$\|A v\|_{Q^{-1}}$ under the measurement constraints $\Phi v=y$. Furthermore, it also holds true that (1) $\Phi \Psi=I_m$ (2) $\Img(\Psi)$ is the orthogonal complement of $\Ker(\Phi)$ with respect to the product $\<\cdot,\cdot\>_{K^{-1}}$ and (3) $\Psi\Phi=P_{\Img(K \Phi^T), K^{-1}}$ and $I_n-\Psi\Phi=P_{\Ker(\Phi), K^{-1}}$.

First observe that implies that $\Phi \Psi=I_m$ where $I_m$ is the identity $m\times m$ matrix. Therefore $\Phi (\Psi w)=w$. Note that implies that for all $z\in \R^m$, $\<\Psi z,v\>_{K^{-1}} =z^T \big(\Phi K \Phi^T\big)^{-1} \Phi v$. Therefore if $v\in \Ker(\Phi)$ then $\<\Psi z,v\>_{K^{-1}}=0$ for all $z\in \R^m$. Conversely if $\<\Psi z,v\>_{K^{-1}}=0$ for all $z\in \R^m$ then $v$ must belong to $\Ker(\Phi)$. Since the dimension of $\Img(\Psi)$ is $m$ and that of $\Ker(\Phi)$ is $n-m$ we conclude that $\Img(\Psi)$ is the orthogonal complement $\Ker(\Phi)$ with respect to the product $\<\cdot,\cdot\>_{K^{-1}}$ and in particular, $$\label{eqn:orthogonalitybis}
 \<\Psi  w,v\>_{K^{-1}} =0, \quad \forall w\in \R^m \text{ and } \forall v\in \R^n \text{ such that }\Phi v=0\,.$$ Let $w\in \R^m$ and $v\in \R^n$ such that $\Phi v=w$. Since $\Psi w -v \in \Ker(\Phi)$, it follows from that $ \<v,v\>_{K^{-1}} = \<\Psi w,\Psi w\>_{K^{-1}} +\<v-\Psi w,v-\Psi w\>_{K^{-1}}$. Therefore $\Psi w$ is the unique minimizer of $\<v,v\>_{K^{-1}}$ over all $v\in \R^n$ such that $\Phi v=w$. Now consider $f\in \R^n$, since 