Three Independent Subspaces that Direct Sum to the Whole Vector Space Three subspaces $L$, $M$, and $N$ of a vector space $V$ are called independent if each one is disjoint from the sum of the other two.  Prove that a necessary and sufficient condition for $V=L\bigoplus(M\bigoplus N)$ (and also for $V=(L\bigoplus M)\bigoplus N$) is that $L$, $M$, and $N$ be independent and that $V=L+M+N$.
I am aware that necessary and sufficient is the same as if and only if.  I'm confused by the question, partially because when it says "disjoint from the sum of the other two" is it talking about the direct sum or the spanning sum?  I think I need some help getting the proof started as well.  Thank you in advance.
 A: Context: the question asks problem 5(a) from Halmos' Finite Dimensional Vector Spaces, section 20.

I'm confused by the question, partially because when it says "disjoint from the sum of the other two" is it talking about the direct sum or the spanning sum?

Yep, this confused me too. What does it mean to say that $L \cap (M \oplus N) = \{ 0 \} $? If we consider $M \oplus N$ the external direct sum, then $M \oplus N$ is the set of all pairs $\langle x, y \rangle$ where $x \in M$ and $y \in N$. It is not clear what subspace of $V$ this direct sum corresponds to, i.e. which vectors of $V$ are in $M \oplus N$. The whole thing makes me uneasy.
Luckily, the theorem of 5(a) follows nicely if we take independent to mean that each subspace is disjoint from the spanning sum of the other two, i.e. $L \cap (M + N) = \{ 0 \}$ and $M \cap (L + N) = \{ 0 \}$ and so on.
Necessary: Suppose $V = L \oplus M \oplus N$. I take this to mean that every vector $x$ in $V$ can be written uniquely as the sum $x_1 + x_2 + x_3$ where $x_1 \in L$, $x_2 \in M$ and $x_3 \in N$.
As an example of the general case, we will show that $L$ is disjoint from $M + N$. Suppose there is a vector $v$ in both $L$ and $M + N$. This gives us two expressions of $v$ in terms of $L$, $M$ and $N$:
$$
v = x_1 + 0 + 0
$$
$$
v = 0 + x_2 + x_3
$$
Since the expression of $v$ must be unique, these combinations can only be $0 + 0 + 0$, so $v = 0$.
Sufficient: Suppose the three subspaces are independent and that $V = L + M + N$. Then every vector $v$ has some expression $x_1 + x_2 + x_3$, with each $x$ coming from one of our subspaces. However, it must be shown that this expression is unique. Suppose there is a different expression, i.e. $v = y_1 + y_2 + y_3$ with $y_1 \in L$, $y_2 \in M$ and $y_3 \in N$. By subtracting one expression from the other, we obtain
$$
(x_1 - y_1) + (x_2 - y_2) + (x_3 - y_3) = 0
$$
At least one of the bracketed expressions on the left does not equal 0 - say it's the first. Then we have:
$$
(x_2 - y_2) + (x_3 - y_3) = y_1 - x_1
$$
But now both sides express a certain vector which belongs to both $M + N$ (left side) and $L$ (right side). So we have arrived at a contradiction with our assumption of independence. That means the expression $x_1 + x_2 + x_3$ must be unique after all.
