# State-dependent positive definite matrix

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 prove this statement without showing that $$\mathbf{M}_\mathbf{x}$$ is always positive definite?

Yes, it is possible; pick a nonzero element of $$\mathbf{x}$$, which must exist since $$\mathbf{x}\neq 0$$; let's say this nonzero element is $$x_j$$. Then define $$\big(\mathbf{M}_{\mathbf{x}}\bigr)_{nm}=-\delta_{nm}(1-\delta_{nj})+\delta_{nm}\delta_{nj}\left(x_{j}^{-2}+x_{j}^{-2}\sum_{k\neq j}x_k^2\right).$$ Now you have the desired result, $$\sum_{n,m}x_nx_m\big(\mathbf{M}_{\mathbf{x}}\bigr)_{nm}=1>0,$$ while $$\mathbf{M}_{\mathbf{x}}$$ is itself not positive definite.
This matrix is symmetric, if you want it to be nonsymmetric, simply add to $$\mathbf{M}_{\mathbf{x}}$$ an arbitrary skew-symmetric matrix $$\mathbf{A}$$ and use that $$\mathbf{x}^T (\mathbf{M}_{\mathbf{x}}+\mathbf{A})\mathbf{x} =\mathbf{x}^T \mathbf{M}_{\mathbf{x}}\mathbf{x}$$.
I don't understand your second question "And how can I prove this statement without showing that $$\mathbf{M}_{\mathbf{x}}$$ is always positive definite?" To prove this you need to examine the function $$f(\mathbf{x})$$, without further knowledge of this function no progress can be made.