if $W_1,W_2,W_3$ are subspaces of a vector space $V$ such that $W_3$ is contained in $W_1$.then which of the followings are true:

a) $W_1 \cap (W_2 + W_3) = W_2 + W_1 \cap W_3.$

b) $W_1 \cap (W_2 + W_3) = W_3 + (W_1 \cap W_2).$

c) $W_1 \cup (W_2 + W_3) = W_2 + W_1 \cup W_3.$

d) none of these.

I tried and found that option b is true can someone tell me about other options. I also want to ask if intersection and addition of subspaces are given together then what should be taken first i.e. what is meant by $W_2 + W_1 \cup W_3$?

  • $\begingroup$ If you have typed out in latex, why not just put a few dollar signs? $\endgroup$ – Inquest Sep 12 '13 at 16:00
  • $\begingroup$ i cannot understand what you are trying to say? $\endgroup$ – user93233 Sep 12 '13 at 16:01
  • $\begingroup$ He's saying that instead of writing \bigcap you should write \$ \bigcap \$, for instance. You should only use \$ \bigcap \$ if you want to put a subscript such as $\bigcap_{y \in X}$ underneath it. Otherwise you should use \$ \cap \$. $\endgroup$ – Patrick Da Silva Sep 12 '13 at 16:02
  • $\begingroup$ [please someone help me $\endgroup$ – user93233 Sep 12 '13 at 16:05
  • 1
    $\begingroup$ @user93233 I'm not aware of any standard order of operations to clarify the right hand side of the first or last expression. One might guess that the addition comes last. $\endgroup$ – rschwieb Sep 12 '13 at 16:08

In the second case we have: \begin{align*} W_1 \cap (W_2+W_3) &= (W_1 \cap W_2)+(W_1 \cap W_3) & \text{distributivity} \\ &=(W_1 \cap W_3)+(W_1 \cap W_2) & \text{commutativity} \\ &=W_1+(W_1 \cap W_2) & \text{by assumption } W_3 \subseteq W_1. \\ \end{align*} So, the second case is always true.

Now, the first and third cases can be true for some $W_1$, $W_2$ and $W_3$ satisfying the conditions, so presumably the task is to show that they're not always true.

We can eliminate the first case by choosing subspaces of $\mathbb{Z}_2$ such that $|W_1|=|W_3|=1$ and $|W_2|=2$, say. This implies $$|W_1 \cap (W_2+W_3)|=1$$ and $$|W_2+(W_1 \cap W_3)|=2$$ which would not be true if the first statement were true in general.

We can eliminate the third case by choosing subspaces of $\mathbb{R}^2$ such that $|W_3|=1$, $W_1=\mathrm{span}((0,1))$ and $W_2=\mathrm{span}((1,0))$. In this situation the identity $$W_1 \cup (W_2+W_3)=W_2+(W_1 \cup W_3)$$ is equivalent to $$W_1 \cup W_2=W_2+W_1$$ which is not true for this example: $(1,1) \in W_2+W_1$, whereas $(1,1) \not\in W_1 \cup W_2$.

  • $\begingroup$ hello rebecca can you tell me first we solve intersection or sum $\endgroup$ – user93233 Sep 12 '13 at 16:34
  • $\begingroup$ I would guess $A+B \cap C=A+(B \cap C)$. But I would argue that $A+B \cap C$ is ill-defined, and it's better to use brackets. (Actually, I noticed I wrote that once; I'll change it.) $\endgroup$ – Rebecca J. Stones Sep 12 '13 at 16:35

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