Linear Algebra Subspaces Basis

In summary: If so, I can try to help you. But it would probably be better if you asked someone in the chat or on the forum.
  • #1
bob258173498
16
0

Homework Statement



a) If U and W are subspaces of R^3, show that it is possible to find a basis B for R^3 such that one subset of B is a basis for U and another subset of B (possibly overlapping) is a basis for W.

b) If U and W are subspaces of a finite-dimensional vector space V, show that it is possible to find a basis for V such that one subset of that basis is a basis for U and another subset of that basis (possibly overlapping) is a basis for W.


Homework Equations



none really

The Attempt at a Solution



B = ( v_1, v_2, v_3 ), such that these vectors are a basis for V.

Then:

V = t_1*v_1 + t_2*v_2 + t_3*v_3

...stuck here
 
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  • #2
help please!
 
  • #3
Please guys
 
  • #4
bob258173498 said:

Homework Statement



a) If U and W are subspaces of R^3, show that it is possible to find a basis B for R^3 such that one subset of B is a basis for U and another subset of B (possibly overlapping) is a basis for W.

b) If U and W are subspaces of a finite-dimensional vector space V, show that it is possible to find a basis for V such that one subset of that basis is a basis for U and another subset of that basis (possibly overlapping) is a basis for W.


Homework Equations



none really

The Attempt at a Solution



B = ( v_1, v_2, v_3 ), such that these vectors are a basis for V.

Then:

V = t_1*v_1 + t_2*v_2 + t_3*v_3

...stuck here
People will help if they wish. They probably will not help you since you keep bumping your thread every few minutes, which is against the rules.
 
  • #5
V = (v_1, v_2, v_3)
U = ? it never told us

Please can someone help.
 
  • #6
Guys please, I'm begging you
 
  • #7
Start off by finding a basis for [itex]U\cap W[/itex].
 
  • #8
U = (v_1, v_2)
W = (v_2, v_3)

U∩W = (v_2)
 
  • #9
please i need to this quickly
 
  • #10
bob258173498 said:
U = (v_1, v_2)
W = (v_2, v_3)

U∩W = (v_2)

No, what makes you think that U and W are two-dimensional?? U and W can be anything!

First construct a basis for [itex]U\cap W[/itex], then extend this basis to a basis of U and extend it to a basis of W.


bob258173498 said:
please i need to this quickly

Messages like this actually incline people to help you less. Just so you know.
 
  • #11
24 hour time out due to ignoring the rules, ignoring the warning and continuing to bump.
 
  • #12
bob258173498 said:
please i need to this quickly

Do you need it quickly because it is part of a midterm exam?
 

1. What is a subspace in linear algebra?

A subspace in linear algebra is a subset of a vector space that satisfies three conditions: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication. In other words, a subspace is a smaller space within a larger space that is still consistent with the rules of vector addition and scalar multiplication.

2. How do you determine if a set of vectors is a basis for a subspace?

To determine if a set of vectors is a basis for a subspace, you must first make sure the set is linearly independent, meaning that none of the vectors in the set can be written as a linear combination of the other vectors. Then, you must also make sure that the set spans the entire subspace, meaning that every vector in the subspace can be written as a linear combination of the basis vectors. If both of these conditions are met, then the set of vectors is a basis for the subspace.

3. Can a subspace have more than one basis?

Yes, a subspace can have more than one basis. This is because there can be multiple sets of linearly independent vectors that span the same subspace. However, all bases for a given subspace will have the same number of vectors, known as the dimension of the subspace.

4. How do you find the dimension of a subspace?

The dimension of a subspace is equal to the number of vectors in any basis for that subspace. To find the dimension, you can either find a basis for the subspace and count the number of vectors in it, or you can use the rank-nullity theorem, which states that the dimension of a subspace is equal to the rank of its associated matrix (the number of linearly independent columns) minus the nullity (the dimension of the null space).

5. Can a subspace contain the zero vector but still not be the entire vector space?

Yes, a subspace can contain the zero vector and still be a proper subset of the entire vector space. This is because the zero vector is a necessary condition for a subspace, but it is not sufficient. In addition to the zero vector, a subspace must also be closed under vector addition and scalar multiplication in order to be considered a subspace.

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