Lemma. condition for a direct sum
Suppose \(V_{1},\dots,V_{m}\) are sub-spaces of \(V\). Then \(V_{1}+\dots+V_{m}\) is a direct sum if and only if the only way to write \(0_{V}\) as a sum \(v_{1}+\dots+v_{m}\) s.t \(\forall1\leq k\le m,\;v_{k}\in V_{k}\) is with \(\forall1\leq k\le m,\;v_{k}=0_{V}\).
Proof. We prove the two implications separately.
First direction:
Suppose \(V_{1}+\dots+V_{m}\) is a direct sum.
Then, in particular, \(0\in V_{1}+\dots+V_{m}\) can be written in only one way as a sum \(v_{1}+\dots+v_{m}=0\) where \(\forall1\leq k\leq m,\ v_{k}\in V_{k}\).
Since \(V_{1},\dots,V_{m}\) are all sub-spaces of \(V\), they all contain \(0\) (i.e. \(0_{V}\)).
And since \(0+\dots+0=0\), we get that this is the one way as needed.
Second direction:
Let us assume \(0+\dots+0=0\) is the only way to write \(0\) as a sum \(v_{1}+\dots+v_{m}\) s.t. \(\forall1\leq k\leq m,\ v_{k}\in V_{k}\).
Then for every \(a\in V_{1}+\dots+V_{m}\), if \(v_{1}+\dots+v_{m}=a\) s.t. \(\forall1\leq k\leq m,\ v_{k}\in V_{k}\) and \(v'_{1}+\dots+v'_{m}=a\) s.t. \(\forall1\leq k\leq m,\ v'_{k}\in V_{k}\), we get that \[v_{1}+\dots+v_{m}-\left(v'_{1}+\dots+v'_{m}\right)=a-a=0\] \[\therefore\quad(v_{1}-v'_{1})+(v_{2}-v'_{2})+\dots+(v_{m}-v'_{m})=0\]
each \(V_{k}\) is a sub-space, hence closed under addition and scalar multiplication, therefore for each \(1\le k\le m\),
\((v_{k}-v'_{k})=v_{k}+(-1)\cdot v'_{k}\in V_{k}\), so the assumption dictates that \[v_{1}-v'_{1}=0\ \land\ v_{2}-v'_{2}=0\ \land\ \dots\ \land\ v_{m}-v'_{m}=0\] \[\therefore\quad v_{1}=v'_{1}\ \land\ v_{2}=v'_{2}\ \land\ \dots\ \land\ v_{m}=v'_{m}\] so there is exactly one way to write \(a\) in this manner (\(a\in V_{1}+\dots+V_{m}\) definitionally means one exists) and \(V_{1}+\dots+V_{m}\) is a direct sum as needed, hence the notation \(V_{1}\oplus\dots\oplus V_{m}\) can be used. ∎