# If $W_1 \subseteq V' \subseteq W_1 \oplus W_2 = V$, then $V' = W_1 \oplus (V' \cap W_2)$

Question: Let $W_1$ and $W_2$ be subspaces of a vector space $V$ such that $V = W_1 \oplus W_2$. Prove that for every subspace $V'$ of $V$ containing $W_1$ one has $V' = W_1 \oplus (V' \cap W_2)$.

Attempt:

Let $W_1$ be a subset of $V'$ spanned by the basis vectors $u_1,\dots,u_r$. Then $W_1$ can similarly be completed to a basis of $V'$, say $u_1,\dots,u_r,u_{r+1},\dots,u_s$. $W_2$ is such that it is completed by the basis vectors $u_{r+1},\dots,u_n$, so a vector $x\in V'\cap W_2$ can therefore be written as a linear combination of the basis vectors $u_{r+1},\dots,u_s$, which are linearly independent. Hence, $u_{r+1},\dots,u_s$ spans $V'\cap W_2$, so $V'=W_1+(V'\cap W_2)$, and clearly $W_1\cap (V'\cap W_2)=\{0\}$, so $V'=W_1\oplus (V'\cap W_2)$.

• Don't mind the "similarly." – Trancot May 22 '13 at 8:12
• Things begin to be problematic from the very beginning: $\,W_1\,$ is a given subspace of $\,V\,$ , so how come you begin with "Let $\,W_1\,$ be a subset of $\,V\,$ ..." ? – DonAntonio May 22 '13 at 8:17
• I think you've got the right idea and you're very close to wrap it all successfully. I'd rather begin with "Let $\,A\,$ be a basis of $\,W_1\le V'\,$ , complete it to a basis $\,B\,$ of $\,V'\,$, and now complete this to a basis $\,C\,$ of the whole space $\,V\,$ . Since $\,V=W_1\oplus W_2\,$ , clearly $\;C-A\,$ must be a basis of $\,W_2\,$ and etc. – DonAntonio May 22 '13 at 8:20

Baisis of $W_1: \ \ u_1,...,u_r:=A$ Basis of $W_2: \ \ v_1,..,v_s:=B$
Since $W_1$ is completly in $V'$ , $A$ can be extendet to a Basis of $V'$. But Now again $V'$ is a subspace of $V:=W_1\oplus W_2$. This means that the basis of $V'$ , call it $C$ can be written as a lin. combination of $A$ and $B$. Hence the only way to extend $A$ to $C$ is by taking the remaing vectors from $B$ that complete $C$ , or equivalently : $C=A+B\cap C$ and since all vectors in $A+B\cap C$ are lin. independet we can write $V'=W_1\oplus (V'\cap W_2)$
It is clear that $W_1\cap(V'\cap W_2)\subseteq W_1\cap W_2 = \{0\}$. Let $v\in V'\subseteq V = W_1\oplus W_2$. Then we can write $v=w_1+w_2$ for some $w_1\in W_1$ and $w_2\in W_2$. Since $v\in V'$ and $w_1\in W_1\subseteq V'$, we obtain $w_2 = v-w_1\in V'$, that is $w_2\in V'\cap W_2$. Hence, $v = w_1+w_2\in W_1+(V'\cap W_2)$. Therefore, $V' = W_1\oplus(V'\cap W_2)$, as required.