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

• squenshl
In summary, we have shown that the set ##T## is a subspace of ##\mathbb{R}^n## by satisfying the three conditions for a subspace.
squenshl
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. Let's 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?

Last edited:
squenshl said:
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. Let's 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.

Last edited by a moderator:
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##.

squenshl said:
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.

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##.

## What is a subspace?

A subspace is a subset of a vector space that satisfies the three properties of closure under addition, closure under scalar multiplication, and contains the zero vector.

## What is the set T:={x∈Rn:Ax∈S}?

The set T is a subset of the vector space Rn, where x is a vector in Rn and A is a matrix. The set T is defined as all vectors x in Rn such that when multiplied by the matrix A, the resulting vector Ax is also in the subspace S.

## How do you prove that a set is a subspace?

To prove that a set T is a subspace, you must show that it satisfies the three properties of closure under addition, closure under scalar multiplication, and contains the zero vector. This can be done by showing that for any two vectors in T, their sum and scalar multiples are also in T, and that the zero vector is also in T.

## Why is it important to prove that a set is a subspace?

Proving that a set is a subspace is important because it allows us to determine whether certain operations, such as addition and scalar multiplication, can be performed on the vectors in the set. If a set is not a subspace, these operations may not be well-defined or may not produce vectors within the set.

## Can a set be a subspace of a vector space without satisfying all three properties?

No, a set must satisfy all three properties of closure under addition, closure under scalar multiplication, and contains the zero vector in order to be considered a subspace of a vector space. If a set does not satisfy all three properties, it is not considered a subspace.

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