Decomposition of vector with respect to direct sum of vector subspaces

Question

$U$ 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 $W$?

Thanks for your help!

Answer 1

To add to Gerben's answer:

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 $i\in\{1,\ldots,m\}$, $U_i\cap(\mathop{\sum}_{i\neq j}U_j) = \{0\}$.

Answer 2

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.

For the general case, you can do this by recurrence, using the basic properties of direct sums (in particular associativity).