linear operator *. It is a special case invariant subspace linear operator A - a linear subspace such that for any vector x ∈ H the vector Ax also belongs to H.

An invariant subspace of a linear operator is also the linear span of any system of its eigenvectors. An invariant subspace of a linear operator not related to its eigenvectors is operator image.