An Injective Composition of Linear Transformation

Question

Suppose that A is a linear transformation from vector spaces U to V and that B is a linear transformation from vector spaces V to W. Suppose further that B composed of A is an injective composition of the aforementioned linear transformations.

For this composition to be injective, do A and B have to be injective? My intuition tells me that B HAS to be injective while it is not required for A to be injective. Also, how would one prove that either A or B have to be injective.

Also, I am quite fairly new to this site and I have no idea how I can use all of the math symbols and notation that everyone else here is using, so if someone can help me out on that then that would be great.

Answer 1

If $A$ is not injective, there are two elements, $x,y \in U$ such that $x \neq y$, but $A(x) = A(y)$. Then $B$ is stuck, it can't tell $A(x)$ and $A(y)$ apart and sends them both to the same place, so $B(A(x)) = B(A(y))$ and the composition is not injective.

Depending on the details of your notation, it's possible for the image of $A$ to not be all of $V$, so that $B$ could fail to be injective, but $A$ can't send two elements to places that $B$ conflates. I doubt that you are using notation this way; I expect that the image of $A$ is all of $V$ and the image of $B$ is all of $W$. In this case, for the composition to be injective, both $A$ and $B$ must be so:

• Were $B$ not injective, there would be two elements of the image of $A$ that were sent to the same element of $W$. Since these came from (at least) two elements $U$, the composition is not injective.
• Were $A$ not injective, the argument above would apply and the composition would not be injective.

Therefore, if the composition is injective, so are $A$ and $B$.