# Explanation of direct sums (linear vector spaces)

I'm a little unclear on what the $\oplus$ operator actually means with regards to vector spaces, subspaces, and spans. The definition I keep coming across is $$U,W\subseteq V\ ,\ U+W=V\ ,\ U\cap W=\{\underline{0}\} \Rightarrow V=U\oplus W$$

I understand what that means with regards to $V$ but it still doesn't really explain to me what the $\oplus$ operator actually does. I guess what I'm asking is if I take two linear vector spaces we'll call $P$ and $Q$ that are over some field $F$, what would the result of $P\oplus Q$ look like?

Here's an example problem I have to solve in this area - please don't post solutions as I'd prefer to try and solve it myself - I'm just not clear on what the direct sum means and I need that clarified to proceed in in the unit on subspaces.

Prove or disprove:

If $S,T\subseteq V$ are non-empty sets and $S\cap T = \phi$, then $Sp(S\cup T)=Sp(S)\oplus Sp(T)$ .

I understand this is a fairly simple problem but I'm not entirely sure what the significance of the direct sum operator is here, which is why I'm asking for some clarification.

Sorry for the long post and thanks in advance :)

Let $P$ and $Q$ be vector spaces over $F$. The direct sum of $P$ and $Q$ is the $F$-vector space $P\oplus Q$ whose underlying set is $P\times Q$ and whose linear operations are \begin{align*} \lambda\cdot (p,q)&=(\lambda\cdot p,\lambda\cdot q) & (p_1,q_1)+(p_2,q_2)&=(p_1+p_2,q_1+q_2) \end{align*} One quickly verifies that $P\oplus Q$ is well-defined.

Now, note that we have canonical linear maps \begin{array}{ccccccc} P&\xrightarrow{i_P}& P\oplus Q & & Q&\xrightarrow{i_Q}& P\oplus Q \\ p & \mapsto & (p,0) & &q & \mapsto & (0,q) \end{array} One easily checks that these maps are injective. This allows us to canonically view $P$ and $Q$ as subspaces of $P\oplus Q$. Note that this identification can be written as \begin{align*} P &= \{(p,q)\in P\oplus Q:q=0\} & Q &= \{(p,q)\in P\oplus Q:p=0\} \end{align*} Of course, we have abuses notation a bit here.

Consequently we see that $P\cap Q=\{0\}$.

• Okay so if I'm understanding you correctly, put in simple terms $P\oplus Q = \{\underline{x}\in V \ |\ \underline{x}=\underline{p}+\underline{q}\ ,\ \underline{p} \in P\ ,\ \underline{q} \in Q\}$ ? Dec 23, 2015 at 7:32
• @OrBairey-Sehayek Not quite. $P\oplus Q = \{(\underline{p},\underline{q}) \ |\ \underline{p} \in P\ ,\ \underline{q} \in Q\}$. When $P, Q$ are subspaces of some other vector space $V$, then there is a natural isomorphism between this space and the one you described. Otherwise, the expression $p+q$ may not even make sense.