Prove that the set T:={x∈Rn:Ax∈S} is a subspace of Rn.

  • Thread starter squenshl
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  • #1
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Homework Statement:
1. Let ##S## be a subspace of ##\mathbb{R}^m## and let ##A## be a ##m\times n## matrix.
Prove that the set ##T:= \left\{\mathbf{x}\in \mathbb{R}^n:A\mathbf{x}\in S\right\}## is a subspace of ##\mathbb{R}^n##.
Relevant Equations:
None
1. Lets show the three conditions for a subspace are satisfied:
Since ##\mathbf{0}\in \mathbb{R}^n##, ##A\times \mathbf{0} = \mathbf{0}\in S##.
Suppose ##x_1, x_2\in \mathbb{R}^n##, then ##A(x_1+x_2) = A(x_1)+A(x_2)\in S##.
Suppose ##x\in S## and ##\lambda\in \mathbb{R}##, then ##A(\lambda x) = \lambda A(x)\in S##.

Is this correct???
 
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Answers and Replies

  • #2
member 587159
Problem Statement: 1. Let ##S## be a subspace of ##\mathbb{R}^m## and let ##A## be a ##m\times n## matrix.
Prove that the set ##T:= \left\{\mathbf{x}\in \mathbb{R}^n:A\mathbf{x}\in S\right\}## is a subspace of ##\mathbb{R}^n##.
Relevant Equations: None

1. Lets show the three conditions for a subspace are satisfied:
Since ##\mathbf{0}\in \mathbb{R}^n##, ##A\times \mathbf{0} = \mathbf{0}\in S##.
Suppose ##x_1, x_2\in \mathbb{R}^n##, then ##A(x_1+x_2) = A(x_1)+A(x_2)\in S##.
Suppose ##x\in S## and ##\lambda\in \mathbb{R}##, then ##A(\lambda x) = \lambda A(x)\in S##.

Is this correct???

Not really. At least the way you write it down does not fully convince me that you have the right idea.

You have to take ##x_1,x_2## in ##T## and then show that ##x_1+x_2\in T##.

Similarly you have to take ##x\in T, \lambda \in \mathbb{R}## and show that ##\lambda x\in T##.

Please also indicate where you use that ##S## is a subspace.

Addendum: Consider the linear transformation $$L:\mathbb{R}^n \to \mathbb{R}^m: x \mapsto Ax $$

This exercice wants you to show that the preimage ##T:=L^{-1}(S)## is a subspace of ##\mathbb{R}^n##.

This occurs in a lot of places in abstract algebra (and other disciplines): the inverse image of a subspace of the codomain is a subspace of the domain.

In abstract algebra, one can prove this result for groups, modules (of which vector spaces are general cases), rings etc. all at once by considering the notion of ##X##-groups.

If you are not familiar with abstract algebra, you can safely ignore the addendum.
 
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  • #3
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Thanks!!!

1.Since ##\mathbf{0}\in T##, ##A(\mathbf{0})\in S## which is non-empty.
2. Suppose ##x_1,x_2\in T##. Then there exists vectors ##x_1,x_2\in S## such that we have ##A(x_1)## and ##A(x_2)##. We then have that ##x_1+x_2\in S## and ##A(x_1+x_2) = A(x_1)+A(x_2)##, i.e. ##x_1+x_2\in T##.
3. Suppose ##\mathbf{x}\in T##, with ##\lambda\in \mathbb{R}##, then ##\lambda \mathbf{x}\in S## and ##A(\lambda \mathbf{x}) = \lambda A(\mathbf{x})## which shows that ##\lambda \mathbf{x}\in T##.
 
  • #4
member 587159
Thanks!!!

1.Since ##\mathbf{0}\in T##, ##A(\mathbf{0})\in S## which is non-empty.
2. Suppose ##x_1,x_2\in T##. Then there exists vectors ##x_1,x_2\in S## such that we have ##A(x_1)## and ##A(x_2)##. We then have that ##x_1+x_2\in S## and ##A(x_1+x_2) = A(x_1)+A(x_2)##, i.e. ##x_1+x_2\in T##.
3. Suppose ##\mathbf{x}\in T##, with ##\lambda\in \mathbb{R}##, then ##\lambda \mathbf{x}\in S## and ##A(\lambda \mathbf{x}) = \lambda A(\mathbf{x})## which shows that ##\lambda \mathbf{x}\in T##.

Your proof for 2 and 3 are again wrong. You claim for example that ##x_1,x_2## live both in ##S## and ##T##, which doesn't make sense. These sets live in (possibly) different vector spaces.
 
  • #5
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Yeah that certainly doesn't make sense!!!

1. Suppose ##\mathbf{0}\in T##, ##A(\mathbf{0})\in S## which is non-empty.
2. Suppose ##x_1,x_2\in T##. We then have that ##A(x_1+x_2) = A(x_1)+A(x_2)\in S##, i.e. ##x_1+x_2\in T##.
3. Suppose ##\mathbf{x}\in T##, with ##\lambda\in \mathbb{R}##, ##A(\lambda \mathbf{x}) = \lambda A(\mathbf{x})\in S## which shows that ##\lambda \mathbf{x}\in T##.
 

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