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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!

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2 Answers 2

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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.

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    $\begingroup$ 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". $\endgroup$ Feb 14, 2016 at 7:24
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  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: \begin{equation} V=W_1\oplus W_2\Leftrightarrow \phi\text{ is a vector space isomorphism.} \end{equation}

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 \begin{equation} \phi(v,w)=\phi(v',w')\Leftrightarrow v+w=v'+w'\Leftrightarrow v-v'=w'-w=:u. \end{equation} Since $v-v'\in W_1$ and $w'-w\in W_2$, $u\in W_1\cap W_2=\{0\}$. Thus, $v=v'$ and $w=w'$.
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