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Linear Transformation/Injective/Surjective |
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| Sep22-06, 09:10 PM | #1 |
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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 |
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Recognitions:
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Do you know the definitions of injective and surjective?
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| Sep22-06, 09:21 PM | #3 |
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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 |
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Recognitions:
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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 |
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Thanks again.
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