# When is $V$ a direct sum of a subspace $S$ and the set $(V-S)\cup\{\mathbf{0}\}$?

In general, for a vector space V, $\dim(V)< \infty$, with subspace $S \subseteq V$ and the set $W :=(V-S) \cup \{ 0\}$ we find that $V \neq S \oplus W$.

For example, if $V=\mathbb{R}^2$ and $S=\{(x,0) : x \in \mathbb{R} \}$. Then take $(a,b), (c, -b) \in W$, for fixed $a, b, c \in \mathbb{R}$ and $a, b, c$ non-zero, and $a \neq -c$. Then $(a,b)+(c, -b) = (a+c, 0) \in S$, proving W is no subspace.

But, some of the time, $V = S \oplus W$. For example, if $S=V$. (But, that seems trivial.)

For what kind of subspaces, $S \subset V$, is this possible?

UPDATE (to clarify)

Okay, then if $V = W_1 \oplus W_2 \oplus \cdots \oplus W_n$ and if $W_1, W_2,..., W_n \neq V$ or { $0$ } then, of course, by definition, we have:

$V = W_1 + W_2 + \cdots + W_n$ and

Wi are pair-wise disjoint (except for the 0 vector)

$W_i \cap (W_1 + W_2 + \cdots + W_{i-1} + W_{i+1} + \cdots + W_n) = \{ 0 \}, \forall i = 1, 2, ..., n$

But you are telling me we also have $V \neq (W_1 \cup W_2 \cup \cdots \cup W_n)$ always.

I could see that $V \neq (W_1 \cup W_2 \cup \cdots \cup W_n)$, some of the time, but it basically never happens unless one of the $W_i$ is $V$!

This really clears things up for me in terms of what a direct sum is really about! It's not like a partition of V it is more like a set of scaffolds for V.

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No, in order for the sum to be direct, we need more than just $\cap_{i=1}^n W_i=\{0\}$. We need for each $i$, $$W_i\cap(W_1+\cdots+W_{i-1}+W_{i+1}+\cdots+W_n) = \{0\}.$$ For a counterexample to your condition, take $V=\mathbb{R}^2$, $W_1$ the $x$-axis, $W_2$ the $y$-axis, $W_3$ the line $x=y$. Then $V=W_1+W_2+W_3$, $W_1\cap W_2\cap W_3 = \{0\}$ but $V$ is not the direct sum of $W_1$, $W_2$, and $W_3$. As for expressing vector spaces as unions of subspaces, see this question, and this link. – Arturo Magidin Mar 3 '12 at 5:03
Thanks, I should have done it pairwise. I'm going to fix it. – futurebird Mar 3 '12 at 5:04
No, pairwise is not enough either: the example I give has pairwise trivial intersection, but it is not a direct sum. You really do need "each $W_i$ intersects the span of the other $W_j$ trivially." – Arturo Magidin Mar 3 '12 at 5:06
"But you are telling me we also have $V\neq W_1\cup\cdots\cup W_n$ always." Well, not always, just not when there are at least two $W_i$ and none of them are zero. – Arturo Magidin Mar 3 '12 at 5:15
The it stuck in my mind many years ago is this: take $V$ to be $\mathbb{R}^2$, and take the two summands to be the $x$-axis and the $y$-axis. – Martin Argerami Mar 3 '12 at 5:15

If $S$ is a subspace of $V$, and $W=(V-S)\cup\{0\}$, then $W$ is a subspace of $V$ if and only if $S=V$ or $S=\{0\}$.

Indeed, if $S=V$ then $W=\{0\}$ is a subspace; if $S=\{0\}$, then $W=V$ is a subspace.

Assume then that $S\neq\{0\}$ and $S\neq V$. Then there exists $\mathbf{s}\in S$, $\mathbf{s}\neq\mathbf{0}$, and there exists $\mathbf{v}\in V-S$, $\mathbf{v}\neq\mathbf{0}$. Note that $\mathbf{s}\notin W$.

Then $\mathbf{v}+\mathbf{s}\notin S$, since $\mathbf{v}\notin S$; hence $\mathbf{v}+\mathbf{s}\in W$. Since $\mathbf{v}\in W$, but $\mathbf{s} = (\mathbf{v}+\mathbf{s})-\mathbf{v}$ is not in $W$, it follows that $W$ is not closed under differences, hence is not a subspace.

This works whether $V$ is finite dimensional or not.

Added. While it is possible to express a vector space as a union of proper subspaces (see in particular Pete Clark's recent Monthly article for the precise answer of how many proper subspaces you need; there's a copy on his website), the union never yields a direct sum; and if $V$ is a direct sum of at least two proper nontrivial subspaces, then it is never equal to the union: take $w_1\in W_1$, $w_2\in W_2$, both nonzero; then $w_1+w_2$ cannot lie in any of the direct summands.

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It can never happen. Note that $V=S\oplus W$ implies that $W$ is a subspace. Take any $s\in S$, $w\in W$ with $w\ne0$. Then $w\not\in S$, and neither is $s+w$. But if $s+w\not\in S$, this means that $s+w\in W$, which implies that $s\in W$, a contradiction. We conclude that $W=\{0\}$.

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Or that $s=0$... – Arturo Magidin Mar 3 '12 at 4:50
@ArturoMagidin: Good point! – Martin Argerami Mar 3 '12 at 5:12