$\Img(\Psi)=\Img(K\Phi^T)$ and $\Img(\Psi)$ is the orthogonal complement of $\Ker(\Phi)$ with respect to the product $\<\cdot,\cdot\>_{K^{-1}}$, there exists a unique $z\in \R^m$ and a unique $g\in \Ker(\Phi)$ such that $f=K \Phi^T  z +g$. Since $\Psi \Phi =K \Phi^T (\Phi K \Phi^T)^{-1} \Phi$, it follows that $\Psi \Phi f= K \Phi^T  z \text{ and }(I_n-\Psi \Phi)f=g$. We conclude by observing that $g=P_{\Ker(\Phi), K^{-1}} f$.

\[thm:gedgdgjdh\] For $v\in \R^n$, $w^*=\Phi v$ is the unique minimizer of $\|v- \Psi w\|_{K^{-1}}$ over all $w\in \R^m$. In particular, $\|v-\Psi \Phi v\|_{K^{-1}}=\min_{z\in \R^m} \|v-K\Phi^T z\|_{K^{-1}}$ and if $x$ is the solution of the original equation , then $\|x-\Psi y\|_{K^{-1}} =\min_{w\in \R^m} \|x-\Psi w\|_{K^{-1}} =\min_{z\in \R^m} \|x-K\Phi^T z\|_{K^{-1}}$.

The proof follows by observing that $v-\Psi \Phi v$ belongs to the null space of $\Phi$ which, from Theorem \[lem:minimizingproperty\], is the orthogonal complement of the image of $\Psi$ with respect to the scalar product defining the norm $\|\cdot\|_{K^{-1}}$. Observe also that the image of $\Psi$ is equal to that of $K\Phi^T$.

\[rmk:accurkm1\] Observe that, from Theorem \[lem:minimizingproperty\], $v-\Psi \Phi v$ spans the null space of $\Phi$, and $\|v\|_{K^{-1}}^2= \big\|v-\Psi \Phi v\big\|_{K^{-1}}^2+\big\|\Psi \Phi v\big\|_{K^{-1}}^2$. Therefore if $D$ is a symmetric positive definite $n\times n$ matrix then $\sup_{v \in \R^n} \big\|v-\Psi 