Linear algebra inner products, self adjoint operator,unitary operation

In summary: Yes. Note that you have a whole eigenspace of functions that are zero on one of these regions. The set of functions that are zero on ##[0, 2/3]## is a vector subspace - and is the eigenspace of your operator corresponding to eigenvalue ##2##.
  • #1
Karl Karlsson
104
12
Homework Statement
Let ##V = \mathbb{C}([0, 1], \mathbb{C})## be the vector space of continuous functions ##f: [0,1] \rightarrow \mathbb{C}## with inner product:
$$\langle f|g \rangle = \int_0^1 \overline {f(t)}g(t)\;\mathrm{d}t$$

For ##h \in V## , let ##L_h: V \rightarrow V## be the operator defined by ##L_h(f)(t)=h(t)f(t)##.
b) Determine the adjoint operator to ##L_h##
c) For which functions is the operator ##L_h## self adjoint?
d) For which functions is the operator ##L_h## unitary?
e) Give example of one function h(t) such that the operator ##L_h## is self adjoint and has at least two different eigenvalues
Relevant Equations
$$\langle f|g \rangle = \int_0^1 \overline {f(t)}g(t)\;\mathrm{d}t$$
##L_h(f)(t)=h(t)f(t)##
b)
Skärmavbild 2020-11-15 kl. 22.21.25.png

c and d):
In c) I say that ##L_h## is only self adjoint if the imaginary part of h is 0, is this correct?
Skärmavbild 2020-11-15 kl. 22.23.27.png

e) Here I could only come up with eigenvalues when h is some constant say C, then C is an eigenvalue. But I' can't find two_Otherwise does b-d above look correct?

Thanks in advance!
 
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  • #2
It's looks all right. For d) perhaps think about writing ##h(t)## in polar form - that might be neater.

For e), remember these functions are only continuous. You have lots of options for where they are zero and non-zero.
 
  • #3
PeroK said:
It's looks all right. For d) perhaps think about writing ##h(t)## in polar form - that might be neater.
Do you mean having ##a(t)=r\cdot cos(t)## and ##b(t)=r\cdot sin(t)## and r=1 ?
PeroK said:
For e), remember these functions are only continuous. You have lots of options for where they are zero and non-zero.
Like what? I can't see how it would work for cos(t) ##e^t## t or any other normal continuous function
 
  • #4
Karl Karlsson said:
Do you mean having ##a(t)=r\cdot cos(t)## and ##b(t)=r\cdot sin(t)## and r=1 ?

Like what? I can't see how it would work for cos(t) ##e^t## t or any other normal continuous function

I was thinking ##e^{ia(t)}## for some real function ##a(t)##. But, what you have is fine.

Try thinking about continuous functions more generally. Something like ##\cos t## is no good. That would never work. What's "normal"?
 
  • #5
PeroK said:
Try thinking about continuous functions more generally. Something like ##\cos t## is no good. That would never work. What's "normal"?
I was just mentioning som elementary continuous functions that came to my mind but none of those seem to work and I can't find one that does.
 
  • #6
Karl Karlsson said:
I was just mentioning som elementary continuous functions that came to my mind but none of those seem to work and I can't find one that does.
What about a continuous step-function?
 
  • #7
PeroK said:
What about a continuous step-function?
Ooh, right! You mean if I for example said h(t)=1 in the intervall 0<t<0.5 and h(t)=2 in the intervall 0.5<t<1. One eigenvector for the eigenvalue 1 could be the function f(t) which is ##t-0.5## on the intervall 0<t<0.5 and f(t)=0 on 0.5<t<1 ? And for for the eigenvalue 2 an eigenvector could be g(t) = 0 on 0<t<0.5 and ##g(t) =t-0.5## on 0.5<t<1?
 
  • #8
Karl Karlsson said:
Ooh, right! You mean if I for example said h(t)=1 in the intervall 0<t<0.5 and h(t)=2 in the intervall 0.5<t<1. One eigenvector for the eigenvalue 1 could be the function f(t) which is ##t-0.5## on the intervall 0<t<0.5 and f(t)=0 on 0.5<t<1 ? And for for the eigenvalue 2 an eigenvector could be g(t) = 0 on 0<t<0.5 and ##g(t) =t-0.5## on 0.5<t<1?
That's not a continuous step function. You need to join the steps. The eigenfunctions would be any function that is zero outside one of the steps.
 
  • #9
PeroK said:
That's not a continuous step function. You need to join the steps. The eigenfunctions would be any function that is zero outside one of the steps.
Yeah, I realized that afterwards. So i can choose h = 1 on 0 <= t <= 1/3 and h=3t on 1/3 < t < 2/3 and h=2 on 2/3 <=t <= 1. An eigenvector to eigenvalue 1 could be a function that is t-1/3 on the intervall 0<=t<=1/3 and 0 on 1/3<t<=1. An eigenvector to eigenvalue 2 could be a function that is 0 on the interval 0<=t<2/3 and t-2/3 on 2/3<=t<=1. Is this correct?
 
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  • #10
Karl Karlsson said:
Yeah, I realized that afterwards. So i can choose h = 1 on 0 <= t <= 1/3 and h=3t on 1/3 < t < 2/3 and h=2 on 2/3 <=t <= 1. An eigenvector to eigenvalue 1 could be a function that is t-1/3 on the intervall 0<=t<=1/3 and 0 on 1/3<t<=1. An eigenvector to eigenvalue 2 could be a function that is 0 on the interval 0<=t<2/3 and t-2/3 on 2/3<=t<=1. Is this correct?
Yes. Note that you have a whole eigenspace of functions that are zero on one of these regions. The set of functions that are zero on ##[0, 2/3]## is a vector subspace - and is the eigenspace of your operator corresponding to eigenvalue ##2##.
 

Related to Linear algebra inner products, self adjoint operator,unitary operation

1. What is a linear algebra inner product?

A linear algebra inner product is a mathematical operation that takes two vectors as inputs and produces a scalar value as output. It is used to measure the magnitude and direction of one vector in relation to another vector.

2. What is a self adjoint operator in linear algebra?

A self adjoint operator is a linear operator that is equal to its own adjoint, or conjugate transpose. This means that the operator and its adjoint have the same matrix representation, and the operator can be considered to be its own inverse.

3. What is a unitary operation in linear algebra?

A unitary operation is a linear transformation that preserves the inner product of vectors. This means that the length and angle between vectors are not changed by the operation. Unitary operations are important in quantum mechanics and signal processing.

4. How are inner products, self adjoint operators, and unitary operations related?

Inner products are used to define the concept of orthogonality, which is a key property of self adjoint operators and unitary operations. Self adjoint operators are also unitary, meaning they preserve the inner product, and unitary operations are self adjoint when they are their own inverse.

5. What are some applications of linear algebra inner products, self adjoint operators, and unitary operations?

These concepts have many applications in mathematics, physics, and engineering. They are used in quantum mechanics to describe the evolution of quantum systems, in signal processing to analyze and manipulate signals, and in optimization problems to find the optimal solution. They are also used in computer graphics and machine learning algorithms.

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