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.
Best Answer
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.