Let X be a Banach space and A a bounded linear operator, defined on X, with range in the Banach space Y. Let x нX and f нY*. Then the value f(Ax) is defined, and the inequalities | f(Ax)| £ ||f ||?||Ax|| £ ||f ||?||A||?||x||.

These inequalities show that the linear functional j(х) defined by the equality j(х) = f(Ax) is a bounded functional. Thus, to each linear bounded functional f нY with the help of the operator A, a linear continuous functional j нХ* is put in correspondence. By changing the element f we will get, generally speaking, different elements j; thus we get the operator

defined on Y*, with range in X* space. This operator A* is related to the operator A by the equality (A*f)(x) = f(Ax). If we apply the notation introduced in Section 2 for the linear functional f(x) = (x, f), then the connection of the operators will look symmetrical:

(Ax, f)=(x, A*f). (one)

The operator A* is uniquely determined by formula (1) and is called the operator adjoint to the operator A.

Indeed, if for all x and y the equalities hold

(Ax, y) = (x, A*y) = (x, A 1 *y),

then it follows from Corollary 4 of the Hahn-Banach theorem that A 1 *y= A*y for all y, which means that A*=A 1 *.

Theorem 11. The adjoint operator A* is linear and .

Proof. Let us prove the additivity of the operator A*. Indeed, if y, z нY*, then the above arguments imply the existence of a unique element (y + z)* нX such that (Ax, y + z)=(x, (y + z)*) for all x нX.

On the other hand, using formula (1), we have

(Ax, y + z) = (Ax, y) + (Ax, z) = (x, A*y) + (x, A*z) = (x, A*y + A*z) = (x , (y+z)*),

those. (y+z)* = A*x + A*y, whence A*(y+z)=A*y+A*z. This proves the additivity of the operator A*. Homogeneity is also easily checked.

To calculate the norm of the operator A*, we make the estimates

This implies that the operator A* is bounded and .

The operator A*, in turn, has an adjoint - A**, defined by an equality similar to (1)

(A*y, x) = (y, A**x) (2).

But, since from (2) A**x is uniquely determined for each xОХ, it follows from a comparison of equalities (1) and (2) that

(Ax, y) = (A**x, y) "хОХ, "yОY.

By Corollary 4 of the Hahn-Banach theorem, the latter means that A**x=Ax for all xнX, i.e., A**= A on the space X. Applying the above inequality for the norm of the adjoint operator to A* and A**, we have , which gives the required equality: . The theorem has been proven.

Theorem. 12. If A and B are linear bounded operators from a Banach space X to a Banach space Y, then

1. (A+B)*=A*+B*

2. (λА)*= λА*

3. Assuming X \u003d Y, the equality (AB) * \u003d B * A * is true.

Proof. The above properties follow from the following relations:

1. ((A+B)x, y) = (Ax, y) + (Bx, y) =(x, A*y) + (x, B*y) = (x, (A* + B* )y);

2. ((λA)x ,y) = λ(Ax ,y) = λ(x, A*y) = (x, (λA*y));

3. ((AB)x, y) = (A(Bx), y) = (Bx, A*y) = (x, B*(A*y)) = (x, (B*A*)y ).

The theorem has been proven.

Example 8. In the space L 2, consider the Fredholm integral operator

with a kernel having an integrable square. We have, using the Fubini theorem,

, where

.

Thus, the transition to the adjoint operator is that the integration is carried out over the first variable. Whereas in the original statement it is carried out according to the second one.

More on the topic 6. Adjoint operator. Conditions for the existence of the adjoint operator. Closedness of the adjoint operator. The adjoint operator to the bounded operator and its norm.:

1. 2. Schauder's theorem on the complete continuity of the adjoint operator. Equations of the first and second kind with completely continuous operators. Theorem on the closed range of an operator
2. 1. Linear operators in linear normed spaces. Equivalence between continuity and boundedness of a linear operator. The concept of the norm of a bounded operator. Various formulas for calculating norms. Examples of linear bounded operators.
3. 4. The core of the operator. Boundedness criterion for the inverse operator. Inverse operator theorems
4. 2. The space of linear continuous operators and its completeness with respect to uniform convergence of operators
5. 5. Examples of inverse operators. Invertibility of operators of the form (I - A) and (A - C).
6. 1. Completely continuous operators and their properties. Fredholm and Hilbert-Schmidt operators
7. 6. Graph of an operator and closed operators. Closing criterion. Banach closed graph theorem. Open mapping theorem

From Wikipedia, the free encyclopedia

General linear space

Let $E,\; \backslash ,\; L$ are linear spaces, and $E^*,\; \backslash ,\; L^*$- conjugate linear spaces (spaces of linear functionals defined on $E,\; \backslash ,\; L$). Then for any linear operator $A\backslash colon\; E\backslash to\; L$ and any linear functional $g\; \backslash in\; L^*$ linear functional is defined $f\; \backslash in\; E^*$- superposition $g$ and $A$: $f(x)=g(A(x))$. Display $g\backslash mapsto\; f$ is called the adjoint linear operator and is denoted $A^*\backslash colon\; L^*\; \backslash to\; E^*$.

