Find the value of the operator on the vector. Eigenvectors of a linear operator

Let - linear transformation n-dimensional linear space V. Nonzero vector \boldsymbol(s) of linear space V satisfying the condition

\mathcal(A)(\boldsymbol(s))=\lambda\cdot \boldsymbol(s),

called eigenvector of linear transformation\mathcal(A) . The number \lambda in equality (9.5) is called eigenvalue transformation\mathcal(A) . The eigenvector is said to correspond to (belong to) the eigenvalue \lambda . If the space V is real (complex), then the eigenvalue \lambda is a real (complex) number.

The set of all eigenvalues of a linear transformation is called its spectrum.

Let us explain the geometric meaning of eigenvectors. A nonzero vector s is an eigenvector of the transformation \mathcal(A) if its image \mathcal(A) (\boldsymbol(s)) is collinear to the inverse image of \boldsymbol(s) . In other words, if \boldsymbol(s) is an eigenvector, then the transformation \mathcal(A) has a one-dimensional invariant subspace. The opposite statement is also true.

Indeed, let the eigenvector \boldsymbol(s) correspond to some eigenvalue \lambda . Any vector \boldsymbol(v) from \operatorname(Lin)(\boldsymbol(s)) looks like \boldsymbol(v)=\alpha \boldsymbol(s), where \alpha is any number from the given field. Let's find the image of this vector

\mathcal(A)(\boldsymbol(v))= \mathcal(A)(\alpha \boldsymbol(s))= \alpha\cdot \mathcal(A)(\boldsymbol(s))= \alpha\cdot \ lambda\cdot \boldsymbol(s)\in \operatorname(Lin) (\boldsymbol(s)).

Hence, \mathcal(A)(\boldsymbol(v))\in \operatorname(Lin)(\boldsymbol(s)) for any vector \boldsymbol(v)\in \operatorname(Lin)(\boldsymbol(s)), i.e. subspace \operatorname(Lin)(\boldsymbol(s)) invariant under the transformation \mathcal(A) . Subspace dimension \operatorname(Lin) (\boldsymbol(s)) is equal to one, since \boldsymbol(s)\ne \boldsymbol(o) a-priory.

The converse statement is proven by reasoning in reverse order.

Relationship between eigenvectors of a linear transformation (operator) and its matrix

Previously, eigenvectors and eigenvalues of a matrix were considered. Recall that the eigenvector of a square matrix A of nth order is called non-zero numeric column s=\begin(pmatrix)s_1&\cdots&s_(n)\end(pmatrix)^T, satisfying condition (7.13):

A\cdot s=\lambda\cdot s.

The number \lambda in (9.6) is called the eigenvalue of the matrix A. It was believed that the eigenvalue \lambda and the numbers s_i~(i=1,\ldots,n) belong to the field of complex numbers.

These concepts are related to eigenvectors and eigenvalues of a linear transformation.

Theorem 9.3 on the eigenvectors of a linear transformation and its matrix. Let \mathcal(A)\colon V\to V is a linear transformation of an n-dimensional linear space V with basis. Then the eigenvalue \lambda and the coordinate column (s) of the eigenvector \boldsymbol(s) of the transformation \mathcal(A) are the eigenvalue and eigenvector of the matrix A of this transformation defined with respect to the basis \boldsymbol(e)_1,\ldots, \boldsymbol(e)_n, i.e.

\mathcal(A)(\boldsymbol(s))=\lambda\cdot \boldsymbol(s)\quad \Rightarrow\quad A\cdot s=\lambda\cdot s, Where \boldsymbol(s)=s_1 \boldsymbol(e)_1+\ldots+s_n \boldsymbol(e)_n,~ s=\begin(pmatrix)s_1&\cdots& s_n\end(pmatrix)^T.

The converse statement is true for additional conditions: if column s=\begin(pmatrix) s_1&\cdots&s_n\end(pmatrix)^T and the number \lambda are the eigenvector and eigenvalue of the matrix A, and the numbers s_1,\ldots,s_n,\lambda belong to the same number field over which the linear space V is defined, then the vector \boldsymbol(s)=s_1 \boldsymbol(e)_1+ \ldots+s_n \boldsymbol(e)_n and number \lambda are the eigenvector and eigenvalue of the linear transformation \mathcal(A)\colon V\to V with matrix A in basis \boldsymbol(e)_1,\ldots,\boldsymbol(e)_n.

