# Linear Algebra: Subspaces

## Homework Statement

Prove that the intersection of any collection of subspaces of V is a subspace of V.

## Homework Equations

To show that a set is a subspace of a vector space, I need to show that there exists an additive identity, and that the set is closed under addition and scalar multiplication.

For notation I will let F denote R or C.

## The Attempt at a Solution

Consider a collection of subspaces U_1, U_2, ..., U_m of a vector space V.

Clearly any intersection of these subspaces will contain 0, and hence the intersection contains the additive identity.

Now consider two elements, u and w, of any arbitrary intersection of U_1, U_2, ..., U_m. Since u and w are in the intersection of U_1, U_2, ..., U_m, u and w are elements of U_1 and U_2 and ... and U_m. Therefore their addition, u + w must be an element of U_1 and U_2 and ... and U_m. So u + w is an element of any intersection of U_1, U_2, ..., U_m and hence this intersection is closed under addition.

Now consider a scalar c in F and an element u of any arbitrary intersection of U_1, U_2, ..., U_m. Since u is in the intersection of U_1, U_2, ..., U_m, u is an element of U_1 and U_2 and ... and U_m. Therefore cu is an element of U_1 and U_2 and ... and U_m. So cu is an element of any intersection of U_1, U_2, ..., U_m and hence this intersection is closed under scalar multiplication.

Any feedback would be greatly appreciated. If it is sloppy, please let me know what I can do to improve my proof writing. Thanks again!

Fredrik
Staff Emeritus
Gold Member
I need to show that there exists an additive identity...
I would rephrase that. You need to prove membership, not existence. So just say that you need to show that 0 (the additive identity of V) is a member of each $U_i$.

Consider a collection of subspaces U_1, U_2, ..., U_m of a vector space V.
That's just a countable collection. You need to consider an arbitrary collection $\{U_i|i\in I\}$, where I might be a "larger" set than the integers.

Clearly any intersection of these subspaces will contain 0, and hence the intersection contains the additive identity.
Right, but instead of saying "clearly", you can explain how you're using the definitions. You should at least say that the intersection contains 0 because each $U_i$ does.

Now consider two elements, u and w, of any arbitrary intersection of U_1, U_2, ..., U_m.
An arbitrary intersection could be the intersection of uncountably many subspaces. You seem to have the right idea about how the rest should be done though, but I didn't check every detail, so don't take this as a guarantee that you did everything right. I like to use logical notation for this type of proofs, but if you prefer you can replace logical symbols with words: $$x,y\in\bigcap_{i\in I}U_i\ \Rightarrow\ \forall i\in I\quad x,y\in U_i\ \Rightarrow\ \cdots\Rightarrow x+y\in\bigcap_{i\in I}U_i$$

(I'll be away from the computer for at least an hour, and probably the rest of the day, but I'll answer follow-up questions tomorrow in the unlikely event that they haven't already been answered by someone else).

That's just a countable collection. You need to consider an arbitrary collection $\{U_i|i\in I\}$, where I might be a "larger" set than the integers.