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

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:

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

and

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

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.

• 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? – Lee White Oct 29 '17 at 12:32
• I will but I advise you that if you cannot prove those things you will not be able to go further. – richarddedekind Oct 29 '17 at 12:37
• thanks for your advise, I'll do it again one week later. And thanks again for your help. – Lee White Oct 29 '17 at 13:19