# Decomposition of vector with respect to direct sum of vector subspaces

$U$ and $W$ are two subspaces of vector space $V$.

If $U \oplus W = V$, then $\forall v \in V$, there exist two unique vectors $u \in U$ and $w \in W$ such that $v = u + w$.

Is the reverse true? That is, if any vector can has such unique decomposition, do we have $U \oplus W = V$?

Can the above statements be generalized to finite number of subspaces $U_1,...,U_n$ instead of just two $U$ and $W$?

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I think you mean \oplus. What definition of direct sum are you working with here? –  Qiaochu Yuan Jan 19 '11 at 0:01
The sum of two subspaces is defined to be the span of the two. When the two subspaces are disjoint, their sum is called direct sum. Is this the definition you use? –  Linda Jan 19 '11 at 0:05
Yes, I mean \oplus. Thanks! –  Linda Jan 19 '11 at 0:07

If every vector $v$ in $V$ has at least one expression of the form $v=u+w$ with $u\in U$ and $w\in W$ ($U$ and $W$ subspaces of $V$), then $V=U+W$.

If every vector $v$ in $V$ has at most one (but possibly none) expression of the form $v=u+w$ with $u\in U$ and $w\in W$, then $U\cap W = \{0\}$.

So if every vector $v$ in $V$ has exactly one expression of the form $v=u+w$ with $u\in U$ and $w\in W$, then $V=U\oplus W$.

For more than two spaces you have to be a bit careful: it is no longer enough for the $U_i$ to be pairwise disjoint (that is, for $U_i\cap U_j$ to equal $\{0\}$ for $i\neq j$). For an example, take $V=\mathbb{R}^2$, $U_1$ the $x$-axis, $U_2$ the $y$-axis, and $U_3$ the line $x=y$. Then $U_1\cap U_2 = U_1\cap U_3 = U_2\cap U_3 = \{0\}$, but $V$ is not the direct sum of $U_1$ and $U_2$. Instead, you have that given subspaces $U_1,\ldots,U_m$ of $V$,

• Each vector $v\in V$ has at least one expression of the form $v=u_1+\cdots + u_m$ with $u_i\in U_i$ if and only if $V=U_1+\cdots+U_m$ (the span of $U_1,\ldots,U_m$).

• Each vector $v\in V$ has at most one (but possibly no) expression of the form $v=u_1+\cdots + u_m$ with $u_i\in U_i$ if and only if for each $i\in\{1,\ldots,m\}$, $U_i\cap(\mathop{\sum}_{i\neq j}U_j) = \{0\}$.

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Yes. In fact it suffices to verify that the uniqueness of the decomposition implies that $U$ and $W$ are disjoint. Together, they also span $V$, since each vector possesses such a decomposition. To make the proof explicit, consider the mapping $V \rightarrow U \times W$, sending each vector $v$ to its decomposition $(u,w)$; then you can verify the axioms for the direct sum.
Just to clarify: $U,V$ should not be disjoint, but satisfy $U \cap V=\{0\}$. –  Fredrik Meyer Jan 19 '11 at 14:56