Proof involving vector subspaces

In summary, to prove that W1 U W2 is a subspace of vector space V, it must be shown that either W1 is a subset of W2 or W2 is a subset of W1. This can be done using an indirect proof, where it is assumed that neither condition is true and then a contradiction is reached. By considering simple examples, such as the x-axis and y-axis in the plane, it can be seen that any sum of vectors in W1 and W2 must be in either W1 or W2, thus satisfying the definition of a subspace.
  • #1
strman
2
0
Let W1 and W2 be subspaces of a vector space V. Prove that W1[itex]\bigcup[/itex]W2 is a subspace of V if and only if W1[itex]\subseteq[/itex]W2 or W2[itex]\subseteq[/itex]W1Well so far, I have proven half of the statement (starting with the latter conditions). Right now I'm struggling to show that the final conditions follow from W1[itex]\bigcup[/itex]W2. I have an idea for the method: assume that W1[itex]\subseteq[/itex]W2 is not true, and then prove that W2[itex]\subseteq[/itex]W1 must follow.

This is my first linear algebra class, and this is from the first problem set due, so I know that nothing that complex is going on here. However, I've been looking at this problem for the past 30 minutes or so, and I'm hoping that someone could push me in the right direction.
 
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  • #2
how about directly doing it? needs a bit of flesh...

W1 U W2 is subspace
take w1 in W1, w2 in W2
w1+w2 is in W1 U W2
but this means w1+w2 is in either W1 or W2
hence either w1 is in W2 or w2 is in W1

The problem with the way you describe might be if W1⊆W2 is not true, then this excludes W1=W2, which means you can only prove a proper subset, but i need to think about it more
 
  • #3
I would use an "indirect proof". suppose it is NOT true that either W1 is a subset of W2 or W2 is a subset of W1. Then there exist a vector w1 that is not in W2 and a vector w2 that is not in W1. If w1+ w2= v is in W1 union W2, it is in either W1 or W2. But w1= v- w2, etc.

It might help to look at a simple example. Suppose W1 is the x-axis (as a subspace of the plane) and W2 is the y- axis. Are there sums of things in W1 and W2 that are not in their union? Generalize that idea.
 

1. What is a vector subspace?

A vector subspace is a subset of a vector space that also follows the same properties as the vector space. This means that it is closed under addition and scalar multiplication, and contains the zero vector.

2. How do you prove that a set is a vector subspace?

To prove that a set is a vector subspace, you must show that it satisfies the three properties of a vector subspace: 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 vectors in the set are also in the set, then it is a vector subspace.

3. What is the difference between a vector space and a vector subspace?

A vector space is a set of vectors that follows specific properties, such as closure under addition and scalar multiplication. A vector subspace is a subset of a vector space that also follows these properties. In other words, a vector subspace is a smaller vector space within a larger vector space.

4. How do you prove that two vector subspaces are equal?

To prove that two vector subspaces are equal, you must show that they have the same elements and the same operations. This means that they have the same vectors and the same rules for addition and scalar multiplication. You can also use the subset test, which states that if the smaller vector subspace is a subset of the larger vector subspace, then they are equal.

5. Can a vector subspace have dimensions?

Yes, a vector subspace can have dimensions. The dimension of a vector subspace is the number of vectors in a basis for the subspace. This means that a vector subspace can have a smaller dimension than the vector space it is a part of, but it cannot have a larger dimension.

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