Finding the conditions for a particular mapping to be a bijection

In summary, the problem is that the necessary condition for T to be a bijection is that there is no colum or line in which all elements = 0 in either A or B. If either A or B are singular, then T is not a bijection.
  • #1
venatorr
4
0

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

what happens if neither one is?
 
  • #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?
 

Related to Finding the conditions for a particular mapping to be a bijection

1. What does it mean for a mapping to be a bijection?

A mapping is considered a bijection if it is both one-to-one and onto. This means that each element in the domain is paired with a unique element in the range, and every element in the range is mapped to by at least one element in the domain.

2. How can I determine if a mapping is a bijection?

To determine if a mapping is a bijection, you can use the horizontal line test. This involves drawing horizontal lines across the graph of the mapping and checking if they intersect the graph at more than one point. If there is only one intersection point for each horizontal line, then the mapping is a bijection.

3. What are the conditions for a mapping to be a bijection?

The conditions for a mapping to be a bijection are that it must be both one-to-one and onto. This means that each element in the domain must be paired with a unique element in the range, and every element in the range must be mapped to by at least one element in the domain.

4. Can a mapping be a bijection if it is not one-to-one?

No, a mapping cannot be a bijection if it is not one-to-one. For a mapping to be a bijection, it must be both one-to-one and onto. If it is not one-to-one, then there will be at least one element in the domain that is mapped to by more than one element in the range, violating the condition of being one-to-one.

5. What happens if a mapping is not onto?

If a mapping is not onto, then there will be at least one element in the range that is not mapped to by any element in the domain. This means that the mapping is not a bijection, as it does not satisfy the condition of being onto. It may still be a one-to-one mapping, but it is not a bijection.

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