Briefly, then $(A^*g,\; x)\; =\; (g,\; Ax)$, where $(B,x)$- functional action $B$ per vector $x$.

Topological linear space

Let $E,\; \backslash ,\; L$ are topological linear spaces, and $E^*,\; \backslash ,\; L^*$- conjugate topological linear spaces (spaces continuous linear functionals defined on $E,\; \backslash ,\; L$). For any continuous linear operator $A\backslash colon\; E\backslash to\; L$ and any continuous linear functional $g\; \backslash in\; L^*$ a continuous linear functional is defined $f\; \backslash in\; E^*$- superposition $g$ and $A$: $f(x)=g(A(x))$. It is easy to check that the mapping $g\backslash mapsto\; f$ linear and continuous. It is called the adjoint operator and is also denoted $A^*\backslash colon\; L^*\; \backslash to\; E^*$.

Banach space

Let $A\backslash colon\; X\backslash to\; Y$ is a continuous linear operator acting from a Banach space $X$ into a Banach space $Y$ let it go $X^*,\; Y^*$- conjugate spaces. Denote $\backslash forall\; x\backslash in\; X,\; f\backslash in\; Y^*\; =f(Ax)$. If a $f$ is fixed, then $$ is a linear continuous functional in $X,\; \backslash in\; X^*$. So for $\backslash forall\; f\backslash in\; Y^*$ a linear continuous functional is defined from $X^*$, so the operator is defined $A^*\backslash colon\; Y^*\backslash to\; X^*$, such that $=$.

$A^*$ called conjugate operator. Similarly, one can define an adjoint operator to an unbounded linear operator, but it will not be defined on the entire space.

For $A^*$ the following properties hold true:

• Operator $A^*$- linear.
• If a $A$ is a linear continuous operator, then $A^*$ also a linear continuous operator.
• Let $O$ is the null operator, and $E$ is the unit operator. Then $O^*=O,\; E^*=E$.
• $(A+B)^*=A^*+B^*$.
• $\backslash forall\backslash alpha\backslash in\backslash mathbb\; C,\; (\backslash alpha\; A)^*=\backslash bar(\backslash alpha)A^*$.
• $(AB)^*=B^*A^*$.
• $(A^(-1))^*=(A^*)^(-1)$.

Hilbert space

In Hilbert space $H$ the Riesz theorem gives an identification of a space with its dual, so for the operator $A\backslash colon\; H\; \backslash to\; H$ equality $(Ax,\; y)\; =\; (x,\; A^*y)$ defines the adjoint operator $A^*\backslash colon\; H\; \backslash to\; H$. Here $(x,\; y)$- scalar product in space $H$.

see also

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Notes

Literature

• Schaefer H. Topological vector spaces. - M .: Mir, 1971.
• Vorovich I.I. , Lebedev L.P. functional analysis and its applications in continuum mechanics. - M .: University book, . - 320 s.
• Trenogin V. A. Functional analysis. - M .: Science, . - 495 p.
• Functional analysis / editor S. G. Krein. - 2nd, revised and supplemented. - M .: Science, . - 544 p. - (Reference mathematical library).
• Halmosh P. Finite-dimensional vector spaces = Finite-dimensional vector spaces. - M .: Fizmatgiz, . - 264 p.
• Shilov G.E. Mathematical analysis(functions of one variable), part 3. - M .: Nauka, . - 352 p.