In fact, condition (9.5) in coordinate form has the form (9.6), which coincides with the definition (7.13) of the matrix eigenvector. On the contrary, equality (9.6) implies equality (9.5) provided that the vectors and \lambda\cdot \boldsymbol(s) defined, i.e. numbers s_1,\ldots,s_n,\lambda belong to the same number field over which the linear space is defined.

Recall that finding the eigenvalues of a matrix reduces to solving its characteristic equation \Delta_A(\lambda)=0, Where \Delta_A(\lambda)=\det(A-\lambda E) is the characteristic polynomial of the matrix A. For linear transformation we introduce similar concepts.

Characteristic polynomial of linear transformation \mathcal(A)\colon V\to V n-dimensional linear space is the characteristic polynomial of the matrix A of this transformation, found with respect to any basis of the space V.

The equation is called characteristic equation of linear transformation.

Conversion \mathcal(A)-\lambda\mathcal(E) called characteristic of a linear transformation \mathcal(A)\colon V\to V.

Notes 9.4

1. The characteristic polynomial of a linear transformation does not depend on the basis in which the transformation matrix is found.

In fact, the matrices \mathop(A)\limits_((\boldsymbol(e))) And \mathop(A)\limits_((\boldsymbol(f))) linear transformation \mathcal(A) in bases (\boldsymbol(e))= (\boldsymbol(e)_1,\ldots, \boldsymbol(e)_n) And (\boldsymbol(f))=(\boldsymbol(f)_1,\ldots,\boldsymbol(f)_n) are, according to (9.4), similar: \nathop(A)\limits_((\boldsymbol(f)))=S^(-1)\mathop(A)\limits_((\boldsymbol(e)))S, where S is the transition matrix from the basis (\boldsymbol(e)) to the basis (\boldsymbol(f)). As shown earlier, the characteristic polynomials of such matrices coincide (see property 3). Therefore, for the characteristic polynomial of the transformation \mathcal(A) we can use the notation \Delta_(\mathcal(A))(\lambda), without specifying the matrix of this transformation.

2. From Theorem 9.3 it follows that any complex (real, rational) root of the characteristic equation is an eigenvalue of the linear transformation \mathcal(A)\colon V\to V linear space V defined over the field of complex (real, rational) numbers.

3. From Theorem 9.3 it follows that any linear transformation of a complex linear space has a one-dimensional invariant subspace, since this transformation has an eigenvalue (see point 2), and therefore eigenvectors. Such a subspace is, for example, the linear span of any eigenvector. A transformation of a real linear space may not have one-dimensional invariant subspaces if all the roots of the characteristic equation are complex (but not real).

Theorem 9.4 on invariant subspaces of a linear operator in a real space. Every linear transformation of a real linear space has a one-dimensional or two-dimensional invariant subspace.

Indeed, let us compose a linear transformation matrix A \mathcal(A)\colon V\to V n-dimensional real linear space V in an arbitrary basis \boldsymbol(e)_1,\ldots,\boldsymbol(e)_n. The elements of this matrix are real numbers. Therefore, the characteristic polynomial \Delta_(\mathcal(A))(\lambda)=\det(A-\lambda E) is a polynomial of degree n with real coefficients. According to Corollaries 3 and 4 of the fundamental theorem of algebra, such a polynomial can have real roots and pairs of complex conjugate roots.

If \lambda=\lambda_1 is a real root of the characteristic equation, then the corresponding eigenvector s=\begin(pmatrix)s_1&\cdots&s_n\end(pmatrix)^T matrix A is also real. Therefore it defines an eigenvector \boldsymbol(s)=s_1 \boldsymbol(e)_1+\ldots+s_n \boldsymbol(e)_n linear transformation (see Theorem 9.3). In this case, there is a one-dimensional subspace invariant under \mathcal(A) \operatorname(Lin)(\boldsymbol(s))(see the geometric meaning of eigenvectors).

