Show Existence of Basis for Vector Space V with No Elements from Subspace M

In summary, the conversation discusses the existence of a basis for a vector space V such that none of its elements belong to a given subspace M, which is not equal to the vector space V itself. The conversation presents different approaches and hints, with one solution being to construct a basis of M and extend it to be a basis of V, ensuring that the added vectors are not in M.
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Let V be a vector space over a field F, and M a subspace of V, where M is not {0}. I need to show there exists a basis for V such that none of its elements belong to M.

Since M is a subspace of V, M must be a subset of V. If M = V, then there does not exist such a basis, so M must be a proper subset of V. Hence, there must exist at least one element b1 from V which is not in M. Further on, the set {b1} is independent. Now, if we consider the span [{b1}], which is a subset of V, we have two options. If [{b1}] = V, then {b1} is a basis, and we proved what we had to. If it is not so, then [{b1}] is a proper subset of V, and there (here's the tricky part) exists (?) at least one element b2 from V \ (M U [{b1}]). If I could proove the existence, I'd know how to carry on. I tried to assume the opposite - there does not exist an element b2 from V \ (M U [{b1}]). This implies that b2 must be in M. But then, {b1} should form a basis for V, which it does not, so we have a contradiction (?).

Directions would be appreciated, thanks in advance.

Interesting that your initial statement doesn't include M=/=V, since that's obviously a requirement.

Even if you find a b2, you have to show it's linearly independent of b1. I wouldn't go the contradiction route, since it seems like far too much work.

You know that if M does not equal V, then dimM < dimV, right (strictly less than)? So you can construct a basis of M, and extend it to be a basis of V. The vectors in the new basis that were just added are not elements of M, which is important. So say the basis of M is {$$m_1, m_2, ... m_k$$} and the basis of V is {$$m_1, m_2, ... ,m_k, v_1,...v_n$$}. Given this, can you find a way of replacing the m's with vectors that aren't in M (which has to be closed under addition, a big hint) such that the set is still a basis?

Yeah, with the hint above, I have an idea:

Suppose dimM < dimV and M has a basis {m1, m2, ..., mk}. Add to this basis some vectors called v1, v2,..., vn such that {m1, m2,...mk, v1, v2,..., vn} becomes a basis of V. Of course, v1, v2,..., vn are not in M.

Define u1 = m1 + v1, u2 = m2 + v1, ..., uk = mk + v1

By using definiton, we can prove the vector system {u1, u2,..., uk, v1, v2,...,vn} is still independent in V, thus it is a basis of V.

However all elements of this new basis of V are not in M. The proff is complete. :)

Office_Shredder said:
Interesting that your initial statement doesn't include M=/=V, since that's obviously a requirement.

It's not interesting, since it isn't my statement.

Office_Shredder said:
Even if you find a b2, you have to show it's linearly independent of b1.

Well, if b2 is not in [{b1}], then it is independent of b1, since there doesn't exist some a from F such that a*b1 = b2.

Thank you both for the other hints, I'll think about it.

Well, if b2 is not in [{b1}], then it is independent of b1, since there doesn't exist some a from F such that a*b1 = b2.

That's true now that I think about it

And by your statement, I meant the statement that you had, i.e. that someone gave you

It's not interesting, since it isn't my statement.
You should find it interesting since it makes your problem simple: the statement is NOT true! Counter-example take M= V.

HallsofIvy said:
You should find it interesting since it makes your problem simple: the statement is NOT true! Counter-example take M= V.

OK, so the 'statement' should be: Let V be a vector space over a field F, and M be a subspace of V, where M is not a trivial subspace, i.e. it is not {0} or V.

this false unless you assume M is not all of V. i came up with the same solution as in 3, which shows clearly how M ≠ V is used.

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1. What is a basis for a vector space?

A basis for a vector space is a set of vectors that are linearly independent and span the entire vector space. This means that any vector in the vector space can be written as a linear combination of the basis vectors.

2. How do you show the existence of a basis for a vector space?

To show the existence of a basis for a vector space, we need to prove that the set of vectors we have chosen is linearly independent and spans the entire vector space. This can be done by showing that the vectors satisfy the definition of a basis, or by using other methods such as the dimension theorem or the rank-nullity theorem.

3. What does it mean for a vector space to have no elements from a subspace?

If a vector space has no elements from a subspace, it means that there is no vector in the vector space that can be written as a linear combination of vectors from the subspace. In other words, the vector space and the subspace are completely separate and do not share any common vectors.

4. Why is it important to show the existence of a basis for a vector space with no elements from a subspace?

Showing the existence of a basis for a vector space with no elements from a subspace is important because it helps us understand the structure of the vector space and the relationship between the vector space and the subspace. It also allows us to perform calculations and make conclusions about the vector space and its properties.

5. Can a vector space have multiple bases?

Yes, a vector space can have multiple bases. This is because there can be different sets of linearly independent vectors that span the same vector space. However, all bases for a given vector space will have the same number of elements, which is known as the dimension of the vector space.

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