Certainly
\(w\) as defined is an element of
\(W\text{,}\) and to see that
\(w^\perp = v-w\) is orthogonal to
\(W\text{,}\) it is sufficient by
Checkpoint 3.2.8 to verify that
\(\la w^\perp, w_i\ra = 0\) for each
\(i=1,
\dots, r.\)
Using the definition of \(w^\perp\) and bilinearity of the inner product we have
\begin{equation*}
\la w^\perp, w_i\ra = \la v-w,w_i\ra = \la v,w_i\ra - \la w,w_i\ra,
\end{equation*}
and since the
\(\{w_j\}\) form an orthogonal basis, the expression for
\(w\) in
(3.2.1) gives
\begin{equation*}
\la w,w_i\ra = \la\frac{\la
v,w_i\ra}{\la w_i,w_i\ra} w_i, w_i\ra = \la\frac{\la
v,w_i\ra}{\la w_i,w_i\ra} \la w_i,w_i\ra = \la
v,w_i\ra.
\end{equation*}
It is now immediate from the first displayed equation that
\begin{equation*}
\la w^\perp, w_i\ra = \la v-w,w_i\ra = \la v,w_i\ra - \la
w,w_i\ra = \la v,w_i\ra - \la v,w_i\ra =0,
\end{equation*}
as desired.
Finally to see that \(w^\perp\) and \(w\) are uniquely determined by these conditions, suppose that as above \(v=
w^\perp + w\text{,}\) and also \(v = w_1^\perp +
w_1\) with \(w_1\in W\) and \(w_1^\perp \in W^\perp\text{.}\)
Setting the two expressions equal to each other and solving gives that
\begin{equation*}
w-w_1 = w_1^\perp - w^\perp.
\end{equation*}
But the left hand side is an element of
\(W\) while the right hand side is an element of
\(W^\perp,\) so by
Proposition 3.2.9, both expressions equal zero, which gives the uniqueness.