If \lambda=\alpha\pm\beta i is a pair of complex conjugate roots (\beta\ne0), then the eigenvector s\ne o of matrix A also has complex elements: s=\begin(pmatrix)x_1+y_1i&\cdots& x_n+y_n i \end(pmatrix)^T. It can be represented as s=x+yi , where x,\,y are real columns. Equality (9.6) will then have the form

A\cdot(x+yi)= (\alpha+\beta i)\cdot(x+yi).

Isolating the real and imaginary parts, we obtain the system

\begin(cases)Ax=\alpha x-\beta y,\\ Ay=\beta x+\alpha y.\end(cases)

Let us show that columns (x) and (y) are linearly independent. Let's consider two cases. If x=o, then from the first equation (9.7) it follows that y=o, since \beta\ne0. Then s=o, which contradicts the condition s\ne o. Let's assume that x\ne o and the x and y columns are proportional, i.e. there is such a thing real number\gamma that y=\gamma x . Then from system (9.7) we obtain \begin(cases)Ax=(\alpha-\beta\gamma)x,\\ \gamma Ax=(\beta-\alpha\gamma)x. \end(cases) Adding the first equation multiplied by (-\gamma) to the second equation, we arrive at the equality [(\beta+\alpha\gamma)-\gamma(\alpha-\beta\gamma)]x=o. Since x\ne o , the expression in square brackets is equal to zero, i.e. (\beta+\alpha\gamma)- \gamma(\alpha- \beta\gamma)= \beta(1+\gamma^2)=0. Since \beta\ne0 , then \gamma^2=-1 . This cannot happen since \gamma is a real number. We got a contradiction. Thus, the x and y columns are linearly independent.

Consider the subspace where \boldsymbol(x)= x_1 \boldsymbol(e)_1+\ldots+x_n \boldsymbol(e)_n,~ \boldsymbol(y)= y_1 \boldsymbol(e)_1+\ldots+ y_n \boldsymbol(y)_n. This subspace is two-dimensional, since the vectors \boldsymbol(x),\boldsymbol(y) are linearly independent (as shown above, their x,y coordinate columns are linearly independent). From (9.7) it follows that \begin(cases)\mathcal(A)(\boldsymbol(x))=\alpha \boldsymbol(x)-\beta \boldsymbol(y),\\ \mathcal(A)(\boldsymbol(y))=\ beta \boldsymbol(x)+\alpha \boldsymbol(y),\end(cases) those. the image of any vector belonging to \operatorname(Lin)(\boldsymbol(x),\boldsymbol(y)), also belongs \operatorname(Lin)(\boldsymbol(x),\boldsymbol(y)). Hence, \operatorname(Lin)(\boldsymbol(x),\boldsymbol(y)) is a two-dimensional subspace invariant under the transformation \mathcal(A) , which is what we needed to prove.

Finding eigenvectors and values of a linear operator (transformation)

To find eigenvectors and eigenvalues of a linear transformation \mathcal(A)\colon V\to V real linear space V, the following steps must be performed.

1. Choose an arbitrary basis \boldsymbol(e)_1,\ldots,\boldsymbol(e)_n linear space V and find the transformation matrix A \mathcal(A) in this basis.

2. Compose the characteristic polynomial of the transformation \mathcal(A)\colon\, \Delta_(\mathcal(A))(\lambda)=\det(A-\lambda E).

3. Find all different real roots \lambda_1,\ldots,\lambda_k characteristic equation \Delta_(\mathcal(A))(\lambda)=0. Complex (but not real) roots of the characteristic equation should be discarded (see paragraph 2 of remarks 9.4).

4. For the root \lambda=\lambda_1 find the fundamental system \varphi_1, \varphi_2,\ldots,\varphi_(n-r) solutions to a homogeneous system of equations (A-\lambda_1E)x=o , where r=\operatorname(rg)(A-\lambda_1E). To do this, you can use either an algorithm for solving a homogeneous system, or one of the methods for finding the fundamental matrix.

