# Understanding the definition of the direct sum of subspaces of a vector space

I have a question regarding the definition of direct sum of a vector space in relation to subspaces.

Definition: A vector space $V$ is called the direct sum of $W_1$ and $W_2$ if $W_1$ and $W_2$ are subspaces of $V$ such that $W_1\cap W_2 = \{0\}$ and $W_1 + W_2 = V$. We denote that $V$ is the direct sum of $W_1$ and $W_2$ by writing $V = W_1\oplus W_2$.

Is this definition saying that any vector in $V$ can be written as a linear combination of the vectors in the set $W_1 + W_2$?

Thanks!

The definition is saying that any vector $v \in V$ can be written as $v = w_1 + w_2$ where $w_1 \in W_1$ and $w_2 \in W_2$ (this is the condition $W_1 + W_2 = V$), and this decomposition is unique (this follows from the condition $W_1\cap W_2 = \{0\}$). You don't need to take linear combinations.

• Right, the key word missing from the OPs statement is "unique". Of course what the OP says is true, it does mean any vector can be written as a combination like that. But it's not equivalent to that without specifying "unique". Feb 14, 2016 at 7:24
1. $$W_1\cap W_2=\{0\}$$ if and only if \begin{align} \phi\colon W_1\times W_2&\to V\\ (v,w)&\mapsto v+w \end{align} is injective and
2. $$V=W_1+W_2$$ if and only if $$\phi$$ is surjective.

In summary: $$$$V=W_1\oplus W_2\Leftrightarrow \phi\text{ is a vector space isomorphism.}$$$$

In other words: $$V=W_1\oplus W_2$$ if and only if for all $$v\in V$$, there is exactly one $$(w_1,w_2)\in W_1\times W_2$$ such that $$v=w_1+w_2$$.

Proof of 1.:

(Reminder: If $$W_1$$ and $$W_2$$ are vector subspaces of $$V$$, $$W_1\cap W_2$$ is also a vector subspace of $$V$$.)

• If $$v\in W_1\cap W_2$$, $$\phi(v,0)=v+0=v=0+v=\phi(0,v)$$. $$\Rightarrow v=0$$ if $$\phi$$ is injective.
• Conversely, if $$W_1\cap W_2=\{0\}$$ and $$(v,w),(v',w')\in W_1\times W_2$$, then $$$$\phi(v,w)=\phi(v',w')\Leftrightarrow v+w=v'+w'\Leftrightarrow v-v'=w'-w=:u.$$$$ Since $$v-v'\in W_1$$ and $$w'-w\in W_2$$, $$u\in W_1\cap W_2=\{0\}$$. Thus, $$v=v'$$ and $$w=w'$$.