Let $V_{1,2,3}$ be Vectorspaces. We want to show the equation above.

My attempt: \begin{align*} (V_1 \cap V_3)+(V_2 \cap V_3) & =\left \{ v: \exists u_1 \in V_1 , \ u_2 \in V_2 \cap V_3, \ v = u_1 + u_2 \right \} \\ & = \left \{v : \exists u_1 \in V_1, \ (u_2 \in V_2 \wedge u_2 \in V_3), \ v = u_1 + u_2 \right \} \\ & = \left \{ v : \exists u_1 \in V_1, \ u_2 \in V_2, \ v = u_1 + u_2 \right \} \\ & \qquad \qquad \cap \left \{ v: \exists u_1 \in V_1, \ u_2 \in V_3, \ v = u_1 + u_2 \right\} \\ & = (V_1 + V_2)\cap(V_1 + V_3) \\ & \overset{V_1 \subset V_3}{=} (V_1+V_2) \cap V_3 \end{align*}

I have a serious doubt that I'm missing something here. Any help is much appreciated.


1 Answer 1


I think your main problem is getting bogged down in symbols, a few words here and there can really make things easier to write and understand.

Let's show both inclusions separately.

First we will show that $(V_1 + V_2)\cap V_3 \subseteq (V_1 \cap V_3) + (V_2 \cap V_3)$.

Take $v \in (V_1 + V_2)\cap V_3$. Then it can be written as $v = v_1 + v_2$ for some $v_1\in V_1$ and $v_2 \in V_2$.

Actually we find that $v_1,v_2\in V_3$ automatically. Why? Well $v_1\in V_1 \subset V_3$ and $v\in V_3$, hence $v_2 = v - v_1 \in V_3$ since $V_3$ is a vector space.

Thus $v \in (V_1 \cap V_3) + (V_2 \cap V_3)$.

Now for the other inclusion, $(V_1 \cap V_3) + (V_2 \cap V_3)\subseteq (V_1 + V_2)\cap V_3 $.

Take $v \in (V_1 \cap V_3) + (V_2 \cap V_3)$. Then $v = v_1 + v_2$ where $v_1\in V_1$, $v_2 \in V_2$ and $v_1,v_2\in V_3$. Now $v_3$ is a vector space hence $v_1+v_2\in V_3$ and clearly $v_1+v_2\in V_1+V_2$. So $v \in (V_1+V_2)\cap V_3$.

  • $\begingroup$ thanks for the answer. My attempt was to show it directly ( without double inclusion). Would that be still possible? $\endgroup$
    – sigmatau
    Dec 30, 2013 at 10:56
  • 1
    $\begingroup$ I would doubt it is straight forward but probably possible. See one direction of the inclusion didn't need the fact that $V_1\subset V_3$ yet the other one did. At some point in this proof you really are forced to write "since $V_1\subset V_3$". $\endgroup$
    – fretty
    Dec 30, 2013 at 11:07

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.