# Existence of a positive definite matrix which satisfies an equation

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.

• Do some reading on Eigen values and Eigen vectors. May 9, 2018 at 13:27
• Thanks for the reply. I've read some but most of the properties can be inferred from an already existing matrix A. Now it seems like I have to construct that matrix A in order to prove its existence. Would you mind showing me which specific part of eigenvalues and eigenvectors should I read? Many thanks! May 9, 2018 at 13:35
• It reads as if $A$ only needs to be able to satisfy the equation for one $u$. If that is the case: If $u=0$ then $b(u)$ should be $=0$. Then any positive-definite (PD) matrix $A$ works. If $b(u)=0$, then $u$ must be $=0$, and then any PD matrix works. If $u,b(u)\neq0$, then you need that the dot product $u^T\cdot b(u)>0$. In that case, there is always a PD matrix $A$ such that $Au=b(u)$. To find one, complete $u$ to an orthonormal basis $u,e_2,e_3$, and also complete $b(u)$ to an orthonormal basis $b(u),f_2,f_3$. Then $A=[b(u),f_2,f_3]\cdot [u,e_2,e_3]^T$ works, where the brackets are ...
– user553213
May 9, 2018 at 13:45
• ... the matrices formed by putting those vectors in columns.
– user553213
May 9, 2018 at 13:46
• Do you need this to hold for all vectors $u$ (in which case $b(u)$ must be a linear function of $u$) or just for one particular vector $u$?
– A.Γ.
May 9, 2018 at 13:56

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.
• As you said indeed, for my problem, (3) can be satisfied for all $u$ in some ball $||u||<u_{max}$. I will try to work on the remaining conditions. Thank you so much. May 9, 2018 at 20:17