Consider a function $f(\mathbf{x})=\mathbf{M}_\mathbf{x}$ that outputs a nonsymmetric matrix $\mathbf{M}_\mathbf{x} \in \mathbb{R}^{N \times N}$ given an input vector $\mathbf{x} \in \mathbb{R}^N$.

Is the following condition possible without requiring a positive definite $\mathbf{M}_\mathbf{x}$? I think so... $$ \mathbf{x}^T f(\mathbf{x}) \, \mathbf{x} = \mathbf{x}^T \, \mathbf{M}_\mathbf{x} \, \mathbf{x} > 0 \hspace{0.4cm} \forall \mathbf{x}, \mathbf{x} \neq \mathbf{0} $$ And how can I proof this statement without showing that $\mathbf{M}_\mathbf{x}$ is always positive definite?