I am working on an optimization problem:
\begin{equation} \min_{v_1,\dots,v_n} \sum_{i=1}^{n} v_i^T Q v_i, \end{equation} where the dynamics is $v_{i+1} = A_iv_i$ and $Q \succeq 0$. $A_i$ is a function of $v_i$, i.e, $A_i = f(v_i)$. And based on $A_i$, the optimization would not be quadratic.
Assume $\exists ~ A$ such that $||A||_2 \succeq ||A_i||_2 ~\forall i$ (specifically $I \succeq A^2 \succeq A_i^2 \succeq 0$, and $A_i$ and $A$ are symmetric). I am trying to see if I can bound the cost $v_i^TQv_i$ using $A$ and solve a sub-optimal convex problem. I am wondering if \begin{equation}\label{eq:cntrl} AQA \geq A_iQA_i ~\forall ~ i. \end{equation} Since $A^2 \succeq A_i^2$, for some vector $u$, I define $x = Au, y = A_iu$, so that $x^Tx \geq y^Ty$.
If the vector dot product $x^Tx \geq y^Ty$ and $Q$ is a symmetric PSD matrix ($ Q \succeq 0$), is $x^TQx \geq y^TQy$? If no, what other conditions must $Q$ satisfy for $x^TQx \geq y^TQy$ to hold? How about if $x^Tx > y^Ty$?
Alternatively, is there a path to directly show $AQA \geq A_iQA_i$? If not, what about when $A^2\succ A_i^2$?
Thank you for your help.
This is true only if $Q$ is a constant multiple of the identity. If not, it has two distinct eigenvalues $\lambda_1 < \lambda_2$. Let $x,y$ be unit eigenvectors corresponding to these respective eigenvalues and you see that $x^T x = y^T y = 1$ while $x^T Q x = \lambda_1 < \lambda_2 = y^T Q y$.