Union and intersection of vector spaces – "Math for Non-Geeks"
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Motivation
[edit | edit source]We know various operations to construct a new set from given sets. If is a family of sets, we can, for example, form the average or the union . Assume that the are also subspaces of a larger vector space . This means that the are non-empty subsets of , which are closed under addition and scalar multiplication. Are the average and the union of then also subspaces of ?
Intersection of vector spaces
[edit | edit source]Is the intersection of subspaces of a vector space again a subspace? To answer this question, let us first consider the case of two subspaces and look at examples in .
- Let us first look at the two planes and (the y-z-plane). In the image we can see that its intersection is the z-axis , i.e. a subspace of .
- The second image shows that the intersection of the line with the y-z plane is also a line, namely .
- If we intersect the y-z plane with the line instead, we see that the intersection only contains the zero vector. This is also a subspace of .
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Intersection of two planes
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Intersection of a plane with a line
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Intersection of a plane with a line
In the examples, the intersection of the two subspaces is always a subspace of . We now show that this also applies to general subspaces of any vector space.
Theorem (Intersection of vector spaces)
Let and be two subspaces of a vector space . The intersection of and is a subspace of .
Proof (Intersection of vector spaces)
First, we have to establish the three subspace criteria:
Proof step: is not empty
Since and are subspaces, we have and . Thus, , and therefore .
Proof step: is closed under addition.
Let now be arbitrary. Then and apply. As the subspaces and are closed under addition, we get and . This also means .
Proof step: is closed under scalar multiplication.
Let and be arbitrary. Then we have and . Since the subspaces and are closed under scalar multiplication, we obtain and . Thus, also .
Hint
We have seen that is a subspace of . Because , the intersection is also a subspace . From this we get . Similarly, follows from . This makes intuitively sense: if we intersect one subspace with another, its dimension cannot possibly become larger.
We have shown that the intersection of two subspaces is again a subspace. However, at no point in the proof is it relevant that there are only two or finitely many subspaces involved. In fact, the statement applies to any family of subspaces.
Math for Non-Geeks: Template:Aufgabe
Union of vector spaces
[edit | edit source]Is the union of subspaces of a vector space a vector space again? Let us first look at an example.
Example (Coordinate axis cross is not a subspace)
Let . We choose the coordinate axes and as subspaces. Their union is the axis cross in . Based on the figure, we can already surmise that this is not a subspace of : There are two "directions", but at the same time is not the two-dimensional plane. And indeed, the two vectors and are in , but their sum is not in the union. Therefore, is not a subspace of .

So we see that the union of two subspaces is generally not a subspace. Is this always the case?
Example (Union of vector spaces is again a vector space)
We consider the two subspaces and of . As we have . Therefore and the union of the two subspaces is again a subspace.

The union of two subspaces is therefore in some cases, but not always, a subspace. In the example, was contained in , so that was a subspace. This always works: If two subspaces are given and one of them is contained in the other, then the union is equal to the larger of the two, i.e. a subspace again.
This is indeed the only case in which the union of two subspaces is again a subspace, as the first example with the coordinate axes makes clear: If and , then the union will not be closed under addition. There will then always exist two vectors with and . The sum thus contains a part that is not in and therefore cannot be in : Otherwise, would also be true. The same applies to .
We therefore have the following criterion for determining when the union of two subspaces is a subspace.
Theorem (Condition for the union of two vector spaces to be a vector space again)
Let be a vector space over a field and let and be two subspaces of . Then is a subspace of if and only if or holds.
Proof (Condition for the union of two vector spaces to be a vector space again)
Proof step: If is a subspace of , then we have or .
We prove the statement by contradiction: Assume that neither nor is true. We show that is then not a subspace of . Instead, we find two elements , so that :
Since , there exists an element that is not contained in . Similarly, since , there exists a that is not contained in .
Thus, . But the sum is neither in nor in : If , then , in contradiction to the choice of . Here we have used the fact that implies also and that is closed under addition. In the same way, you can see that does not lie in .
Therefore, applies. This means that the union is not closed under addition, i.e. it is not a subspace.
Proof step: If or , then is a vector space
If , then , so the union is a subspace. Similarly, it follows from that is a subspace.
The proof of the theorem shows that the property of being a subspace fails due to addition. The scalar multiplication on was not relevant in the proof. In fact, is always closed under scalar multiplication, even if the union is not a subspace: If and , say , then holds, since is closed under scalar multiplication as a subspace. The case is analogous.
Since is a vector space and are subspaces, forms a group and subgroups. We have thus effectively shown that is a subgroup of if and only if or holds. There is a more general statement about (not necessarily commutative) groups. The proof is quite analogous to the proof for subspaces that we have seen above.
Theorem (Union of subgroups)
Let be a group and subgroups. Then is a subgroup of if and only if or holds.
The union of subspaces and is generally not a subspace. However, you can define the smallest subspace that contains . This subspace is the subspace sum .
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