# Union of two subspaces versus intersection of two subspaces

According to the definition, the union of two subspaces is not a subspace. That is easily proved to be true. For instance,

Let $U$ contain the general vector $(x,0)$, and $W$ contain the general vector $(0,y).$
Clearly, the union of these two subspaces would not be in either of the subspaces as it will violate closure axioms.

As for the intersection of the two subspaces, I believe I understand the concept. However, I want to be sure of that, and I believe it comes down to the difference between union and intersection as applied to vector/subspaces.

Basically, union - in this context - is being used to indicate that vectors can be taken from both subspaces, but when operated upon they have to be in one or the other subspace. Intersection, on the other hand, also means that vectors from both subspaces can be taken. But, a new subspace is formed by combining both subspaces into one.

To explain, I'll use the same subspaces as above. Let $U$ contain the general vector $(x,0)$, and $W$ contain the general vector $(0,y).$

So, the intersection of $U$ and $V$ would contain the general vector $(x,y)$ (I should say based on what I said above). Therefore, the closure axioms are fulfilled.

Am I correct in my reasoning?

Any feedback is appreciated.

• Your reasoning is wishy-washy, to put it nicely. What does "combining both subspaces into one" actually mean? If $(x,0)\in U$ and $(0,y)\in W$, it still doesn't follow that $(x,y)\in U\cap V$. You may be confusing the intersection with the span or sum of subspaces, $\langle V,W\rangle=V+W$, which is incidentally the subspace spanned by their set-theoretic union. If you want to know why the intersection of subspaces is itself a subspace, you need to get your hands dirty with the actual vector space axioms. Ultimately it boils down to logical and operations and implications. – anon Oct 11 '12 at 20:53

## 4 Answers

No, the intersection is the set of all vectors that are in both subspaces. In your example, that would only be the vector $(0,0)$ For a less trivial example, in $\mathbb R^3$ let $U$ be all vectors of the form $(x,y,0)$ and $V$ be all vectors of the form $(0,y,z)$. Then $U \cap V$ is all vectors of the form $(0,y,0)$.

The intersection $A \cap B$ of two sets $A$ and $B$ contains only elements which are in both subsets. To verify that the intersection $W := U \cap V$ of two subspaces $U,V$ is a subspace, you need to check if for any two elements $a,b \in W$ you also have $a+b \in W$, and similarly that for any element $a \in W$ and every scalar $\lambda \in K$ you have $\lambda a \in W$.

You do that by observing that if $a,b \in W$ then $a,b \in U$ and $a,b \in V$ because the intersection contains only elements that are in both sets. Now, since $a,b \in U$ you get $a+b \in U$ because $U$ is a subspace, and simiarly $a+b \in V$ because $V$ is also a subspace. Which then of course means $a+b \in W$, since $a+b$ lies in both of the intersected sets. The same line of reasoning works for the $\lambda a$ case.

Note that this a a very general principle - a lot of mathematical structures have the property that if $A$ and $B$ fulfill the structure's axioms then $A \cap B$ does too. And the proof very often basically works as in the case of subspaces above.

No. The intersection of two sets is the set of elements that are in both of them. In your example the intersection is $\{(0,0)\}$. The union is the set of elements that are in one, or the other, or both.

Let's prove that the intersection of two subspaces is also a subspace.

Assume that we have a vector space $$V$$ which has two subspaces $$S$$ and $$T$$ inside. We would like to prove that $$S\ \cap\ T$$ is also a subspace of $$V$$. So we can list the assumptions as given below:

1. $$S$$ is a subspace of $$V$$.
2. $$T$$ is a subspace of $$V$$.
3. $$S\ \cap\ T$$ is a subset (not subspace) of $$V$$.

we want to prove that that:

• $$S\ \cap\ T$$ is a subspace of $$V$$.

Based on assumptions 1 and 2 we can imply that all linear combinations of vectors inside $$S$$ reside in $$S$$, and that all linear combinations of vectors inside $$T$$ reside in $$T$$. Formally speaking:

$$\alpha x + \beta y \in S$$ $$\gamma m + \theta n \in T$$ where $$x$$ and $$y$$ are two random vectors in $$S$$ and $$m$$ and $$n$$ are two random vectors in $$T$$. $$\alpha$$, $$\beta$$, $$\gamma$$ and $$\theta$$ are real numbers. Based on the third assumption we can choose a set of vectors $$x$$ and $$y$$ which reside in both sets $$S$$ and $$T$$ and we can rewrite the above formalization as: $$\alpha x + \beta y \in S$$ $$\gamma x + \theta y \in T$$ Simplifying the above formula, we can rewrite the case in which $$\alpha$$ is equal to $$\gamma$$ and $$\beta$$ is equal to $$\theta$$. In this case, we'll have: $$\alpha x + \beta y \in T$$ $$\alpha x + \beta y \in S$$ This shows that linear combinations of two random vectors $$x$$ and $$y$$ reside in both sets $$S$$ and $$T$$. So these linear combinations will be definitely inside the intersection of $$S$$ and $$T$$ as well. $$\alpha x + \beta y \in S\ \cap\ T$$ hence $$S\ \cap\ T$$ is just is a cute little subspace.