Suppose that $W_1, W_2, ..., W_n$ is a family of subspaces of V. Prove that the following set is a subspace of $V$:

$\cap_{i=1}^n W_i = ${$z | z \in W_i, 1 \leq i \leq n$}

The question counts for 4 marks. I'm just a bit confused as to how exactly this must be "proven"

My thoughts:

Since $W_1, ..., W_n$ are all subspaces, vector addition and scalar multiplication are closed under each of them.

Since any one of these subspaces is a vector space itself, the intersection of any of these subspaces must also be a vector space because vector addition and scalar multiplication are closed under all of them individually and, therefore, together as well.

$W_1 \cap W_2 \cap ... \cap W_n \subseteq V$ since they are a family of subspaces/subsets of $V$.

The set is a subspace of $V$ because it is a vector space under the vector addition and scalar multiplication defined on $V$

  • 1
    $\begingroup$ I think you're on the right idea, but I would write down the second step in a more rigorous way. That is, take $v,w \in \bigcap_{i=1}^nW_i$ and $c$ a scalar and check that $v+w \in \bigcap_{i=1}^nW_i$ and $cv \in \bigcap_{i=1}^nW_i$ $\endgroup$
    – Reveillark
    Apr 5, 2015 at 18:07

2 Answers 2


Your reasoning is correct but I guess you should provide more details for this task. So to prove for example that this intersection is closed under addition, a proper proof would look like this :

Take $x,y\in \cap_{i=1}^n W_i$, then for each $i$, we have $x,y\in W_i$. Since $W_i$ is a vector space, $x+y\in W_i$. Since this holds for each $i$, we have that $x+y\in \cap_{i=1}^n W_i$. Hence $\cap_{i=1}^n W_i$ is closed under addition.

  • $\begingroup$ I don't understand why for each $i$, we have $x,y \in W_i$. Surely $x,y$ are elements of $W_i$ for some $i$ but "each"? $\endgroup$ Apr 5, 2015 at 18:18
  • $\begingroup$ That's the definition of an intersection : If $x\in \cap_{j\in J} U_j$, then $x\in U_j$ for all $j\in J$. $\endgroup$ Apr 5, 2015 at 18:19
  • $\begingroup$ What I mean is: Say we have $W_1, W_2, W_3$ then a vector from $W_2$ is an element of the intersection of all three but it is not an element of each of the 3 individually? Am I reading it wrong? $\endgroup$ Apr 5, 2015 at 18:23
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    $\begingroup$ No, if $x\in W_2$, then this does not imply that $x\in W_1\cap W_2 \cap W_3$. I think you're confusing intersection with unions. Btw, in general it's not true that the union of subspaces is again a subspace. $\endgroup$ Apr 5, 2015 at 18:25
  • $\begingroup$ Oh yeah. How stupid of me lol. It's where they all intersect $\endgroup$ Apr 5, 2015 at 18:26

That's the idea!

For example, showing that $\cap W_i$ is closed under scalar multiplication would go as follows:

Let $\in \cap W_i$, $c\in \mathbb{F}$. Then $v\in W_i$ for all $i$. Since each $W_i$ is a subspace, $cv\in W_i$ for each $i$. Therefore, $cv\in \cap W_i$, and we see that it is closed under scalar multiplication.

The other properties follow from a similar argument.


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