Let $M$ be an SPD matrix and let $\Pi=QQ^T$ be the orthogonal projection onto the range of $Q$ (a "tall" matrix with orthonormal columns). I have an expression in the form
$$\tag{1}
K=\max\_{v}\frac{v^T(I-\Pi)v}{v^T(I-\Pi)M^{-1}(I-\Pi)v},
$$
which I would like to express as a maximum of something which has $M$ in the numerator and such that neither the denominator nor the (sub)space over which the maximisation is made depend on $M$.


Actually I know that the solution should be
$$\tag{2}
K=\max\_v\frac{v^T(I-\tilde{\Pi})^TM(I-\tilde{\Pi})v}{v^T(I-\Pi)v},
$$
where $\tilde{\Pi}=Q(Q^TMQ)^{-1}Q^TM$ (this pie is the $M$-orthogonal projection onto the range of $Q$). I would appreciate any hint on how to get from (1) to (2) (or vice versa).




---


I've tried to get from (2) to (1). First, 
$$(I-\tilde{\Pi})^TM(I-\tilde{\Pi})=M^{1/2}(I-\overline{\Pi})M^{1/2},$$ 
where now 
$$\overline{\Pi}=M^{1/2}Q(Q^TMQ)^{-1}Q^TM^{1/2}$$
is the orthogonal projection onto the range of $M^{1/2}Q$. So putting this to (2) and taking $w=M^{1/2}v$ gives
$$
K=\max\_w\frac{w^T(I-\overline{\Pi})w}{w^TM^{-1/2}(I-\Pi)M^{-1/2}w},
$$
which has already a sort of $M^{-1}$ in the denominator, but still not in the form of (1) and with a different projector in the numerator.




---


P.S.: I've asked a [related question](https://mathoverflow.net/questions/167362/maximising-a-rayleigh-quotient-over-a-subspace) already, but the simply stated solution there does not much help me to get to (2) as the subspace over which the maximum is evaluated depends on $M$. What I'm actually looking for is a characterisation of (1) which does not change the subspace over which we maximise. In fact, (1) and (2) give that
$$
K=\max\_{w\in\mathcal{R}(I-\Pi)}\frac{w^Tw}{w^TM^{-1}w}=\max\_{w\in\mathcal{R}(I-\Pi)}\frac{w^T(I-\tilde{\Pi})^TM(I-\tilde{\Pi})w}{w^Tw}.
$$



Maximising a Rayleigh quotient over a subspace II
After some playing with this problem, I think I've found a solution.


Let $U$ be an orthonormal basis of the range of $I-\Pi$, that is, $[Q,U]$ is a square orthogonal matrix such that $U^TQ=0$ and $I-\Pi=UU^T$. Then
$$
\begin{split}
K&=\max\_v\frac{v^T(I-\Pi)v}{v^T(I-\Pi)M(I-\Pi)v}=\max\_v\frac{v^TUU^Tv}{v^TUU^TM^{-1}UU^Tv}
=\max\_v\frac{v^T(U^TM^{-1}U)^{-1}v}{v^Tv}\\
&=\max\_v\frac{v^T(U^TM^{-1}U)^{-1}U^TM^{-1}U(U^TM^{-1}U)^{-1}v}{v^Tv}\\
&=\max\_v\frac{v^TU^TU(U^TM^{-1}U)^{-1}U^TM^{-1}MM^{-1}U(U^TM^{-1}U)^{-1}U^TUv}{v^TU^TUv}\\
&=\max\_{v\in\mathcal{R}(I-\Pi)}\frac{v^T[U(U^TM^{-1}U)^{-1}U^TM^{-1}]M[M^{-1}U(U^TM^{-1}U)^{-1}U^T]v}{v^Tv}.
\end{split}
$$
What remains to show is that $$\tag{1}I-\tilde{\Pi}=I-Q(Q^TMQ)^{-1}Q^T\quad\text{and}\quad\Phi=M^{-1}U(U^TM^{-1}U)^{-1}U^T$$ are equal.
It is easy to show that the square matrix $[U,MQ]$ is nonsingular. So $I-\Pi=\Phi$ if and only if $[U,MQ]^T(I-\Pi-\Phi)=0$. But this is easy to verify using $U^TQ=0$ and the expressions in (1).


