Show that linear transformation is surjective but not injective

[pearl0993]

Hi,

My question is to show that the linear transformation T: M_2x2(F) -> P₂(F) defined by

T (a b c d) = (a-d) | (b-d)x | (c-d)x²

is surjective but not injective.

thanks in advance.

 

[chiro]

Hi,

My question is to show that the linear transformation T: M_2x2(F) -> P₂(F) defined by

T (a b c d) = (a-d) | (b-d)x | (c-d)x²

is surjective but not injective.

thanks in advance.

Hey pearl0993 and welcome to the forums.

What is the space of your answer (i.e. P₂(F))? Is it a 2x2 column vector?

Also it would be helpful to show any working out or any ideas that you have for answering the questions.

Also if you are completely lost with these ideas, this web-page might help you get started:

http://en.wikipedia.org/wiki/Bijection,_injection_and_surjection

 

[Deveno]

i think what the intended mapping was (this is just an educated guess) is:

[a b]
[c d], maps to:

(a-d) + (b-d)x + (c-d)x².

to prove surjectivity, it suffices to exhibit 3 matrices:

one whose image is 1, another whose image is x, and a 3rd whose image is x².

to prove it is not injective, it suffices to exhibit a non-zero matrix that maps to the 0-polynomial.

 

[conquest]

another way to handle the problem is by noting that the space of 2x2 matrices is of dimension three and the space of up to second order polynomials is of dimension 2.

Then proving surjectivity in the way described by Devenoe automatically shows the map is not injective.

Since a linear bijection is an isomorphism of vector spaces and vector spaces that are isomorphic have to at least have equal dimension.

 

[blue_raver22]

Sure, let's first start by defining what it means for a linear transformation to be surjective and injective. A linear transformation is said to be surjective if every element in the codomain (in this case, P2(F)) has at least one preimage in the domain (M2x2(F)). In other words, every element in the range of the transformation can be mapped back to by at least one element in the domain. On the other hand, a linear transformation is said to be injective if every element in the codomain has at most one preimage in the domain. This means that no two distinct elements in the domain map to the same element in the codomain.

Now, let's apply these definitions to the given linear transformation T. We can see that T is surjective because for any polynomial in P2(F), say ax2 + bx + c, we can find a corresponding element in the domain, (a b c 0), such that T(a b c 0) = ax2 + bx + c. This shows that every element in the range of T has at least one preimage in the domain, making T surjective.

However, T is not injective because there are elements in the domain that map to the same polynomial in the codomain. For example, consider the elements (1 1 1 1) and (2 2 2 2) in the domain. Both of these elements map to the polynomial x2, meaning that T(1 1 1 1) = T(2 2 2 2) = x2. This shows that T is not injective since two distinct elements in the domain are mapping to the same element in the codomain.

In summary, we have shown that the given linear transformation T is surjective but not injective. This means that while every element in the codomain can be mapped back to by at least one element in the domain, there are elements in the domain that map to the same element in the codomain, violating the injectivity property.