Finding the conditions for a particular mapping to be a bijection

  • Thread starter venatorr
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  • #1
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Homework Statement


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


Homework Equations





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.
 

Answers and Replies

  • #2
Deveno
Science Advisor
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what happens if either A or B is singular?

what happens if neither one is?
 
  • #3
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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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