An excerpt characterizing the adjoint operator

Adjutants galloped ahead of him into the yard. Kutuzov, impatiently pushing his horse, which was ambling under his weight, and constantly nodding his head, put his hand to the misfortune of the cavalry guard (with a red band and without a visor) cap that was on him. Having approached the guard of honor of the young grenadiers, mostly cavaliers, who saluted him, for a minute he silently, carefully looked at them with a commanding stubborn look and turned to the crowd of generals and officers standing around him. His face suddenly took on a subtle expression; he shrugged his shoulders with a gesture of bewilderment.
- And with such good fellows, everything retreats and retreats! - he said. “Well, goodbye, general,” he added, and touched the horse through the gate past Prince Andrei and Denisov.
- Hooray! Hurrah! Hurrah! shouted from behind him.
Since Prince Andrei had not seen him, Kutuzov had grown fat, flabby and swollen with fat. But the familiar white eye, and the wound, and the expression of weariness in his face and figure were the same. He was dressed in a uniform frock coat (a whip on a thin belt hung over his shoulder) and in a white cavalry guard cap. He, heavily blurring and swaying, sat on his cheerful horse.
“Fu… fu… fu…” he whistled almost audibly as he drove into the yard. His face expressed the joy of reassuring a man who intends to rest after the representation. He pulled his left leg out of the stirrup, falling down with his whole body and grimacing from the effort, with difficulty brought it to the saddle, leaned on his knee, grunted and went down on his hands to the Cossacks and adjutants who supported him.
He recovered, looked around with his narrowed eyes, and looking at Prince Andrei, apparently not recognizing him, walked with his diving gait to the porch.
“Fu… fu… fu,” he whistled and looked back at Prince Andrei. The impression of Prince Andrei's face only after a few seconds (as is often the case with old people) was associated with the memory of his personality.
“Ah, hello, prince, hello, my dear, let’s go ...” he said wearily, looking around, and heavily entered the porch, creaking under his weight. He unbuttoned and sat down on a bench on the porch.
- Well, what about the father?
“Yesterday I received news of his death,” said Prince Andrei shortly.
Kutuzov looked at Prince Andrei with frightened open eyes, then took off his cap and crossed himself: “Kingdom to him in heaven! May the will of God be over all of us! He sighed heavily, with all his chest, and was silent. “I loved and respected him and I sympathize with you with all my heart.” He embraced Prince Andrei, pressed him to his fat chest and did not let go for a long time. When he released him, Prince Andrei saw that Kutuzov's swollen lips were trembling and there were tears in his eyes. He sighed and grabbed the bench with both hands to stand up.
“Come, come to me, we’ll talk,” he said; but at this time Denisov, as little shy before his superiors as before the enemy, despite the fact that the adjutants at the porch stopped him in an angry whisper, boldly, banging his spurs on the steps, entered the porch. Kutuzov, leaving his hands resting on the bench, looked displeasedly at Denisov. Denisov, having identified himself, announced that he had to inform his lordship of a matter of great importance for the good of the fatherland. Kutuzov began to look at Denisov with a tired look and with an annoyed gesture, taking his hands and folding them on his stomach, he repeated: “For the good of the fatherland? Well, what is it? Speak." Denisov blushed like a girl (it was so strange to see the color on that mustachioed, old and drunken face), and boldly began to outline his plan for cutting the enemy's line of operations between Smolensk and Vyazma. Denisov lived in these parts and knew the area well. His plan seemed undoubtedly good, especially in terms of the force of conviction that was in his words. Kutuzov looked at his feet and occasionally looked back at the yard of a neighboring hut, as if he was expecting something unpleasant from there. Indeed, during Denisov's speech, a general appeared from the hut he was looking at with a briefcase under his arm.
- What? - in the middle of Denisov's presentation, Kutuzov said. - Ready?
“Ready, your grace,” the general said. Kutuzov shook his head, as if to say: "How can one person do all this," and continued to listen to Denisov.
“I give you an honest noble word from a Hussian officer,” said Denisov, “that I am g“ azog ”wu of Napoleon’s messages.

