# Show that $W=(W\cap W_1)\oplus(W\cap W_2)$

Statement: If $V=W_1\oplus W_2$ and $W$ is any subspace such that $W_1\subseteq W$. Show that $W=(W\cap W_1)\oplus (W\cap W_2)$.

Proof: Let $V=W_1\oplus W_2$ and W is any subspace such that $W_1\subseteq W$.

Case 1: $W=W_1$ and $W_1\subseteq W$. Since $w\in W$ iff $w\in W_1$, then if $w\in W_1$ then $w\notin W_2$. Which implies $W\cap W_2=\emptyset$. $\Rightarrow$ $(W_1\cap W)\oplus(W_2\cap W)=W\oplus\emptyset=W$.

Case 2: $W\neq W_1$ and $W_1\subseteq W$. $\Rightarrow$ $\exists w\in W$ such that $w\notin W_1$. However, if $w'\in W_1$ then $w'\in W$. Suppose $\omega\in W$ such that $\omega\notin W_1$. But $W\subset V$, thus $\omega\in V$ which forces $\omega\in W_2$ and implying that $W\cap W_2\neq\emptyset$. Let $U=\{\mu\in V\mid \mu\notin W_1\land\mu\in W_2\}$. Thus, $\omega\in U$, $\forall\omega\in V$ such that $\omega\notin W_1$ and $\omega\in W_2$. Therefore, $W_1\subset W$ and $W\cap W_2=U\neq\emptyset\Rightarrow U\subset W\Rightarrow (W\cap W_1)\oplus(W\cap W_2)=W_1\oplus U\subset W$. Since it is also the case that $W\subset W_1\oplus U$, then $W=(W\cap W_1)\oplus(W\cap W_2)$.

• The argument that $w\notin W_1 \implies w\in W_2$ is very problematic, you need a better understanding of direct sum of subspaces. The fact that $V=W_1\oplus W_2$ does not mean $V=W_1\cup W_2$. – Mathematical Feb 6 '18 at 16:31
• The book I'm using discussed it in a paragraph. I'm getting a different book then. thanks! – TheLast Cipher Feb 6 '18 at 16:35
• what would be a valid way of arguing this? What i do know now is since this is a direct sum then $W_1\cap W_2=\emptyset$. Also are you referring to this argument "But $W\subset V$, thus $\omega\in V$ which forces $\omega\in W_2$" – TheLast Cipher Feb 6 '18 at 16:37
• Yes, I was referring to that sentence :) – Mathematical Feb 6 '18 at 16:59

Here is a detailed proof so that you understand the basics. The proof is split in two parts, following the definition of a direct sum: first prove that $\forall w\in W$, there exist $w_1\in W\cap W_1$ and $w_2\in W\cap W_2$, such that $w=w_1+w_2$; second prove that $(W\cap W_1)\cap(W\cap W_2) = \{\mathbf{0}\}$.
$\forall w\in W$, since $w\in V$, there exist $w_1\in W_1$ and $w_2\in W_2$ such that $w=w_1+w_2$. Now since $W_1\subseteq W$, we get $w_1 \in W_1 = W\cap W_1$. Since $W$ is a subspace and $w,w_1\in W$, we get $w_2 = w - w_1 \in W$, so $w_2 \in W\cap W_2$. The first part is thus proved.
The second part follows immediately from the fact that $V = W_1 \oplus W_2$ is a direct sum (so that $W_1\cap W_2=\{\mathbf{0}\}$) and that $\mathbf{0} \in W\cap W_1 \subseteq W_1$, $\mathbf{0} \in W\cap W_2 \subseteq W_2$.
• Okay. I understand the proof, but not so sure about the motivation of the first part as to why this is what needs to be shown. The definition I've read is that $W$ is a direct sum of $W_1$ and $W_2$; $W=W_1\oplus W_2$ if the following conditions are met: (a)$W_1$ and $W_2$ are independent of each other (I think this is equivalent to saying that every element of one subspace is independent of the complement subspace), (b)the intersection of the subspaces is the zero vector(as what you've shown), (c) the order basis $\beta_i$ of all subspace $W_i$ is independent (this directly follows from a.) – TheLast Cipher Feb 7 '18 at 4:47
• @TheLastCipher I cannot quite understand the first definition containing (a), (b) and (c), actually I think it is not very clearly stated. And yes, I followed a very well known definition of direct sums in my answer. Other equivalent definitions exist and the essence is that $\forall w$, $w = w_1 + w_2$ and $w_1$ and $w_2$ are unique. – Mathematical Feb 7 '18 at 15:14