Proving the Intersection of Subspaces is a Subspace: A Vector Space Proof

In summary, the question is asking for a proof that the intersection of K and H is a subspace of V. The student attempted to show that the zero vector is in the intersection but was not able to continue after that.
  • #1
Benny
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Hi can someone please help me with the following question. Such questions always trouble me because I don't know where to start and/or cannot continue after starting.

Q. Let H and K be subspaces of a vector space V. Prove that the intersection of K and H is a subspace of V.

By the way that the question is set out I figure that all I need to show is that the intersection of K and H is non-empty, closed under scalar multiplication and addition. So here is what I've tried.

H and K are subspaces of the vector space V so they both contain the zero vector. So it follows that the intersection contains the zeor vector so that [tex]H \cap K \ne \emptyset [/tex].

That's all I can think of. I'm not sure if I can make any other assumptions about vectors which are common to H and K and so I cannot continue.
 
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  • #2
Now show that a lineair combination of vectors of [tex]H \cap K[/tex] is still in [tex]H \cap K[/tex].
 
  • #3
Thanks for your response but that's the sort of thing that I'm having trouble with. All I've been able to show is that the zero vector is in the intersection. I don't know which are vectors are in the intersection. I cannot figure out what else I extract from the stem of the question to assist me. It's probably just a conceptual thing but I can't really see what I can and can't use.
 
  • #4
Well you already showed it's not empty. Now take the scalars [itex]\alpha ,\beta \in \mathbb{R}[/itex] (or any other field of course) and the vectors [itex]\vec x,\vec y \in H \cap K[/tex]. Now, since the vectors are in both subspaces, we can say that:
[tex]\alpha \vec x + \beta \vec y \in H[/tex]
[tex]\alpha \vec x + \beta \vec y \in K[/tex]

And thus: [tex]\alpha \vec x + \beta \vec y \in H \cap K[/tex]
 
  • #5
Ok thanks for the help.
 

FAQ: Proving the Intersection of Subspaces is a Subspace: A Vector Space Proof

What is the intersection of subspaces?

The intersection of two subspaces is the set of all elements that are in both subspaces. In other words, it is the collection of all vectors that satisfy the conditions of both subspaces.

How do you find the intersection of subspaces?

To find the intersection of two subspaces, you can use the method of elimination. This involves setting up a system of equations with the basis vectors of each subspace, and then solving for the common variables to find the vectors that satisfy both sets of equations.

What are the properties of the intersection of subspaces?

The intersection of subspaces is itself a subspace. This means that it follows all the properties of a subspace, such as closure under addition and scalar multiplication. In addition, the dimension of the intersection is equal to the sum of the dimensions of the two original subspaces.

Can the intersection of subspaces be empty?

Yes, it is possible for the intersection of subspaces to be empty. This occurs when the two subspaces do not have any common vectors, meaning there is no element that satisfies both sets of conditions. In this case, the intersection would be the trivial subspace consisting of only the zero vector.

How is the intersection of subspaces related to linear independence?

If the intersection of two subspaces is non-empty, it means that there is at least one vector that satisfies both sets of conditions. This vector must also be in the span of both subspaces, which indicates that the two subspaces are not linearly independent. Therefore, the intersection of two subspaces being empty implies that the two subspaces are linearly independent.

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