1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vector Spaces and Correspondence

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    This question came out of a section on Correspondence and Isomorphism Theorems

    Let [itex]V[/itex] be a vector space and [itex]U \neq V, \left\{ \vec{0} \right\} [/itex] be a subspace of [itex]V[/itex]. Assume [itex]T \in L(V,V)[/itex] satisfies the following:
    a) [itex]T(\vec{u} ) = \vec{u}[/itex] for all [itex]\vec{u} \in U[/itex]
    b) [itex]T(\vec{v} + U) = \vec{v} + U[/itex] for all [itex]\vec{v} \in V[/itex]
    Set [itex]S=T-I_{V}[/itex]. Prove that [itex]S^{2}=\vec{0}_{V \rightarrow V}[/itex]

    2. Relevant equations
    [itex]I_{V}[/itex] is the identity map
    [itex]L(V,V)[/itex] is the map of all linear operators on V

    3. The attempt at a solution
    I have trouble understanding the question.
    Since [itex]T \in L(V,V)[/itex] then how is [itex]T(\vec{v} + U) = \vec{v} + U[/itex] for all [itex]\vec{v} \in V[/itex]?
    Wouldn't that mean [itex]T:V/U \rightarrow V/U[/itex]?
    I don't understand, what is [itex]T(\vec{v})[/itex] equal to?
    Does [itex]T(\vec{v})=\vec{v}[/itex] or [itex]T(\vec{v}) = [\vec{v}]_W[/itex] or something else?

    I'm sorry if this is a silly question.
     
  2. jcsd
  3. Oct 8, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    With [itex]T(v+U)=v+U[/itex], they mean that the set v+U is mapped to the set v+U.
    So

    [tex]T(\{v+u~\vert~u\in U\})=\{v+u~\vert~u\in U\}[/tex]

    Or in another way: for each [itex]v\in V[/itex] and [itex]u\in U[/itex], there exists [itex]u^\prime\in U[/itex] such that [itex]T(v+u)=v+u^\prime[/itex].
     
  4. Oct 8, 2012 #3
    This is where I confuse myself. T is a linear transformation from V to V. If it maps the set v+U to the set v+U, then wouldn't that be mapping cosets of V mod U to cosets of V mod U instead of mapping V to V?
     
  5. Oct 9, 2012 #4
    I think I get it now. The function maps the coset of V mod U to the same coset of V mod U by mapping each individual element to another element in that coset.

    I'm sorry if it seemed I brushed over your post and didn't completely read it. I did. I just didn't understand it. I'm having trouble with my Linear Algebra 2 class, and I'm glad that you responded. Thank you a lot!

    So then I get
    Let [itex]\vec{v} \in V[/itex] be arbitrary
    [itex]S^2(\vec{v} + \vec{u}), \vec{u} \in U[/itex]
    [itex]=S(T(\vec{v} + \vec{u}) - I(\vec{v} + \vec{u}))[/itex]
    [itex]=S(\vec{v} + \vec{u}` - ( \vec{v} + \vec{u}))[/itex] where [itex]\vec{u}` \in U[/itex]
    [itex]=S(\vec{u}` - \vec{u})[/itex]
    [itex]=S(\vec{y})[/itex] where [itex]\vec{y} = \vec{u}`- \vec{u} \in U[/itex]
    [itex]=T(\vec{y}) - I(\vec{y})[/itex]
    [itex]=\vec{y} - \vec{y}=\vec{0}[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Vector Spaces and Correspondence
  1. Vector Space (Replies: 1)

  2. Vector space (Replies: 8)

  3. Vector Space (Replies: 3)

  4. Not a Vector Space (Replies: 3)

  5. Vector Space (Replies: 10)

Loading...