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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


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    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?
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