If \(V\) contains only the zero vector, we have \(V^{\bot}=W\), and so the dimension formula is valid in this case:

\(\dim(W)\)

\(=\)

\(0 + \dim(V^{\bot})\)

Consider now the case where \(V\) contains a nonzero vector. Then we have

1. \(\span(V\cup V^{\bot}) = W\), because every \(\Vect{x}\in W\) can be written as \(\Vect{x} = \Vect{x}_V + \Vect{x}_{\bot}\) which is a linear combination of a vector in \(V\) and one in \(V_{\bot}\).
2. \(V\) and \(V^{\bot}\) have only the zero vector in common. – If not, the decomposition of \(\Vect{x}\in W\) as \(\Vect{x} = \Vect{x}_{V} + \Vect{x}_{\bot}\) would not be unique.

This means that \(V\) and \(V^{\bot}\) form a splitting of \(W\). Now the formula

\(\dim(W)\)

\(=\)

\(\dim(V) + \dim(V^{\bot})\)

follows from the dimension formula for arbitrary splittings of a vector space.