# Show that linear transformation is surjective but not injective

1. Feb 5, 2012

### pearl0993

Hi,

My question is to show that the linear transformation T: M2x2(F) -> P2(F) defined by

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

is surjective but not injective.

2. Feb 6, 2012

### chiro

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

3. Feb 6, 2012

### 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.

4. Feb 9, 2012

### 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.