Given a subspace \(W\) of \(\RNrSpc{n}\), together with a splitting \(W=U\dotplus V\), each \(\Vect{w}\in W\) can be expressed in exactly one way as \(\Vect{w}=\Vect{u}+\Vect{v}\), with \(\Vect{u}\in U\) and \(\Vect{v}\in V\).
Given a subspace \(W\) of \(\RNrSpc{n}\), together with a splitting \(W=U\dotplus V\), each \(\Vect{w}\in W\) can be expressed in exactly one way as \(\Vect{w}=\Vect{u}+\Vect{v}\), with \(\Vect{u}\in U\) and \(\Vect{v}\in V\).
From the property that \(W=\span(U\cup V)\), we deduce that \(\Vect{w}\in W\) can be expressed in some way as \(\Vect{w}=\Vect{u}+\Vect{v}\) with \(\Vect{u}\in U\) and \(\Vect{v}\in V\). It remains to show that this is the only way to achieve such an expression. So consider the situation where there are \(\Vect{u},\Vect{u}'\in U\) and \(\Vect{v},\Vect{v}'\in V\) with
\[\Vect{u} + \Vect{v} = \Vect{w} = \Vect{u}' + \Vect{v}'\]We then have
| \(\Vect{u} - \Vect{u}'\) | \(=\) | \(\Vect{v}' - \Vect{v}\) |
Now \((\Vect{u}-\Vect{u}')\in U\) because \(U\) is closed under vector addition, and \((\Vect{v}'-\Vect{v})\in V\) because \(V\) is closed under vector addition. This means that \((\Vect{u}-\Vect{u}')\) belongs to both, \(U\) and \(V\) and, therefore,
| \(\Vect{u} - \Vect{u}'\) | \(=\) | \(\Vect{0}\) |
| \(\Vect{u}\) | \(=\) | \(\Vect{u}'\) |
The identity \(\Vect{v}=\Vect{v}'\) follows with a similar argument. This proves the claim.