Linear Transformation/Injective/Surjective

[embury]

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?

 

[AKG]

Do you know the definitions of injective and surjective?

 

[embury]

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.

 

[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?

 

[embury]

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.