Is the ellipsoid $x'Qx < \alpha$ equivalent to $\alpha Q^{-1} - x x' \succ 0$? Suppose we have an ellipsoid defined by $x'Qx < \alpha$, where $x$ is a column vector. Can we also write $\alpha Q^{-1}- xx' \prec 0$?
I say this because if $Q$ is symmetric and positive definite (thus, invertible) and $\alpha > 0$ is a scalar, then this can be written as
$$\left[ {\begin{array}{*{20}{c}}
\alpha &{x'}\\
x&{{Q^{ - 1}}}
\end{array}} \right] \succ 0$$
and the Schur complement can be used to write
$$Q^{-1} \succ 0, \qquad \alpha - x' Q x > 0$$
which is equivalent to
$$\alpha > 0, \qquad Q^{-1} - x \alpha^{-1} x' \succ 0$$
which directly leads to the proposition since $\alpha$ is a scalar. Please correct me if I am wrong.
 A: I assume that $Q$ is nonsingular and Hermitian. When $Q\succ0$, the two inequalities are equivalent. Note that
\begin{align}
x^\ast Qx<\alpha
&\Leftrightarrow \alpha-x^\ast Qx>0 \text{ and } Q^{-1}\succ0\\
&\Leftrightarrow \pmatrix{\alpha&x^\ast\\ x&Q^{-1}}\succ0\\
&\Leftrightarrow \alpha>0 \text{ and } Q^{-1}-\alpha^{-1} xx^\ast\succ0.\tag{1}\\
\end{align}
Obviously, $(1)$ implies that $\alpha Q^{-1}-xx^\ast\succ0$. Conversely, suppose $\alpha Q^{-1}-xx^\ast\succ0$. Then $\alpha Q^{-1}\succ xx^\ast\succeq0$. Since $Q$ is positive definite, $\alpha$ must be positive and hence $(1)$ holds. Thus $\alpha Q^{-1}-xx^\ast\succ0$ is equivalent to $(1)$, which in turn is equivalent to $x^\ast Qx<\alpha$.
When $Q\not\succ0$, neither inequality implies the other. E.g. consider $Q=-I_2$.

*

*When $\alpha=1$, we have $x^\ast Qx<\alpha$ for every vector $x$, but $\alpha Q^{-1}-xx^\ast=-(I_2+xx^\ast)$ is never positive definite.

*When $\alpha=-2$ and $x=(1,0)^T$, we have $\alpha Q^{-1}-xx^\ast=\operatorname{diag}(1,2)\succ0$ but $x^\ast Qx=-1>-2=\alpha$.

