Let \(W\) be a finite-dimensional subspace of an inner product space \(V\), and let \(E\) be the orthogonal projection of \(V\) on \(W\). Prove that \((E\alpha \mid \beta) = (\alpha \mid E\beta)\) for all \(\alpha, \beta\) in \(V\).
Solution
For arbitrary \(\alpha,\beta\in V\), write their orthogonal decompositions as \[ \alpha=w+u, \qquad \beta=w'+u', \] where \[ w,w'\in W, \qquad u,u'\in W^\perp. \] By definition of the orthogonal projection, \[ E\alpha=w, \qquad E\beta=w'. \] Therefore \[ \begin{aligned} (E\alpha\mid\beta) &=(w\mid w'+u')\\ &=(w\mid w') \end{aligned} \] because \(w\perp u'\). Similarly, \[ \begin{aligned} (\alpha\mid E\beta) &=(w+u\mid w')\\ &=(w\mid w') \end{aligned} \] because \(u\perp w'\). Hence \[ \boxed{(E\alpha\mid\beta)=(\alpha\mid E\beta)} \] for all \(\alpha,\beta\), so the orthogonal projection is self-adjoint [@HK71].