5. Write linearly independent eigenvectors of the transformation \mathcal(A) corresponding to the eigenvalue \lambda_1:

\begin(matrix) \boldsymbol(s)_1=\varphi_(1\,1)\boldsymbol(e)_1+ \ldots+ \varphi_(n\,1)\boldsymbol(e)_n,\\ \boldsymbol(s) _2=\varphi_(1\,2)\boldsymbol(e)_1+ \ldots+ \varphi_(n\,2)\boldsymbol(e)_n,\\ \vdots\\ \boldsymbol(s)_(n-r)=\ varphi_(1\,n-r) \boldsymbol(e)_1+ \ldots+\varphi_(n\,n-r)\boldsymbol(e)_n. \end(matrix)

To find the set of all eigenvectors corresponding to the eigenvalue \lambda_1, form non-zero linear combinations

\boldsymbol(s)= C_1 \boldsymbol(s)_1+C_2 \boldsymbol(s)_2+\ldots+ C_(n-r)\boldsymbol(s)_(n-r),

Where C_1,C_2,\ldots,C_(n-r)- arbitrary constants, not equal to zero simultaneously.

Repeat steps 4, 5 for the remaining eigenvalues \lambda_2,\ldots,\lambda_k linear transformation \mathcal(A) .

To find the eigenvectors of a linear transformation of a complex linear space, you need to determine in step 3 all the roots of the characteristic equation and, without discarding the complex roots, perform steps 4 and 5 for them.

Examples of eigenvectors of linear operators (transformations)

1. For zero conversion \mathcal(O)\colon V\to V any nonzero vector is an eigenvector corresponding to a zero eigenvalue \lambda=0 , since \mathcal(O)(\boldsymbol(s))=0\cdot \boldsymbol(s)~ \forall \boldsymbol(s)\in V.

2. For identity transformation \mathcal(E)\colon V\to V any non-zero vector \boldsymbol(s)\in V is the eigencorresponding to the identity eigenvalue \lambda=1 , since \mathcal(E) (\boldsymbol(s))=1\cdot \boldsymbol(s)~ \forall \boldsymbol(s)\in V.

3. For central symmetry \mathcal(Z)_(\boldsymbol(o))\colon V\to V any non-zero vector \boldsymbol(s)\in V \mathcal(Z)_(\boldsymbol(o)) (\boldsymbol(s))=(-1)\cdot \boldsymbol(s)~ \forall \boldsymbol(s)\in V.

4. For homothety \mathcal(H)_(\lambda)\colon V\to V any non-zero vector \boldsymbol(s)\in V is an eigencorresponding to the eigenvalue \lambda (the homothety coefficient), since \mathcal(H)_(\lambda) (\boldsymbol(\boldsymbol(s)))= \lambda\cdot \boldsymbol(s)~ \forall \boldsymbol(s)\in V.

5. To turn \mathcal(R)_(\varphi)\colon V_2\to V_2 plane (at ) there are no eigenvectors, since when rotated by an angle not a multiple of \pi, the image of each non-zero vector is non-collinear to the inverse image. Here we consider the rotation of the real plane, i.e. two-dimensional vector space over the field of real numbers.

6. For the differentiation operator \mathcal(D)\colon P_n(\mathbb(R))\to P_n(\mathbb(R)) any nonzero polynomial of degree zero (not identically zero) is an eigenvector corresponding to the zero eigenvalue \lambda=0 , since \mathcal(D)(s(x))=0\cdot s(x) \forall s(x)\equiv \text(const). Any polynomial of non-zero degree is not an eigenvector, since the polynomial is not proportional to its derivative: \mathcal(D)(s(x))=s"(x)\ne \lambda\cdot s(x), since they have different degrees.

