Find $\min_{\|u\|=1}\frac{ u^T x}{\sqrt{ u^T Au}}$ where $A$ symmetric positive define matrix Given a symmetric positive define matrix $A \in \mathbb{R}$ how do we minimize the following expression:
\begin{align}
\min_{\|u\|=1}\frac{ u^T x}{\sqrt{ u^T Au}}
\end{align}
For the scalar case this is easy as we get
\begin{align}
\min_{ u \in \{+1, -1\} }\frac{ u x}{\sqrt{ A}}= -\frac{  |x|}{\sqrt{ A}}
\end{align}
where the minimum is attained by setting $u=-{\rm sign}(x)$.
Also, if $A$ is identity, then the optimal $u$ is given by  $-\frac{x}{\|x\|}$.
 A: Firstly, notice you can drop the condition $\|u\|=1$. To show that, consider $v=\frac{u}{\|u\|}$. Then
$$
\frac{v^Tx}{\sqrt{v^TDv}} = \frac{u^Ty}{\sqrt{u^TDu}}
$$
So the maximum over both sets is the same.
As $A$ is simetric, it can be diagonalized orthogonally. Let $A = O^T D O$. Your problems then becomes to minimize $\frac{v^Ty}{\sqrt{v^TDv}}$, where $v=Ou$ and $y = Ox$. That is, you can assume your matrix to be diagonal and with positive eigenvalues.
Then, consider $D = \sqrt{D}^2 = \sqrt{D}^T \sqrt{D}$, where $P = \sqrt{D}$. Then, $P$ is bijetive, and the problem becomes minimizing $\frac{\langle w, z \rangle}{\|w\|}$, where $w = Pv = POu$ and $z = P^{-1}y = P^{-1}Ox$. Now apply Cauchy-Schwarz, to get
$$
\frac{\langle w, z \rangle}{\|w\|} \leq \|z\| = \|\sqrt{D}^{-1}Ox\| 
$$
But
$$
\sqrt{D}^{-1} = \sqrt{D^{-1}} = \sqrt{OA^{-1}O^T} = O\sqrt{A^{-1}}O^T
$$
Then
$$
\|\sqrt{D}^{-1}Ox\|  = \| O\sqrt{A^{-1}} O^TOx\| = \| O\sqrt{A^{-1}}x\| = \|\sqrt{A^{-1}}x\| = \sqrt{x^T A^{-1} x}
$$
Equality hold if $w$ is proportional to $z \Rightarrow POu = \lambda P^{-1}Ox$. Then
$u = \lambda O^t D^{-1} O x = \lambda A^{-1}x$. So minimum occur at $\frac{A^{-1}x}{\|A^{-1}x\|}$
