complementary subspace
Complimentary
2002-07-26 08:08:07
2008-06-01 19:46:22
Definition
complementary
direct sum
decomposition
orthogonal complement
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\paragraph{Direct sum decomposition.}
Let $U$ be a vector space, and $V,W\subset U$ subspaces. We say that
$V$ and $W$ span $U$, and write
$$U=V+W$$ if
every $u\in U$ can be expressed as a sum
$$u=v+w$$
for some $v\in V$ and $w\in W$.
If in addition, such a decomposition is unique for all $u\in U$, or
equivalently if
$$V\cap W=\{ 0\},$$
then we say that $V$ and $W$ form a \emph{direct sum decomposition} of $U$
and write
$$U=V\oplus W.$$
In such circumstances, we also say that $V$ and $W$
are \emph{complementary} subspaces, and also say that $W$ is an \emph{algebraic complement} of $V$.
Here is useful characterization of complementary subspaces if $U$ is
finite-dimensional.
\begin{proposition}
Let $U, V, W$ be as above, and suppose that $U$ is
finite-dimensional. The subspaces $V$ and $W$ are complementary if
and only if for every basis $v_1,\ldots, v_m$ of $V$ and
every basis
$w_1,\ldots,w_n$ of $W$, the combined list
$$v_1,\ldots,v_m,w_1,\ldots,w_n$$
is a basis of $U$.
\end{proposition}
\textbf{Remarks}.
\begin{itemize}
\item
Since every linearly independent subset of a vector space can be extended to a basis, every subspace has a complement, and the complement is necessarily unique.
\item
Also, direct sum decompositions of a vector space $U$ are in a one-to correspondence fashion with projections on $U$.
\end{itemize}
\paragraph{Orthogonal decomposition.}
Specializing somewhat, suppose that the ground field $\kf$ is either
the real or complex numbers, and that $U$ is either an inner product
space or a unitary space, i.e. $U$ comes equipped with a
positive-definite inner product
$$\langle,\rangle:U\times U\rightarrow \kf.$$
In such circumstances,
for every subspace $V\subset U$ we define the orthogonal complement of
$V$, denoted by $V^\perp$ to be the subspace
$$V^\perp = \{ u\in U: \langle v,u\rangle = 0,\text{ for all }
v\in V\}.$$
\begin{proposition}
Suppose that $U$ is finite-dimensional and $V\subset U$ a subspace.
Then, $V$ and its orthogonal
complement $V^\perp$ determine a direct sum decomposition of $U$.
\end{proposition}
Note: the Proposition is false if either the finite-dimensionality
or the positive-definiteness assumptions are violated.