Existence of a positive definite matrix which satisfies an equation

Question

While doing control analysis, I've bumped into a mathematical problem that looks like below:

$Au = b(u)$

Under what conditions of $b,u$ (both are a $3 \times 1$ vector) that there exists a positive-definite matrix $A$ (a $3 \times 3$ matrix) which satisfies the above equation? Vector $b$ is a function of $u$, which in turns, $u$ may be constrained by the condition $||u||<u_{max}$. I'm really not strong in mathematics, and I don't know where to get started. I thought I will write every component of $A$ and solve for a bunch of inequalities (maybe using Sylvester criterion) but I wonder whether it could be done more efficiently.

Your help is greatly appreciated.

Answer 1

You have the conditions that

1. $b(u)=Au$ for some matrix $A$ in some ball around the origin.
2. $A=A^T$.
3. $u^TAu>0$ for all $u\ne 0$.

2 and 3 means that $A$ is positive definite. Now those conditions are equivalent to the following

1. The first condition is equivalent to $b(u)$ being a (locally) linear map, i.e. $b(u+\alpha v)=b(u)+\alpha b(v)$, for all $u$, $v$ and $w=u+\alpha v$ that satisfy the norm bound. The matrix of the linear map $A$ is defined uniquely (in the given basis).
2. The symmetry is equivalent to $v^TAu=u^TAv$, i.e. $v^Tb(u)=u^Tb(v)$, for all $u$, $v$ in the ball. It is sufficient to test only the basic vectors $e_k$ (vector with all zeros except one identity at the position $k$), that is, $v^Tb(u)=u^Tb(v)$ for $u=\alpha e_i$, $v=\alpha e_j$ for all $i\ne j$, where $\alpha\ne 0$ is a scalar chosen such that $u$, $v$ satisfy the norm bound.
3. $u^Tb(u)>0$ for all $u$ in the ball.