7. Consider the operator \Pi_(L_1)\colon V\to V projection onto the subspace L_1 parallel to the subspace L_2. Here V=L_1\oplus L_2, \Pi_(L_1)(\boldsymbol(v)_1+ \boldsymbol(v)_2)=\boldsymbol(v)_1 For \Pi_(L_1)(\boldsymbol(v)_1)=1\cdot \boldsymbol(v)_1, and any non-zero vector is an eigenvector corresponding to the eigenvalue \lambda=0 , since \Pi_(L_2)(\boldsymbol(v)_2)=0\cdot \boldsymbol(v)_2 \Pi_(L_1)(\boldsymbol(v)_1+\boldsymbol(v)_2)= \boldsymbol(v)_1= \lambda(\boldsymbol(v)_1+\boldsymbol(v)_2) it is possible either at or at .

8. Consider the operator \mathcal(Z)_(L_1)\colon V\to V reflections onto the subspace L_1 parallel to the subspace L_2. Here V=L_1\oplus L_2 \mathcal(Z)_(L_1)(\boldsymbol(v)_1+\boldsymbol(v)_2)= \boldsymbol(v)_1- \boldsymbol(v)_2, For \boldsymbol(v)=\boldsymbol(v)_1+\boldsymbol(v)_2, \boldsymbol(v)_1\in L_1,~ \boldsymbol(v)_2\in L_2. For this operator, any nonzero vector \boldsymbol(v)_1\in L_1 is the eigenvalue corresponding to the eigenvalue \lambda=1 , since \mathcal(Z)_(L_1) (\boldsymbol(v)_1)= 1\cdot \boldsymbol(v)_1, and any nonzero vector \boldsymbol(v)_2\in L_2 is the eigenvalue corresponding to the eigenvalue \lambda=-1 since \mathcal(Z)_(L_2) (\boldsymbol(v)_2)= (-1)\cdot \boldsymbol(v)_2. Other vectors are not eigenvectors, since the equality \mathcal(Z)_(L_1)(\boldsymbol(v)_1+\boldsymbol(v)_2)= \boldsymbol(v)_1- \boldsymbol(v)_2= \lambda(\boldsymbol()_1+ \boldsymbol(v )_2) possible either with \boldsymbol(v)_1=\boldsymbol(o), or at \boldsymbol(v)_2= \boldsymbol(o).

9. In the space V_3 of radius vectors of space, plotted from a fixed point O, consider a rotation by an angle \varphi\ne\pi k,~ k\in\mathbb(Z), around the \ell axis defined by the radius vector \vec(\ell) . Any nonzero vector collinear to the vector \vec(\ell) is an eigencorresponding to the eigenvalue \lambda=1 . This transformation has no other eigenvectors.

Example 9.1. Find the eigenvalues and eigenvectors of the differentiation operator \mathcal(D)\colon T_1\to T_1, transforming the space of trigonometric polynomials (frequency \omega=1):

a) with real coefficients T_1=T_1(\mathbb(R))= \operatorname(Lin) (\sin(t),\cos(t));

b) with complex coefficients T_1=T_1(\mathbb(C))= \operatorname(Lin) (\sin(t),\cos(t)).

Solution. 1. Let's choose a standard basis e_1(t)=\sin(t),~ e_2(t)=\cos(t) and in this basis we compose the matrix D of the operator \mathcal(D):

D=\begin(pmatrix)0&-1\\ 1&0 \end(pmatrix)\!.

2. Let's compose the characteristic polynomial of the transformation \mathcal(D)\colon\, \Delta_(\mathcal(D))(\lambda)= \begin(vmatrix)-\lambda&-1\\ 1&-\lambda\end(vmatrix)= \lambda^2+ 1..

3. The characteristic equation \lambda^2+1=0 has complex conjugate roots \lambda_1=i,~ \lambda_2=-i. There are no real roots, therefore the transformation \mathcal(D) of the real space T_1(\mathbb(R)) (case (a)) has no eigenvalues, and therefore no eigenvectors. The transformation \mathcal(D) of the complex space T_1(\mathbb(C)) (case (b)) has complex eigenvalues \lambda_1,\,\lambda_2.

4(1). For the root \lambda_1=i we find the fundamental system \varphi_1 of solutions to the homogeneous system of equations (D-\lambda_1 E)x=o:

\begin(pmatrix)-i&-1\\ 1&-i\end(pmatrix)\!\cdot\! \begin(pmatrix) x_1\\x_2 \end(pmatrix)= \begin(pmatrix)0\\0 \end(pmatrix)\!.

