I have a quadratic form that looks like,
$$ (x-v)^TA(x-v) < 1 $$
$x,v$ are column vectors, $A$ is diagonal, and $v$ is constant.
I would like to determine what happens if $x$ is replaced with $Px$, how does it relate to the inequality above? $P$ is an arbitrary matrix. Let's say I use $A = \tilde{A}^T\tilde{A}$, since $A$ is diagonal, and
$$ (x-v)^T\tilde{A}^T\tilde{A} (x-v) < 1 $$
Replacing $x$ with $Px$,
$$ [\tilde{A} (Px- Pv)]^T [\tilde{A} (Px-Pv)] $$
$$ = [\tilde{A}Px- \tilde{A}Pv)]^T [\tilde{A}Px-\tilde{A}Pv)] $$
Not sure where to go from here. Or,
$$ [\tilde{A} P (x-v) ]^T [ \tilde{A} P (x-v) ] $$
$$ = (x-v)^T P^T \tilde{A}^T\tilde{A} P (x-v) $$
$$ = (x-v)^T P^T A P (x-v) $$
What to do after this..? I thought of using diagonalization to replace $P^TAP$ with $S\Lambda S^{-1}$, $S$ being a matrix of eigenvectors, $\Lambda$ being a diagonal matrix of eigenvalues. This would be doable since $P^TAP$ is symmetric, positive definite. Then I have
$$ (x-v)^T S \Lambda S^{-1} (x-v) $$
$S$ is orthogonal, then maybe I could argue it does not qualitatively change $x-v$, then the only difference between the two formulas is $A$ and $\Lambda$ which are both diagonal. But how quantify this into a relation to $<1$ inequality?
Thanks,
For the sake of simplifying notations, Let $w=x-v$ and $x\ne v$. Defining $X=\frac{1}{\Vert w\Vert ^2}I$ yields: $$w^T A w<w^T X w$$ If $x$ is arbitrary, it means that the above inequality holds for all $w$. Therefore $A-X$ is negative definite. $$A-X<0\implies P^T(A-X)P<0\implies P^T A P<P^T X P$$ The rest seems straightforward: $$w^T P^T A P w<w^T P^T X P w=\frac{1}{\Vert w\Vert ^2} w^T P^T P w= (P u)^T P u$$ where $u=\frac{w}{\Vert w\Vert}$ is the unit vector along the direction of $x-v$.
Remember how we define the euclidean norm of a matrix: $$\Vert P\Vert=\sup_{\Vert u\Vert=1} \Vert P u\Vert$$ Then it is safe to say that: $$(x-v)^T A(x-v)<1\implies (x-v)^T P^T A P(x-v)<{\Vert P\Vert}^2$$