A non-zero element x G V is called an eigenelement of the linear operator A: V V, if there is such a number A - an eigenvalue of the linear operator A, such that Eigenvalues and own elements. Associated operator. does not have its own elements. Let some trigonometric polynomial a cos t + 0 sin t become proportional after differentiation: This means that or, which is the same, The last equality is satisfied if and only if it follows that a = p = 0 and, hence, the polynomial can only be zero. Theorem 6. A real number A is an eigenvalue of a linear operator A if and only if this number is the root of its characteristic polynomial: x(A) = 0. Necessity. Let A be an eigenvalue of the operator A. Then there is a non-zero element x for which Ax = Ax. Let be the basis of the space. Then the last equality can be rewritten in an equivalent matrix form or, which is the same, And this, that x is an eigenelement, it follows that its coordinate column x(c) is nonzero. This means that the linear system (1) has a nonzero solution. The latter is possible only under the condition that or, which is the same, Sufficiency. A way to build your own element. Let A be the root of a polynomial Consider a homogeneous linear system with a matrix A(c) - AI: Due to condition (2), this system has a nonzero solution. Let us construct the element x according to the rule The coordinate column x(c) of this element satisfies the condition or, which is also, The latter is equivalent to or, in more detail, Consequently, x is an eigenelement of the linear operator A, and A is the corresponding eigenvalue. Comment. To find all eigenelements corresponding to a given eigenvalue A, it is necessary to construct the FSR of system (3). Example 1. Find the eigenvectors of a linear operator acting according to the rule (projection operator) (Fig. 6). M Consider the actions of the linear operator P on the basis vectors. We have Write down the operator matrix: Eigenvalues ​​and eigenelements. Associated operator. construct a characteristic polynomial and find its roots. We have Construct homogeneous linear systems with matrices: We get, respectively: Let's find the fundamental systems of solutions for each of these systems. We have 1 Thus, the eigenvectors of this projection operator are: the vector k with eigenvalue 0 and any vector with eigenvalue basis I, t, O has the form the characteristic polynomial -A3 has exactly one root A = 0. The solution of the system is the set 1,0,0, which corresponds to a polynomial of degree zero. §5. Adjoint operator In the Euclidean space over linear operators, one can introduce one more action - the operation of conjugation. Let V be an n-dimensional Euclidean space. With every linear operator acting in this space; another linear operator adjoint to the given one is naturally connected. Definition. A linear operator (read: “a with a star”) is called adjoint to a linear operator A: V - * V, if for any elements x and y from the space V the equality is satisfied Linear operator A *, adjoint this operator Ah, it always exists. Let c = (et,..., en) be the orthobasis of the space V and A = A(c) = (o^) be the matrix of the linear operator A in this basis, i.e., by direct calculations one can verify that for linear operator A": V -> V, defined by the rule, equality (1) holds for any x and y. Recall that, according to Theorem 1, in order to construct a linear operator, it is enough to specify its action on the basic elements. Example. We introduce in linear space M\ polynomials with real coefficients of degree not higher than the first operation of scalar multiplication by next rule. Let Let Thus, M\ be a two-dimensional Euclidean space. Let V: M\ - M\ be the differentiation operator: V(a + d»f) = b. Let us construct an adjoint operator. The operator matrix V in this basis has the form. Then is the matrix of the adjoint operator V, acting according to the rule: For an arbitrary polynomial, we obtain Properties of the conjugation operation 1. For each linear operator, there is exactly one operator conjugate to it. Let B and C be operators conjugate to a given uoperator A. This means that for any elements x and y from the space V, equalities are satisfied. It follows that Eigenvalues ​​and eigenelements. Associated operator. and, further, By virtue of the arbitrariness of the choice of the element x, we conclude that the element Vu-Su is orthogonal to any element of the space V and, in particular, to itself. The latter is possible only in the case when By - Cy = 0 and, therefore, By = C y. Due to the fact that y is an arbitrary element, we obtain B ~ C. 2. (a.4)* = aL*, where a is an arbitrary real number. Let A:V -+ V and B:V -+ V be linear operators. Then Properties 2-5 follow easily from the uniqueness of the adjoint operator. 6. Let c be an orthobasis of the space V. In order for the operators A: V V and B: V -* V to be mutually conjugate, the equalities B = A", A = B* are satisfied, it is necessary and sufficient that their matrices A = A(c) and B = B(c) are obtained from one another by transposition. Remark. We emphasize that property 6 is valid only for the matrix 7. If a linear operator A is non-degenerate, then its conjugate operator A* is also non-degenerate, and the equality

Let us study additional properties of linear operators related to the concept of orthogonality in Euclidean space. Let us first prove the following property: if A and B are linear operators acting in n-dimensional Euclidean space V, and ( x , Ay ) = (x , By ), x , y V, then A = B .

Indeed, putting in equality ( x , Ay ) = (x , By ) Û ( x , (A – B )y ) = 0 vector x = (A – B )y , we get (( A –B )y , (A – B )y ) = ||(A – B )y || 2 = 0, y V, which is equivalent to ( A – B )y = 0 , y V, i.e. A – B = O , or A = B .

Definition 11.1. Linear operator A * called conjugate operator A , if

(Ax , y ) = (x , A * y ), x , y V. (11.1)

The question naturally arises: does there exist for a given operator A conjugate?

Theorem 11.1. Each line operator A has a single adjoint operator A * .

