If $V=V_1\oplus V_2$, and $W$ subspace of $V$, then $W=(W\cap V_1)\oplus(W\cap V_2)$ if and only if there is a subset $S\subset W$ such that $W=\text{span}(S)$ and $S=(S\cap V_1)\cup(S\cap V_2)$.


$\implies$: $W=(W\cap V_1)\oplus(W\cap V_2)$, thus any vector $w\in W$ can be expressed as $w=w_1+w_2$, where $w_1\in W\cap V_1$ and $w_2\in W\cap V_2$. Consider any spanning set of $W$, $R=\{w_k\mid k\in K\}$. Again, $w_k=w_{k,1}+w_{k,2}$. We define $S=\{w_{k,1}\mid k\in K\}\cup \{w_{k,2}\mid k\in K\}$. Clearly $\langle S\rangle=W$. Additionally, $(S\cap V_1)\cup(S\cap V_2)=\{w_{k,1}\mid k\in K\}\cup \{w_{k,2}\mid k\in K\}=S$.

$\impliedby$: From $S=(S\cap V_1)\cup(S\cap V_2)$ follows that, if $0\neq w\in S$, then either $w\in V_1$ or $w\in V_2$ ($V=V_1\oplus V_2$ implies $V_1\cap V_2=\{0\}$). We define $S_1:=S\cap V_1$ and $S_2:=S\cap V_2$. Any vector $w\in W$ can be expressed as $w=w_1+\cdots+w_m+w'_1+\cdots+w'_n$, where $w_i\in S_1$ and $w'_i\in S_2$. $w_1+\cdots+w_m\in\langle S_1\rangle$ and $w'_1+\cdots+w'_n\in\langle S_2\rangle$. Now $\langle S_1\rangle=W\cap V_1$ and $\langle S_2\rangle=W\cap V_2$ ($\color{red}{why?}$), therefore $(W\cap V_1)\oplus(W\cap V_2)=W$. $\tag*{$\square$}$

$\color{red}{why?}$ Clearly $\langle S_1\rangle\subset W\cap V_1$


That is because $V_1$ and $V_2$ are in direct sum, and the elements of $S$ which are not in $V_1$ are all in $V_2$.

Edit: pick $v\in W\cap V_1$. Since $W=$ Span $S$, and since $S=S_1\cup S_2$, $v$ can be written $v_1+v_2$ with $v_1\in \left\langle S_1\right\rangle\subset V_1$ and $v_2\in \left\langle S_2\right\rangle\subset V_2$.

Now $v-v_1\in V_1\cap V_2$, therefore $v=v_1$.

  • $\begingroup$ Thank you for your reply. I used the fact that the elements of $S$ are either in $V_1$ or in $V_2$ previously in the proof. Now the question is why $\langle S_1\rangle=W\cap V_1$ and particularly why $v\in W\cap V_1$ implies $v\in\langle S_1\rangle$. I'm not able to get the connection so directly. $\endgroup$ – PeptideChain May 8 '18 at 20:03
  • $\begingroup$ @LiPo I understood your question. What is not clear in my answer? $\endgroup$ – Arnaud Mortier May 8 '18 at 20:06
  • $\begingroup$ you speak about $S$, I'm speaking about $\langle S_1 \rangle$. I honestly do not see how your answer is connected with my question (the connection with the rest is clear). $\endgroup$ – PeptideChain May 8 '18 at 20:08
  • $\begingroup$ @LiPo I elaborated a bit, is it clearer now? $\endgroup$ – Arnaud Mortier May 8 '18 at 20:12
  • $\begingroup$ Thank you! I needed some time... now it is very clear $\endgroup$ – PeptideChain May 8 '18 at 20:18

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