Let $V$ be a vector space over a field $K$. Let $W_1,W_2,W_1',W_2'$ be subspaces of $V$ such that $W_1=\left(W_1\cap W_2\right)\oplus W_1'$ and $W_2=\left(W_1\cap W_2\right)\oplus W_2'$, then why $\left(W_1\cap W_2\right)+W_1'+W_2'=\left(W_1\cap W_2\right)\oplus W_1'\oplus W_2'$? I've tried to prove it using the equivalent statement, that $W_1'\cap \left(\left(W_1\cap W_2\right)+ W_2'\right)=W_2'\cap \left(\left(W_1\cap W_2\right)+ W_1'\right)=\left(W_1\cap W_2\right)\cap\left(W_1'+W_2'\right)=\{0\}$ and could prove it for $W_1'\cap \left(\left(W_1\cap W_2\right)+ W_2'\right)$ and $W_2'\cap \left(\left(W_1\cap W_2\right)+ W_1'\right)$ but I got stuck with $\left(W_1\cap W_2\right)\cap\left(W_1'+W_2'\right)$. Is this the right way? Any help is highly appreciated.


That is indeed the correct way.

For $(W_1 \cap W_2) \cap (W_1' + W_2') = \{0\}$ assume there is a vector $x$ contained in there. Then $x \in W_1' + W_2'$, $x \in W_1$, and $x \in W_2$. As $x \in W_1' + W_2'$ write $x = w_1 + w_2$ with $w_i \in W_i'$. Look at $w_1 = x - w_2$. As $x \in W_2$ and $w_2 \in W_2' \subseteq W_2$ this means $w_1 \in W_2$. In particular, $w_1 \in W_1 \cap W_2$. But $W_1 = (W_1 \cap W_2) \oplus W_1'$ so $(W_1 \cap W_2) \cap W_1' = \{0\}$. We have $w_1$ in this intersection so $w_1 = 0$.

The same argument with $1$ and $2$ reversed gives $w_2 = 0$. Thus $x = 0$.

  • $\begingroup$ Thank you so much, your answer is great! +1 $\endgroup$ – Redundant Aunt Feb 7 '15 at 18:27
  • $\begingroup$ I'm just wondering; can you think of another way to prove it? $\endgroup$ – Redundant Aunt Feb 8 '15 at 21:56

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