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Linear Transformation/Injective/Surjective

 
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Sep22-06, 09:10 PM   #1
 

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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Sep22-06, 09:15 PM   #2
AKG
 
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Do you know the definitions of injective and surjective?
 
Sep22-06, 09:21 PM   #3
 
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.
 
Sep22-06, 09:26 PM   #4
AKG
 
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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?
 
Sep22-06, 09:40 PM   #5
 
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.
 
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