Let's reduce the system matrix to stepwise form by multiplying the first equation by (i) and subtracting it from the second equation:

\begin(pmatrix)-i&-1\\ 1&-i \end(pmatrix)\sim \begin(pmatrix)1&-i\\ 1&-i \end(pmatrix)\sim \begin(pmatrix)1&-i\ \0&0\end(pmatrix)\!.

We express the basic variable x_1 in terms of the free variable: x_1=ix_2. Assuming x_2=1, we get x_1=i, i.e. \varphi=\begin(pmatrix)i&1 \end(pmatrix)^T.

5(1). We write down the eigenvector corresponding to the eigenvalue \lambda_1= i\colon\, s_1(t)=i\cdot\sin(t)+1\cdot\cos(t). The set of all eigenvectors corresponding to the eigenvalue \lambda_1=i form non-zero functions proportional to s_1(t) .

4(2). For the root \lambda_2=-i we similarly find the fundamental system (consisting of one vector) \varphi_2=\begin(pmatrix)-i&1 \end(pmatrix)^T solutions to a homogeneous system of equations (D-\lambda_2E)x=o:

\begin(pmatrix)i&-1\\ 1&i \end(pmatrix)\!\cdot\! \begin(pmatrix) x_1\\x_2 \end(pmatrix)= \begin(pmatrix)0\\0 \end(pmatrix)\!.

5(2). We write down the eigenvector corresponding to the eigenvalue \lambda_2=-i\colon\, s_2(t)=-i\cdot\sin(t)+1\cdot\cos(t). The set of all eigenvectors corresponding to the eigenvalue \lambda_2=-i form non-zero functions proportional to s_2(t) .

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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.

Diagonal matrices have the simplest structure. The question arises whether it is possible to find a basis in which the matrix of the linear operator would have a diagonal form. Such a basis exists.
Let us be given a linear space R n and a linear operator A acting in it; in this case, operator A takes R n into itself, that is, A:R n → R n .

Definition. A non-zero vector is called an eigenvector of the operator A if the operator A translates into a collinear vector, that is. The number λ is called the eigenvalue or eigenvalue of the operator A, corresponding to the eigenvector.
Let us note some properties of eigenvalues and eigenvectors.
1. Any linear combination of eigenvectors operator A corresponding to the same eigenvalue λ is an eigenvector with the same eigenvalue.
2. Eigenvectors operator A with pairwise different eigenvalues λ 1 , λ 2 , …, λ m are linearly independent.
3. If the eigenvalues λ 1 =λ 2 = λ m = λ, then the eigenvalue λ corresponds to no more than m linearly independent eigenvectors.

So, if there are n linearly independent eigenvectors , corresponding to different eigenvalues λ 1, λ 2, ..., λ n, then they are linearly independent, therefore, they can be taken as the basis of the space R n. Let us find the form of the matrix of the linear operator A in the basis of its eigenvectors, for which we will act with the operator A on the basis vectors: Then .
Thus, the matrix of the linear operator A in the basis of its eigenvectors has a diagonal form, and the eigenvalues of the operator A are along the diagonal.
Is there another basis in which the matrix has a diagonal form? The answer to this question is given by the following theorem.

Theorem. The matrix of a linear operator A in the basis (i = 1..n) has a diagonal form if and only if all the vectors of the basis are eigenvectors of the operator A.

Rule for finding eigenvalues and eigenvectors

Let a vector be given

, where x 1, x 2, …, x n are the coordinates of the vector relative to the basis

and is the eigenvector of the linear operator A corresponding to the eigenvalue λ, that is. This relationship can be written in matrix form

. (*)

Equation (*) can be considered as an equation for finding , and , that is, we are interested in non-trivial solutions, since the eigenvector cannot be zero. It is known that nontrivial solutions of a homogeneous system of linear equations exist if and only if det(A - λE) = 0. Thus, for λ to be an eigenvalue of the operator A it is necessary and sufficient that det(A - λE) = 0.
If equation (*) is written in detail in coordinate form, we obtain a system of linear homogeneous equations:

(1)
Where - linear operator matrix.

