## Linear Transformation/Injective/Surjective

I am extremely confused when it comes to linearly transformations and am not sure I entirely understand the concept. I have the following assignment question:

Consider the 2x3 matrix
A=
1 1 1
0 1 1
as a linear transformation from R3 to R2.
a) Determine whether A is a injective (one-to-one) function.
b) Determine whether A is a surjective (onto) function.

For a) I said that we need to solve Ax=0 and the matrix then looks like:

1 1 1 : 0
0 1 1 : 0

Since x3 is a free variable A cannot be injective.

For b) I have the matrix:
1 1 1 : *
0 1 1 : *
(note that it doesn't matter what * is)

This matrix is consistent so the matrix A is surjective.

Am I understanding this question correctly?

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 Recognitions: Homework Help Science Advisor Do you know the definitions of injective and surjective?

 Quote by AKG Do you know the definitions of injective and surjective?
The definitions we were given are:

Injective: A linear transformation T: R^p -->R^m is injective (one to one) if and only if the equation Tx=0 has only the solution x=0.

Surjective: If T:R^p --> R^m is linear then T is surjective if and only if the system Tx=b is consistent for all vectors b in all real numbers m.

Recognitions:
Homework Help

## Linear Transformation/Injective/Surjective

By "the system Tx = b is consistent" you mean "the equation Tx = b has a solution" i.e. "there exists x such that Tx = b"?

For part a), find a nonzero vector x such that Ax = 0.

For part b), given a vector b = (b1 b2)T (T denotes transpose), can you find a vector x = (x1 x2 x3)T such that Tx = b?

 Quote by AKG By "the system Tx = b is consistent" you mean "the equation Tx = b has a solution" i.e. "there exists x such that Tx = b"? For part a), find a nonzero vector x such that Ax = 0. For part b), given a vector b = (b1 b2)T (T denotes transpose), can you find a vector x = (x1 x2 x3)T such that Tx = b?
Thank you for your help, I think I understand. If not, I'll be back. Thanks again.