1. Limited time only! Sign up for a free 30min personal 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!

Finding the conditions for a particular mapping to be a bijection

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data
    here's the problem:

    Let A and B be n x n matrix with coefficient in K (any field), let Mn(K) be the set of all n x n matrix with coefficient in K . T is a linear map defined like this
    T : Mn(K)---> Mn(K)
    T(Y) = AYB

    what are the necessary conditions for T to be a bijection (isomorphism). Prove it


    2. Relevant equations



    3. The attempt at a solution
    For now I think I'm pretty sure I found that the conditions would be that there is no colum or line in which all elements = 0 in A or B. Because if it was the case then you would know that there exists a vector x of K^n such that Ax=0 or Bx = 0 and If X and Y were non equals matrix with colums being scalar multiple of x Then AXB = 0 = AYB which mean that T is not injective.

    Now I think this prove that my condition is necessary but I don't know how to prove that this is the only condition needed. Also I have difficulty proving the injection and the surjection of T once this condition is applied.
     
  2. jcsd
  3. Oct 14, 2011 #2

    Deveno

    User Avatar
    Science Advisor

    what happens if either A or B is singular?

    what happens if neither one is?
     
  4. Oct 14, 2011 #3
    If A or B is singular you get that AXB can be = 0 even if X is not 0 and so obviously T is not a bijection.

    If neither are, I'd say that the only way to get AXB = 0 would be to have X = 0
    So if AXB= AYB then AX=AY → X=Y. Is that right? and so T is injective .

    if Y is in Mn(K). Suppose that T[X] = Y then Y= AXB → (A^-1)Y(B^-1) = X (A and B are invertible since they are not singular) which means that X is in Mn(K) (since the inverse of A and B are in Mn(K) and Y is in Mn(K) ) and so T is surjective and so T is a bijection.

    Am I right?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding the conditions for a particular mapping to be a bijection
  1. Bijective Maps (Replies: 9)

Loading...