Let $W_1$, $W_2$, $W_3$ be subspaces of a finite-dimensional vector space $V$.

Why $\left( {{W_1} \cap {W_3}} \right){\rm{ + }}\left( {{W_2} \cap {W_3}} \right) \subset \left( {{W_1}{\rm{ + }}{W_2}} \right) \cap {W_3} $?

And When $\left( {{W_1} \cap {W_3}} \right){\rm{ + }}\left( {{W_2} \cap {W_3}} \right){\rm{ = }}\left( {{W_1}{\rm{ + }}{W_2}} \right) \cap {W_3}$?


The first inclusion is easy to prove. You have to observe that if $w_1\in W_1\cap W_3$ and $w_2\in W_2\cap W_3$ then for $w=w_1+w_2$ apply that:

$$w_1,w_2\in W_3\Rightarrow w=w_1+w_2\in W_3$$


$$w_1\in W_1, w_2\in W_2 \Rightarrow w=w_1+w_2\in W_1+W_2$$

So $w\in (W_1+W_2)\cap W_3$.

Now for the equality. Those two are equals if $W_1\subseteq W_3$ or $W_1\subseteq W_2$, which can be proven in the same way as before. Suppose without loss of generality that $W_1\subseteq W_3$. Then, take $w\in (W_1+W_2)\cap W_3$.

$$w\in (W_1+W_2)\Rightarrow \exists w_1\in W_1\exists w_2\in W_2 : w=w_1+w_2$$

We have that $w_1\in W_1\subseteq W_3$ and $w\in W_3$, hence $w_2=w-w_1\in W_3$ and so it applies that

$w_2\in W_2\cap W_3$ and $w_1\in W_1=W_1\cap W_3$ which is the desirable.

  • $\begingroup$ Thank you for your answer, It's a clever way. But I still can't understand the equality case, could you please give me a simply proving? $\endgroup$ – Lee White Oct 29 '17 at 12:32
  • $\begingroup$ I will but I advise you that if you cannot prove those things you will not be able to go further. $\endgroup$ – richarddedekind Oct 29 '17 at 12:37
  • $\begingroup$ thanks for your advise, I'll do it again one week later. And thanks again for your help. $\endgroup$ – Lee White Oct 29 '17 at 13:19

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.