2
$\begingroup$

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 $V$

$\endgroup$
  • 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 '15 at 18:07
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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"? $\endgroup$ – StephanCasey Apr 5 '15 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 $j\in J$. $\endgroup$ – Mathematician 42 Apr 5 '15 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? $\endgroup$ – StephanCasey Apr 5 '15 at 18:23
  • 1
    $\begingroup$ 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. $\endgroup$ – Mathematician 42 Apr 5 '15 at 18:25
  • $\begingroup$ Oh yeah. How stupid of me lol. It's where they all intersect $\endgroup$ – StephanCasey Apr 5 '15 at 18:26
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.