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Linear operator, linear functional difference?

  1. Dec 13, 2014 #1
    What is a difference between linear operator and linear functional?

    Do I understand it correctly that linear operator is any operator that when applied on a vector from a vector space, gives again a vector from this vector space. And also obeys linearity conditions.

    And linear functional is a vector, dot product of which with the vector from the vector space gives again a vector from this vector space. And also linearity conditions are met.

    So, linear functional is a vector and linear operator is not a vector?
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  3. Dec 13, 2014 #2


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    Let V be a vector space defined on a field F(don't worry about the word "field", just think about real numbers). A linear operator is a linear map from V to V. But a linear functional is a linear map from V to F. So linear functionals are not vectors. In fact they form a vector space called the dual space to V which is denoted by [itex] V^* [/itex]. But when we define a bilinear form on the vector space, we can use it to associate a vector with a functional because for a vector v, [itex] \langle v, () \rangle: V \rightarrow F [/itex] is a functional in the dual space.
    Last edited: Dec 13, 2014
  4. Dec 13, 2014 #3


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  5. Dec 13, 2014 #4
    Suppose we have a vector space V(x1,x2,x3) where x1,x2,x3 ∈ R
    And a vector u = <1, 2, 3>

    Then for instance:

    Linear operator: D(u) = 2 * u , u ∈ V
    Linear functional: D(u) = <1, 1, 1> ⋅ u , u ∈ V

  6. Dec 13, 2014 #5


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    Yeah, that's right!
    But you should note that functionals don't have to be defined in terms of vectors. What I explained is a way of making a correspondence(which technically is called isomorphism) between vectors and functionals.
    I can define a functional as [itex] \omega(u)=u_1+u_2+u_3 [/itex] which, by your construction, is associated with the vector <1,1,1>, because we have [itex] \omega(u)=D(u) [/itex].
    I also recommend you to read the pages that PeroK gave the links to. Because I just answered the question while there are more things to it!
  7. Dec 13, 2014 #6
    Thank you,

    Is it by definition that the inner product of functional with the vector from V has to be a value from the field of the vector space (in our case real number)?

    Or these can also be considered as functional for the space V, since they outcome a real number which is an instance from the filed of V?

    1. ω(u) = u1
    2. ω(u) = u2 + 5
    3. ω(u) = 0
  8. Dec 13, 2014 #7


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    You take the inner product of two vectors, not a functional and a vector(See the link bilinear form above!).
    Yes, its the definition of inner product to take two vectors and give a scalar(a member of the field)!
    These are functionals too.
    I don't see any connection between your two questions that can make you use that "Or" between them!
  9. Dec 13, 2014 #8
    As far as I understand.

    By definition, dual space for vector space V is a vector space consisting of all linear functionals for vector space V.

    If a vector space given by components x1,x2 ... xn, then linear functional can be written as a function

    Or it can be written in a matrix form:

    From here, we have a vector space of row vectors [a1 ... an] which form a dual space. (Am I correct?)
    Which implies that number of vector dimensions n must be the same for dual vector space and actual vector space.

    If n = 3, then

    In this case:

    1. ω(u) = u1 + u2 + u3

    We have <1, 1, 1> vector in a dual vector space, right?

    But what are the vectors in the dual vector space formed by these functionals then?

    1. ω(u) = u1
    2. ω(u) = u2 + 5
    3. ω(u) = 0
    Last edited: Dec 13, 2014
  10. Dec 13, 2014 #9


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    Yeah, that's correct.
    Yes, this is correct too. But not completely. As I said, you should read some detailed text.

    1. (1 \ \ 0 \ \ 0) \\
    3. (0 \ \ 0 \ \ 0)
    The second one is not a linear functional:
    \omega(u+v)=(u+v)_1+5=u_1+v_1+5 \neq u_1+5+v_1+5=\omega(u)+\omega(v)
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