Serlo: EN: Union and intersection of vector spaces

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Motivation

We know various operations to construct a new set from given sets. If $(U_i{)}_{i\in I}$ is a family of sets, we can, for example, form the average $\bigcap _{i\in I}{U}_i$ or the union $\bigcup _{i\in I}{U}_i$. Assume that the $U_i$ are also subspaces of a larger vector space $V$. This means that the $U_i$ are non-empty subsets of $V$, which are closed under addition and scalar multiplication. Are the average and the union of $U_i$ then also subspaces of $V$?

Intersection of vector spaces Vorlage:Anker

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 ${ℝ}^3$.

1. Let us first look at the two planes $U_1=\mathrm{span}\{(2,1,0{)}^T,(0,0,1{)}^T\}$ and $W=\mathrm{span}\{(0,2,0{)}^T,(0,0,-1{)}^T\}$ (the y-z-plane). In the image we can see that its intersection is the z-axis $\mathrm{span}\{(0,0,1{)}^T\}$, i.e. a subspace of ${ℝ}^3$.
2. The second image shows that the intersection of the line $U_2=\mathrm{span}\{(0,1,2{)}^T\}$ with the y-z plane $W$ is also a line, namely $U_2$.
3. If we intersect the y-z plane $W$ with the line $U_3=\mathrm{span}\{(1,1,1{)}^T\}$ instead, we see that the intersection only contains the zero vector. This is also a subspace of ${ℝ}^3$.

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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 ${ℝ}^3$. We now show that this also applies to general subspaces of any vector space.

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.

Union of vector spaces

Is the union of subspaces of a vector space a vector space again? Let us first look at an example.

So we see that the union of two subspaces is generally not a subspace. Is this always the case?

The union of two subspaces is therefore in some cases, but not always, a subspace. In the example, $U$ was contained in $W$, so that $U\cup W=W$ 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 $U⊈W$ and $W⊈U$, then the union will not be closed under addition. There will then always exist two vectors $u,w$ with $u\in U\setminus W$ and $w\in W\setminus U$. The sum $u+w$ thus contains a part that is not in $U$ and therefore cannot be in $U$: Otherwise, $w=u+w-u\in U$ would also be true. The same applies to $u+w\notin W$.

We therefore have the following criterion for determining when the union of two subspaces is a subspace. Mathe für Nicht-Freaks: Vorlage:Satz The proof of the theorem shows that the property of being a subspace fails due to addition. The scalar multiplication on $V$ was not relevant in the proof. In fact, $U\cup W$ is always closed under scalar multiplication, even if the union is not a subspace: If $\lambda \in K$ and $x\in U\cup W$, say $x\in U$, then $\lambda x\in U\subseteq U\cup W$ holds, since $U$ is closed under scalar multiplication as a subspace. The case $x\in W$ is analogous.

Since $V$ is a vector space and $U,W$ are subspaces, $(V,+)$ forms a group and $(U,+),(W,+)$ subgroups. We have thus effectively shown that $U\cup W$ is a subgroup of $V$ if and only if $U\subseteq W$ or $W\subseteq U$ 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.

The union of subspaces $U$ and $W$ is generally not a subspace. However, you can define the smallest subspace that contains $U\cup W$. This subspace is the subspace sum $U+W$.

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