Vector space/Direct sum/Introduction/Section

Definition

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space. Let ${}U_{1},\ldots ,U_{m}$ be a family of linear subspaces of ${}V$. We say that ${}V$ is the direct sum of the ${}U_{i}$ if the following conditions are fulfilled.

1. Every vector ${}v\in V$ has a representation

   ${}v=u_{1}+u_{2}+\cdots +u_{m}\,,$

   where ${}u_{i}\in U_{i}$.

2. ${}U_{i}\cap {\left(\sum _{j\neq i}U_{j}\right)}=0$ for all ${}i$.

If the sum of the ${}U_{i}$ is direct, then we also write ${}U_{1}\oplus \cdots \oplus U_{m}$ instead of ${}U_{1}+\cdots +U_{m}$. For two linear subspaces

${}U_{1},U_{2}\subseteq V\,,$

the second condition just means ${}U_{1}\cap U_{2}=0$.

Example

Let ${}V$ denote a finite-dimensional ${}K$-vector space together with a basis ${}v_{1},\ldots ,v_{n}$. Let

${}\{1,\ldots ,n\}=I_{1}\uplus \ldots \uplus I_{k}\,$

be a partition of the index set. Let

${}U_{j}=\langle v_{i},\,i\in I_{j}\rangle \,$

be the linear subspaces generated by the subfamilies. Then

${}V=U_{1}\oplus \cdots \oplus U_{k}\,.$

The extreme case ${}I_{j}=\{j\}$ yields the direct sum

${}V=Kv_{1}\oplus \cdots \oplus Kv_{n}\,$

with one-dimensional linear subspaces.

Lemma

Let ${}V$ be a finite-dimensional ${}K$-vector space, and let ${}U\subseteq V$ be a linear subspace. Then there exists a linear subspace ${}W\subseteq V$ such that we have the direct sum decomposition

${}V=U\oplus W\,.$

Proof

Let ${}v_{1},\ldots ,v_{k}$ denote a basis of ${}U$. We can extend this basis, according to fact, to a basis ${}v_{1},\ldots ,v_{k},v_{k+1},\ldots ,v_{n}$ of ${}V$. Then

${}W=\langle v_{k+1},\ldots ,v_{n}\rangle \,$

fulfills all the properties of a direct sum.

$\Box$

In the preceding statement, the linear subspace ${}W$ is called a direct complement for ${}U$ (in ${}V$). In general, there are many different direct complements.