Finding the conditions for a particular mapping to be a bijection

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SUMMARY

The discussion focuses on determining the necessary conditions for the linear map T: Mn(K) → Mn(K), defined by T(Y) = AYB, to be a bijection. It concludes that both matrices A and B must be non-singular, meaning they cannot have any rows or columns consisting entirely of zeros. This ensures that the map is injective, as AXB = 0 implies X = 0 when A and B are invertible. The proof of surjectivity follows from the existence of inverses for A and B, confirming that T is indeed a bijection when both matrices are non-singular.

PREREQUISITES
  • Understanding of linear maps and bijections in linear algebra
  • Knowledge of matrix operations, specifically multiplication and inversion
  • Familiarity with concepts of injectivity and surjectivity
  • Basic understanding of fields in mathematics, particularly K
NEXT STEPS
  • Study the properties of linear transformations and their representations
  • Learn about the implications of matrix singularity on linear mappings
  • Explore the concept of matrix rank and its relation to bijections
  • Investigate the use of determinants in determining matrix invertibility
USEFUL FOR

Students of linear algebra, mathematicians studying linear transformations, and anyone interested in the properties of matrix mappings and their bijective conditions.

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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.
 
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what happens if either A or B is singular?

what happens if neither one is?
 
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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