Proof. Let's choose in space V orthonormal basis u 1 , u 2 ,…, u n. Every linear operator A : V® V in this basis corresponds the matrix BUT = , i, j = 1, 2,..., n. Let be the matrix obtained from the matrix BUT transposition. It corresponds to the linear operator B . Then

(Au j, u i) = (a 1 ju 1 + a 2 ju 2 +…+ and nju n, u i) = and ij;

(u j, Bu i) = (u j, a i 1 u 1 + a i 2 u 2 +…+ and inu n) = and ij.

(Au j, u i) = (u j, Bu i), i, j = 1, 2,..., n. (11.2)

Let further x = x 1 u 1 + x 2 u 2 +…+ x nu n and y = at 1 u 1 + at 2 u 2 +…+ at nu n are any two vectors from V. Consider the scalar products ( Ax , y ) and ( x , By ):

(Ax , y ) = (Au j, u i),

(x , By ) = (u j, Bu i).

Comparing these expressions, taking into account equality (11.2) and the property noted above, we obtain the equality ( Ax , y ) = (x , By ), x , y V, i.e. B = A * .

Thus, we have proved that for every linear operator A in a finite-dimensional Euclidean space there is an operator adjoint to it A * whose matrix in any orthonormal basis is transposed with respect to the operator matrix A . Operator uniqueness A * follows from the definition of the adjoint operator and the property proved above.¨

It is easy to verify that the operator A * adjoint to the linear operator A , is linear.

So the operator A * is linear and has a corresponding matrix A*. Therefore, the matrix relation corresponding to formula (11.1) has the form

(BUTx , y ) = (x , A * y ), x , y V.

The adjoint operators have the following properties:

1°. E * = E .

2°. ( A *) * = A .

3°. ( A + B ) * = A * + B * .

4°. ( BUT ) * = A * , R.

5°. ( AB ) * = B * A * .

6°. ( A –1) * = (A *) –1 .

The validity of properties 1°–5° follows from the properties of matrix transposition.

Let us verify the validity of property 6°. Let A -1 exists. Then from the equalities AA –1 = A –1 A = E and properties 1°, 5° it follows that ( AA –1) * = (A –1 A ) * = E * = = E and ( AA –1) * = (A –1) * A * , (A –1 A ) * = A * (A –1) * , i.e. that ( A –1) * = (A *) -one . From this we obtain another important property of matrix transposition:

(A –1) * = (A *) –1 .

Example 1 Let A – rotation of the Euclidean plane R 2 per corner j with matrix

in an orthonormal basis i , j . Then the matrix of the adjoint operator in this basis is

= .

Consequently, A * - rotation of the plane by an angle j in the opposite direction.·

Reverse operator

Let V be a linear space over a field P, and let A be an operator (not necessarily linear) acting on V.

Definition. An operator A is said to be invertible if there exists an operator B acting on V such that BA = AB = I.

Definition. The operator B that satisfies the condition BA = AB = I is called the inverse of A and is denoted.

Thus, the operator inverse to the operator A satisfies the condition A = A = I. For an invertible operator A, the equalities Ax = y and y = x are equivalent. Indeed, let Ax = y, then y = (Ax) = (A)x = Ix = x.

If y = x, then

Ax \u003d A (y) \u003d (A) y \u003d Iy \u003d y.

Theorem. If a linear operator is invertible, then its inverse operator is also linear.

Proof. Let A be an invertible linear operator acting in a linear space V over a field P, let A be an operator inverse to A. Take arbitrary vectors and numbers. Let . Then A=, A=. Due to the linearity of the operator A

From here we get:

= = ,

That is, the operator is linear.

Adjoint Linear Operator

Let two unitary spaces X, Y be given.

Definition. The operator A*, acting from Y to X, is called adjoint with respect to the operator A, acting from X to Y, if for any vectors xX, yY the equality

(Ax, y) = (x, A*y). (one)

Theorem. For any linear operator A there is an adjoint operator A*, and only one.

Proof. Let us choose some orthonormal basis in X. For any vector xX, there is a decomposition

If the operator A* exists, then, according to this formula, for any vector yY we have

Or by definition

But this means that if the operator A* exists, then it is unique.

The operator A* constructed in this way is linear. It also satisfies the equality (Ax, y) = (x, A*y). Indeed, taking into account the orthonormality of the system and taking into account (1), (2), we obtain for any vectors xX, yY

(Ax, y) = (A) =,

(x, A*y) = (A) =

The theorem has been proven.

The adjoint operator A* is related to the operator A by certain relations. Let's note some of them:

Proof. Consider an arbitrary operator A and its adjoint operator A*. In turn, for the operator A* the operator (A*)* will be adjoint. Now for any xX, yY we have

(y, (A*)*x) = (A*y, x) == = (y, Ax).