# Proving Subspaces of Vector Spaces: Evaluating A Vector x

• Supernova123
In summary, when determining if a vector space is a subspace of another one, it is important to ensure that the basis vectors of the subspace can be formed from a linear combination of the basis vectors of the vector space. In the given example, the linear space spanned by e1 and e2 (denoted by V) is only two-dimensional and therefore cannot contain all three rows of matrix A, leading to the failure of the attempted method.
Supernova123

## Homework Statement

How would one determine if a vector space is a subspace of another one? I think that the basis vectors of the subspace should be able to be formed from a linear combination of the basis vectors of the vector space.

However, that doesn't seem to be true for this question: Let matrix A consist of column vectors (1,2,-3), (-4,-4,4) and (6,2,-8) with eigenvalues -2 and -5. I found e1=(2,3,1) and e2=(1,0,-1).

The linear space spanned by e1 and e2 is denoted by V. Prove that, for any vector x belonging to V, the vector Ax also belongs to V. I tried forming any of the column vectors in matrix A through a linear combination of e1 and e2 but that seems to fail. Instead, the answer is this: Let x=ae1+be2. Ax=A(ae1+be2)=aAe1+bAe2=-2ae1-5ae2. I understand this but why doesn't my method work?

## The Attempt at a Solution

Supernova123 said:

## Homework Statement

How would one determine if a vector space is a subspace of another one? I think that the basis vectors of the subspace should be able to be formed from a linear combination of the basis vectors of the vector space.

However, that doesn't seem to be true for this question: Let matrix A consist of column vectors (1,2,-3), (-4,-4,4) and (6,2,-8) with eigenvalues -2 and -5. I found e1=(2,3,1) and e2=(1,0,-1).

The linear space spanned by e1 and e2 is denoted by V. Prove that, for any vector x belonging to V, the vector Ax also belongs to V. I tried forming any of the column vectors in matrix A through a linear combination of e1 and e2 but that seems to fail. Instead, the answer is this: Let x=ae1+be2. Ax=A(ae1+be2)=aAe1+bAe2=-2ae1-5ae2. I understand this but why doesn't my method work?

## The Attempt at a Solution

Isn't Ax = -2ae1-5be2 a linear combination of the vectors e1 and e2?

Ax=a(1,2,-3)+b(-4,-4,4)+c(6,2,-8) and V contains Ax. So wouldn't de1+fe2=(1,2,-3) or (-4,-4,4) or (6,2,-8)?

Supernova123 said:
Ax=a(1,2,-3)+b(-4,-4,4)+c(6,2,-8) and V contains Ax. So wouldn't de1+fe2=(1,2,-3) or (-4,-4,4) or (6,2,-8)?

Think about it: as x ranges over S = R3 (the whole space), Ax ranges over S as well (because for any y in S the equation Ax = y has a unique solution). However, your defined subspace V (= set of all linear combinations of e1 and e2) is only two-dimensional, so cannot contain all three rows of A. In other words, a two-dimensional space is not a three-dimensional space.

Supernova123

## 1. What is a subspace of a vector space?

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

## 2. How do you prove that a subset is a subspace of a vector space?

To prove that a subset is a subspace of a vector space, you must show that it satisfies the three properties of closure under addition, closure under scalar multiplication, and contains the zero vector. You can also use the subspace test, which states that if all linear combinations of the vectors in the subset are also in the subset, then the subset is a subspace.

## 3. What is the process for evaluating a vector x in a subspace of a vector space?

To evaluate a vector x in a subspace of a vector space, you must first check if x satisfies the three properties of a subspace. If it does, then x is in the subspace. If it does not, then x is not in the subspace.

## 4. Can a subspace of a vector space contain only one vector?

Yes, a subspace of a vector space can contain only one vector. This vector must satisfy the three properties of a subspace, including being closed under addition and scalar multiplication, and containing the zero vector.

## 5. How is the dimension of a subspace of a vector space related to the dimension of the vector space?

The dimension of a subspace of a vector space can be equal to or less than the dimension of the vector space. The dimension of the subspace is determined by the number of linearly independent vectors in the subspace. If the subspace has fewer linearly independent vectors than the vector space, then its dimension will be lower.

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