[proofplan]
We use the definition of the internal [direct sum](/page/Direct%20Sum): every vector $v \in V$ must have a unique decomposition $v = u+w$ with $u \in U$ and $w \in W$. In the forward direction, existence gives $V = U+W$, while uniqueness applied to the two decompositions $0_V = 0_V+0_V = x+(-x)$ forces every $x \in U \cap W$ to vanish. In the reverse direction, $V=U+W$ gives existence, and the hypothesis $U \cap W=\{0_V\}$ forces two decompositions of the same vector to be equal term by term.
[/proofplan]
[step:Use uniqueness in the direct sum to force the intersection to be zero]
Assume $V = U \oplus W$. By the definition of internal direct sum, every $v \in V$ has a unique expression
\begin{align*}
v = u+w
\end{align*}
with $u \in U$ and $w \in W$. The existence part of this definition gives $V = U+W$.
It remains to prove $U \cap W = \{0_V\}$. Since $U$ and $W$ are subspaces of $V$, both contain $0_V$. Hence $\{0_V\} \subset U \cap W$.
For the reverse inclusion, let $x \in U \cap W$. Then $x \in U$ and $x \in W$. Since $W$ is a subspace, $-x \in W$. Since $U$ and $W$ both contain $0_V$, the zero vector has two decompositions of the required type:
\begin{align*}
0_V = 0_V + 0_V
\end{align*}
and
\begin{align*}
0_V = x + (-x).
\end{align*}
Here the first summands $0_V$ and $x$ lie in $U$, and the second summands $0_V$ and $-x$ lie in $W$. By uniqueness of the direct-sum decomposition, the first summands are equal, so $x = 0_V$. Therefore $U \cap W \subset \{0_V\}$, and hence
\begin{align*}
U \cap W = \{0_V\}.
\end{align*}
[/step]
[step:Use the zero intersection to prove uniqueness of decompositions]
Assume now that $V = U+W$ and $U \cap W = \{0_V\}$. The equality $V=U+W$ means that for every $v \in V$ there exist $u \in U$ and $w \in W$ such that
\begin{align*}
v = u+w.
\end{align*}
Thus the required decompositions exist.
We prove uniqueness. Let $v \in V$, and suppose $u,u' \in U$ and $w,w' \in W$ satisfy
\begin{align*}
v = u+w
\end{align*}
and
\begin{align*}
v = u'+w'.
\end{align*}
Then
\begin{align*}
u+w = u'+w'.
\end{align*}
Subtracting $u'$ and $w$ in the [vector space](/page/Vector%20Space) $V$ gives
\begin{align*}
u-u' = w'-w.
\end{align*}
Because $U$ is a subspace and $u,u' \in U$, the vector $u-u'$ lies in $U$. Because $W$ is a subspace and $w,w' \in W$, the vector $w'-w$ lies in $W$. Since these two vectors are equal, their common value lies in $U \cap W$. By the hypothesis $U \cap W = \{0_V\}$, we get
\begin{align*}
u-u' = 0_V.
\end{align*}
Therefore $u=u'$. Substituting $u=u'$ into $u+w=u'+w'$ gives $w=w'$. Hence the decomposition of $v$ is unique.
[guided]
We now start from the two algebraic conditions and recover the definition of an internal direct sum. That definition has two parts: every vector must have a decomposition, and that decomposition must be unique.
The existence part is exactly the equality $V = U+W$. By the definition of the sum of subspaces, this means that for each vector $v \in V$ there are vectors $u \in U$ and $w \in W$ such that
\begin{align*}
v = u+w.
\end{align*}
The only remaining issue is uniqueness. Fix a vector $v \in V$, and suppose it has two decompositions of the required form. Thus there are vectors $u,u' \in U$ and $w,w' \in W$ such that
\begin{align*}
v = u+w
\end{align*}
and
\begin{align*}
v = u'+w'.
\end{align*}
Equating the two expressions for $v$ gives
\begin{align*}
u+w = u'+w'.
\end{align*}
We isolate the difference between the $U$-parts on one side and the difference between the $W$-parts on the other side. Subtracting $u'$ and $w$ inside the vector space $V$ gives
\begin{align*}
u-u' = w'-w.
\end{align*}
Now the subspace hypotheses become essential. Since $U$ is closed under subtraction and $u,u' \in U$, the vector $u-u'$ belongs to $U$. Since $W$ is closed under subtraction and $w,w' \in W$, the vector $w'-w$ belongs to $W$. The equality above says these are the same vector, so that common vector lies in both subspaces:
\begin{align*}
u-u' \in U \cap W.
\end{align*}
By the hypothesis $U \cap W = \{0_V\}$, the only vector that can lie in both subspaces is $0_V$. Hence
\begin{align*}
u-u' = 0_V.
\end{align*}
Therefore $u=u'$. Returning to the equality $u+w=u'+w'$ and substituting $u=u'$ gives
\begin{align*}
w=w'.
\end{align*}
Thus the two decompositions are identical term by term, which proves uniqueness.
[/guided]
[/step]
[step:Conclude the equivalence with the definition of internal direct sum]
Under the hypotheses $V=U+W$ and $U \cap W=\{0_V\}$, the previous step proves that every $v \in V$ has a unique decomposition $v=u+w$ with $u \in U$ and $w \in W$. By the definition of internal direct sum, this is precisely
\begin{align*}
V = U \oplus W.
\end{align*}
Together with the forward implication, this proves the equivalence.
[/step]