System (1) has a non-zero solution if its determinant D is equal to zero

We received an equation for finding eigenvalues.
This equation is called the characteristic equation, and its left side- the characteristic polynomial of the matrix (operator) A. If the characteristic polynomial has no real roots, then the matrix A does not have eigenvectors and cannot be reduced to diagonal form.
Let λ 1, λ 2, …, λ n be the real roots of the characteristic equation, and among them there may be multiples. Substituting these values in turn into system (1), we find the eigenvectors.

Example 12. The linear operator A acts in R 3 according to the law, where x 1, x 2, .., x n are the coordinates of the vector in the basis , , . Find the eigenvalues and eigenvectors of this operator.
Solution. We build the matrix of this operator:
.
We create a system for determining the coordinates of eigenvectors:

We compose a characteristic equation and solve it:

.
λ 1,2 = -1, λ 3 = 3.
Substituting λ = -1 into the system, we have:
or
Because , then there are two dependent variables and one free variable.
Let x 1 be a free unknown, then We solve this system in any way and find common decision this system: Fundamental system solutions consists of one solution, since n - r = 3 - 2 = 1.
The set of eigenvectors corresponding to the eigenvalue λ = -1 has the form: , where x 1 is any number other than zero. Let's choose one vector from this set, for example, putting x 1 = 1: .
Reasoning similarly, we find the eigenvector corresponding to the eigenvalue λ = 3: .
In the space R 3, the basis consists of three linearly independent vectors, but we received only two linearly independent eigenvectors, from which the basis in R 3 cannot be composed. Consequently, we cannot reduce the matrix A of a linear operator to diagonal form.

Example 13. Given a matrix .
1. Prove that the vector is an eigenvector of matrix A. Find the eigenvalue corresponding to this eigenvector.
2. Find a basis in which matrix A has a diagonal form.
Solution.
1. If , then is an eigenvector

.
Vector (1, 8, -1) is an eigenvector. Eigenvalue λ = -1.
The matrix has a diagonal form in a basis consisting of eigenvectors. One of them is famous. Let's find the rest.
We look for eigenvectors from the system:

Characteristic equation: ;
(3 + λ)[-2(2-λ)(2+λ)+3] = 0; (3+λ)(λ 2 - 1) = 0
λ 1 = -3, λ 2 = 1, λ 3 = -1.
Let's find the eigenvector corresponding to the eigenvalue λ = -3:

The rank of the matrix of this system is two and equal to the number unknowns, so this system has only the zero solution x 1 = x 3 = 0. x 2 here can be anything other than zero, for example, x 2 = 1. Thus, the vector (0,1,0) is an eigenvector, corresponding to λ = -3. Let's check:
.
If λ = 1, then we obtain the system
The rank of the matrix is two. We cross out the last equation.
Let x 3 be a free unknown. Then x 1 = -3x 3, 4x 2 = 10x 1 - 6x 3 = -30x 3 - 6x 3, x 2 = -9x 3.
Assuming x 3 = 1, we have (-3,-9,1) - an eigenvector corresponding to the eigenvalue λ = 1. Check:

.
Since the eigenvalues are real and distinct, the vectors corresponding to them are linearly independent, so they can be taken as a basis in R 3 . Thus, in the basis , , matrix A has the form:
.
Not every matrix of a linear operator A:R n → R n can be reduced to diagonal form, since for some linear operators there may be less than n linear independent eigenvectors. However, if the matrix is symmetric, then the root of the characteristic equation of multiplicity m corresponds to exactly m linearly independent vectors.

Definition. A symmetric matrix is a square matrix in which the elements symmetric about the main diagonal are equal, that is, in which .
Notes. 1. All eigenvalues of a symmetric matrix are real.
2. The eigenvectors of a symmetric matrix corresponding to pairwise different eigenvalues are orthogonal.
As one of the many applications of the studied apparatus, we consider the problem of determining the type of a second-order curve.

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 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 - 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 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, is equal to the number of 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, let us 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 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 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).

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 for a real number λ to be 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)x = 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 see 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.