Is every linear projection orthogonal with respect to some inner product?

Question

For example, for any given idempotent matrix $P^2=P\in \mathbb{R}^{n\times n}$, is there always a symmetric, positive-definite matrix $A$ such that \begin{aligned} \langle Px,y\rangle_A=x'P'Ay=x'APy=\langle x,Py\rangle_A \end{aligned} for all $x,y\in \mathbb{R}^n$? Does it generalize to arbitrary inner product spaces?

Answer 1

Yes. Let $V = \ker P$ and $W = P(\mathbb{R}^n)$. Let $\{v_1,\dots,v_k\}$ be a basis of $V$ and $\{w_1,\dots,w_m\}$ be a basis of $W$. Then $\{v_1,\dots,v_k,w_1,\dots,w_m\}$ is a basis of $\mathbb{R}^n$. Define the inner product $\langle\langle.,.\rangle\rangle$ satisfying that \begin{eqnarray*} \langle\langle v_i,v_j\rangle\rangle &=& \delta_{ij}\\ \langle\langle w_i,w_j\rangle\rangle &=& \delta_{ij}\\ \langle\langle v_i,w_j\rangle\rangle &=& 0. \end{eqnarray*} We know there is a matrix $A$ such that $\langle\langle.,.\rangle\rangle = \langle.,.\rangle_{A}$. Because $\langle\langle.,.\rangle\rangle$ is an inner product we must have that $A$ is symmetric and positive-definite. Moreover it follows that $$ \langle Px,y\rangle_A = \langle x,Py\rangle_A $$ as can be easily checked.