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Discusses basic properties and classifications of operators.

We start with a simple but useful property.

Lemma 1 If A B ( X , X ) and x , A x = 0 for all x X , then A = 0 .

Let a C , then for any x , y X , we have that x + a y X . Therefore, we obtain

0 = x + a y , A ( x + a y ) , = x + a y , A x + a A y , = x , A x + a ¯ x , A y + a y , A x + | a | 2 y , A y .

Since x , y X , we have that x , A x = y , A y = 0 ; therefore, 0 = a ¯ x , A y + a y , A x .

If we set a = 1 , then 0 = x , A y + y , A x . So in this case, x , A y = - y , A x .

If we set a = i , then 0 = - i x , A y + i y , A x . So in this case, x , A y = y , A x .

Thus, x , A y = 0 for all x , y X , which means A y = 0 for all y X . So we can come to the conclusion that A = 0 .

Solutions to operator equations

Assume X and Y are two normed linear spaces and A B ( X , Y ) is a bounded linear operator. Now pick y Y . Then we pose the question: does a solution x ^ X to the equation A x = y exist?

There are three possibilities:

  1. A unique solution exists;
  2. multiple solutions exist; or
  3. no solution exists.

We consider these cases separately below.

  1. Unique solution : Assume x and x 1 are two solutions to the equation. In this case we have A x = A x 1 . So A ( x - x 1 ) = 0 . Therefore, x - x 1 N ( A ) . If the solution x is unique then we must have x = x 1 and x - x 1 = 0 . Therefore, N ( A ) = { 0 } . Since the operator has a trivial null space, then A - 1 exists. Thus, the solution to the equation is given by x ^ = A - 1 y ;
  2. Multiple solutions : In this case, we may prefer to pick a particular solution. Often, our goal is to find the solution with smallest norm (for example, to reduce power in a communication problem). Additionally, there is a closed-form expression for the minimum-norm solution to the equation A x = y . Theorem 1 Let X , Y be Hilbert spaces, y Y , and A B ( X , Y ) . Then x ^ is the solution to A x = y if and only if x ^ = A * z , where z Y is the solution of A A * z = y . Let x 1 be a solution to A x = y . Then all other solutions can be written as x ^ = x 1 - u , where u N ( A ) . We therefore search for the solution that achieves the minimum value of | | x 1 - u | | over u N ( A ) , i.e., the closest point to x 1 in N ( A ) . Assume u ^ is such a point; then, x 1 - u ^ N ( A ) . Now, recall that ( N ( A ) ) = R ( A * ) , so x ^ = x 1 - u ^ R ( A * ) . Thus x ^ = A * z for some z Y , and y = A x ^ = A A * z . Note that if A A * is invertible, then x ^ = A * z = A * ( A A * ) - 1 y .
  3. No solution : In this case, we may aim to find a solution that minimizes the mismatch between the two sides of the equation, i.e., y - A x . This is the well-known projection problem of y into R ( A ) . Theorem 2 Let X , Y be Hilbert spaces, y Y , and A B ( X , Y ) . The vector x ^ minimizes | | x - A y | | if and only if A * A x ^ = A * y . Denote u = A x so that u R ( A ) . We need to find the minimum value of | | y - u | | over u R ( A ) . Assume u ^ is the closest point to y in R ( A ) , then y - u ^ R ( A ) , which means y - u ^ ( R ( A ) ) . Recall that ( R ( A ) ) = N ( A * ) , which implies that y - u ^ N ( A * ) . So A * ( y - u ^ ) = 0 . Thus A * y = A * u ^ . Now, denoting u ^ = A x ^ , we have that A * y = A * A x ^ . Note that if A * A is invertible, then we have x ^ = ( A * A ) - 1 A * y .

Unitary operator

Definition 1 An operator A B ( X , X ) is said to be unitary if A A * = A * A = I .

This implies that A * = A - 1 . Unitary operators have norm-preservation properties.

Theorem 3 A B ( X , X ) is unitary if and only if R ( A ) = X and | | A x | | = | | x | | for all x X .

Let A be unitary, then for any x X ,

| | x | | = x , x = x , A A * x = A x , A x = | | A x | | .

Since I : X X and A A * : R ( A ) R ( A ) , then X = R ( A ) . So if A B ( X , X ) is unitary, then R ( A ) = X and | | A x | | = | | x | | for all x X . From [link] we can find that

0 = | | x | | 2 - | | A x | | 2 = x , x - A x , A x = x , x - x , A * A x = x , x - A * A x .

Since this is true for all x X we have that x - A * A x = 0 for all x X , which means that x = A * A x for all x X . Therfore, we must have A * A = I . Additionally, since the operator is unitary, we find that for any x , y X we have that | | A x - A y | | = | | A ( x - y ) | | = | | x - y | | . So x = y if and only if A x = A y , implying that A is one-to-one. Since R ( A ) = X , then A is onto as well. Thus, A is invertible, which means A - 1 A = I . So A - 1 A = I = A * A , and therefore A - 1 = A * . The result is that A is unitary. We have shown that if R ( A ) = X and | | A x | | = | | x | | for all x X , then A B ( X , X ) is unitary.

Corollary 4 If X is finite-dimensional, then A is unitary if and only if | | A x | | = | | x | | for all x X .

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Source:  OpenStax, Introduction to compressive sensing. OpenStax CNX. Mar 12, 2015 Download for free at http://legacy.cnx.org/content/col11355/1.4
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