Self-Adjoint and Normal Operators
[32/0%]Suppose is a positive integer. Define by
Find a formula for .
Suppose . Prove that
Suppose and . Prove that
Suppose and is a subspace of . Prove that
Suppose . Suppose is an orthonormal basis of and is an orthonormal basis of . Prove that
The numbers in the equation above depend on the orthonormal basis , but the right side does not depend on . Thus the equation above shows that the sum on the left side does not depend on which orthonormal basis is used.
Suppose . Prove that
is injective is surjective;
is surjective is injective.
Prove that if , then
;
.
Suppose is an -by- matrix with entries in . Use (b) in Exercise 7A.7 to prove that the row rank of equals the column rank of .
This exercise asks for yet another alternative proof of a result that was previously proved in 3.57 and 3.133.
Theorems 3.57 and 3.133 (column rank equals row rank): suppose . Then the column rank of equals the row rank of .
Prove that the product of two self-adjoint operators on is self-adjoint if and only if the two operators commute.
Suppose and . Prove that is self-adjoint if and only if
for all .
Define an operator by .
Find a formula for .
Show that is normal but not self-adjoint.
Find all eigenvalues of .
If , then is the operator on of counterclockwise rotation by .
An operator is called skew if
Suppose that . Prove that is normal if and only if there exist commuting operators and such that is self-adjoint, is a skew operator, and .
Suppose . Define by for all .
Find all eigenvalues of .
Find the minimal polynomial of .
Define an inner product on by . Define an operator by
Show that with this inner product, the operator is not self-adjoint.
The matrix of with respect to the basis is
This matrix equals its conjugate transpose, even though is not self-adjoint. Explain why this is not a contradiction.
Suppose is invertible. Prove that
is self-adjoint is self-adjoint;
is normal is normal.
Suppose .
Show that the set of self-adjoint operators on is a subspace of .
What is the dimension of the subspace of in (a) (in terms of )?
Suppose . Show that the set of self-adjoint operators on is not a subspace of .
Suppose . Show that the set of normal operators on is not a subspace of .
Suppose and for every . Prove that is normal.
This exercise fails on infinite-dimensional inner product spaces, leading to what are called hyponormal operators, which have a well-developed theory.
Suppose is such that . Prove that the following are equivalent.
-
is self-adjoint.
-
is normal.
-
There is a subspace of such that .
Suppose is the differentiation operator defined by . Prove that there does not exist an inner product on that makes a normal operator.
Give an example of an operator such that is normal but not self-adjoint.
Suppose is a normal operator on . Suppose also that satisfy the equations
Show that .
Suppose and
is the minimal polynomial of . Prove that the minimal polynomial of is
This exercise shows that the minimal polynomial of equals the minimal polynomial of if .
Suppose . Prove that is diagonalizable if and only if is diagonalizable.
Fix . Define by for every .
Prove that if is a real vector space, then is self-adjoint if and only if the list is linearly dependent.
Prove that is normal if and only if the list is linearly dependent.
Suppose is normal. Prove that
for every positive integer .
Suppose is normal. Prove that if , then the minimal polynomial of is not a polynomial multiple of .
Prove or give a counterexample: If and there is an orthonormal basis of such that for each , then is normal.
Suppose that is normal and . Suppose . Prove that .
Fix a positive integer . In the inner product space of continuous real-valued functions on with inner product , let
Define by . Show that . Conclude that is normal but not self-adjoint.
Define by . Show that is self-adjoint.
Suppose is a linear map. Show that under the standard identification of with (see 6.58) and the corresponding identification of with , the adjoint map corresponds to the dual map . More precisely, show that
for all , where and are defined as in 6.58.
6.58 (Riesz representation theorem, revisited): suppose is finite-dimensional. For each , define by for each . Then is a one-to-one function from onto .