# Matrix equation

Let $A$ be $k\times n$ matrix i.e., $A=(a_{1},\ldots, a_{n})$ where $a_{j} \in \mathbb{R}^{k}$, $rank(A)=k$ and $1\leq k \leq n$. Let $q=(q_{1},\ldots, q_{n})\in\mathbb{R}^{n}$ be such that $0<q_{j}\leq 1$.

For which $A$ and $q$ there exists symmetric matrix $C$ such that \begin{align*} A\left\{ \frac{q_{j}}{\langle Ca_{j},a_{j}\rangle}\right\}A^{T}C=I_{k\times k} \end{align*} so that $\langle Ca_{j},a_{j}\rangle >0$ (here $\langle\cdot, \cdot\rangle$ denotes scalar product in Euclidian space), and $\left\{ \frac{q_{j}}{\langle Ca_{j},a_{j}\rangle}\right\}$ denotes $n\times n$ diagonal matrix.

• I'm curious why this has been tagged with convex-optimization. Are you aware of a link for this problem to that field? – Michael Grant Nov 30 '14 at 0:11

A necessary condition. Note that $D=diag(\dfrac{q_i}{a_j^TCa_j})$ is symmetric $>0$.

$C^{-1}=ADA^T$ is symmetric $>0$; then $C$ is symmetric $>0$. One has $A\sqrt{D}\sqrt{D}A^T\sqrt{C}\sqrt{C}=I_k$, that implies $(\sqrt{C}A\sqrt{D})(\sqrt{D}A^T\sqrt{C})=I_k$, that is equivalent to say that the $k$ rows of $\sqrt{C}A\sqrt{D}$ form an orthonormal system.

Note that $A\sqrt{D}=[a_1\dfrac{\sqrt{q_1}}{||\sqrt{C}a_1||},\cdots]$ and $\sqrt{C}A\sqrt{D}=[\sqrt{q_1}\dfrac{\sqrt{C}a_1}{||\sqrt{C}a_1||},\cdots]$. In particular, if $k=1$, then there is a sole condition: $q_1+\cdots+q_n=1$.

1. yes, $tr(I_k)=||\sqrt{C}A\sqrt{D}||_F^2$ implies $k=q_1+\cdots q_n$.
2. Yet, if $n=k$, $D$ is not necessarily $I_n$. Indeed, the columns of $\sqrt{C}A\sqrt{D}$ form an orthonormal basis, that is equivalent to $q_i=1$ and for $i\not= j$, $<\sqrt{C}a_i,\sqrt{C}a_j>=0$, that is $a_j^TCa_i=0$. In particular, there are no conditions upon $a_i^TCa_i$.
• Right. And for another boundary case $k=n$ one can see (by equating traces) that $q_{j}=1$ and $C=(AA^{T})^{-1}$. – Paata Ivanishvili Aug 30 '15 at 20:05
• Again by the same reason one can get necessary condition $\sum q_{j}=k$. – Paata Ivanishvili Aug 30 '15 at 20:19