Let \(V\) be an inner product space and \(W\) a finite-dimensional subspace of \(V\). There are (in general) many projections which have \(W\) as their range. One of these, the orthogonal projection on \(W\), has the property that \(||E\alpha|| \leq ||\alpha||\) for every \(\alpha\) in \(V\). Prove that if \(E\) is a projection with range \(W\), such that \(||E\alpha|| \leq ||\alpha||\) for all \(\alpha\) in \(V\), then \(E\) is the orthogonal projection on \(W\).
Solution
1. Let \(E\) be a projection with range \(W\) and \(\|E\alpha\| \le \|\alpha\|\) for all \(\alpha\).
hypotheses.
2. It suffices to show \(\ker E = W^\perp\).
a projection with range \(W\) is the orthogonal projection iff its kernel is \(W^\perp\).
3. Let \(\alpha \in \ker E\) and \(w \in W\). For any \(t \in \mathbb{R}\),\[\|w\|^2 = \|E(w + t\alpha)\|^2 \le \|w + t\alpha\|^2 = \|w\|^2 + 2t\operatorname{Re}\langle w, \alpha \rangle + t^2\|\alpha\|^2.\]
\(E(w) = w\) (since \(w \in W = \operatorname{range} E\)) and \(E\alpha = 0\), so \(E(w + t\alpha) = w\); then apply the norm bound and expand.
4. Hence \(2t\operatorname{Re}\langle w, \alpha \rangle + t^2\|\alpha\|^2 \ge 0\) for all \(t \in \mathbb{R}\).
3, subtracting \(\|w\|^2\).
5. This forces \(\operatorname{Re}\langle w, \alpha \rangle = 0\) (otherwise choosing \(t\) of the appropriate small sign makes the left side negative).
4 (a linear term in \(t\) with nonzero coefficient dominates the quadratic term for small \(t\)).
6. Replacing \(\alpha\) by \(i\alpha\) (in the complex case) gives \(\operatorname{Im}\langle w, \alpha \rangle = 0\) as well, so \(\langle w, \alpha \rangle = 0\).
5 applied to \(i\alpha\).
7. Hence \(\alpha \in W^\perp\), so \(\ker E \subseteq W^\perp\).
6 (for all \(w \in W\)).
8. Conversely, if \(\alpha \in W^\perp\) then \(E\alpha \in W\) and \(\langle E\alpha, w \rangle = \langle \alpha, E^* w \rangle\); more directly, \(\|E\alpha\| \le \|\alpha\|\) and \(E\alpha \in W\) with \(\alpha - E\alpha \in \ker E\); since \(\ker E \subseteq W^\perp\) and \(\alpha \in W^\perp\), we get \(E\alpha = 0\).
7 and the fact that \(E\alpha \in W\) while \(\alpha \in W^\perp\) forces \(E\alpha = 0\) (as \(E\alpha = \alpha - (\alpha - E\alpha)\) with \(\alpha - E\alpha \in \ker E \subseteq W^\perp\)).
9. Hence \(\ker E = W^\perp\), so \(E\) is the orthogonal projection onto \(W\).
7, 8, and 2.
10. Q.E.D.
9.