Return to Question

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 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 $$ K=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^Tw}{w^TM^{-1}w}=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^T(I-\tilde{\Pi})^TM(I-\tilde{\Pi})w}{w^Tw}. $$

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 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 $$ K=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^Tw}{w^TM^{-1}w}=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^T(I-\tilde{\Pi})^TM(I-\tilde{\Pi})w}{w^Tw}. $$

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 $\tilde{\Pi}=Q(Q^TM^{-1}Q)^TQ^TM$ (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 $$\overline{\Pi}=M^{1/2}Q(Q^TM^{-1}Q)^TQ^TM^{1/2}$$ 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 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 $$ K=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^Tw}{w^TM^{-1}w}=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^T(I-\tilde{\Pi})^TM(I-\tilde{\Pi}))w}{w^Tw}. $$

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 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 $$ K=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^Tw}{w^TM^{-1}w}=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^T(I-\tilde{\Pi})^TM(I-\tilde{\Pi})w}{w^Tw}. $$

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 the denominator does not 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 $\tilde{\Pi}=Q(Q^TM^{-1}Q)^TQ^TM$ (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 $$\overline{\Pi}=M^{1/2}Q(Q^TM^{-1}Q)^TQ^TM^{1/2}$$ 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 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 $$ K=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^Tw}{w^TM^{-1}w}=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^T(I-\tilde{\Pi})^TM(I-\tilde{\Pi}))w}{w^Tw}. $$

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 $\tilde{\Pi}=Q(Q^TM^{-1}Q)^TQ^TM$ (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 $$\overline{\Pi}=M^{1/2}Q(Q^TM^{-1}Q)^TQ^TM^{1/2}$$ 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 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 $$ K=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^Tw}{w^TM^{-1}w}=\max_{w\in\mathcal{R}(I-\Pi)}\frac{w^T(I-\tilde{\Pi})^TM(I-\tilde{\Pi}))w}{w^Tw}. $$