ThmDex – An index of mathematical definitions, results, and conjectures.
Anova partition of orthogonal projection
Formulation 0
Let $N$ be an D1247: Inner product-normed vector space such that
(i) $\Vert \cdot \Vert$ is the D504: Inner product norm in $N$
(ii) $P : N \to N$ is an D36: Orthogonal projection operator in $I$
(iii) $P^{\perp} : N \to N$ is the D2029: Coprojection operator of $P$
(iv) $x \in N$ is a D1129: Vector in $N$
Then \begin{equation} \Vert x \Vert^2 = \Vert P x \Vert^2 + \Vert P^{\perp} x \Vert^2 \end{equation}
Proofs
Proof 0
Let $N$ be an D1247: Inner product-normed vector space such that
(i) $\Vert \cdot \Vert$ is the D504: Inner product norm in $N$
(ii) $P : N \to N$ is an D36: Orthogonal projection operator in $I$
(iii) $P^{\perp} : N \to N$ is the D2029: Coprojection operator of $P$
(iv) $x \in N$ is a D1129: Vector in $N$
Result R1396: Image elements of orthogonal projection and coprojection are orthogonal shows that $P x, P^{\perp} x$ is an orthogonal collection in $N$. We can write \begin{equation} x = x + P x - P x = P x + x - P x = P x + (I - P) x = P x + P^{\perp} x \end{equation} Thus, applying R26: Pythagorean theorem the the collection $P x, P^{\perp} x$ yields \begin{equation} \Vert x \Vert^2 = \left\Vert P x + P^{\perp} x \right\Vert^2 = \left\Vert P x \right\Vert^2 + \left\Vert P^{\perp} x \right\Vert^2 \end{equation} $\square$