Lower bound on quadratic form Suppose I have a non-symmetric matrix $A$ and I can prove that $x^T A x = x^T \left(\frac{A+A^T}{2}\right) x>0$ for any $x \ne 0$? Can I then say that $x^T A x \ge \lambda_{\text{min}}(A) \|x\|^2 > 0$?

Hint: If $A$ is positive definite you can diagonalize the Matrix
$$A = V^{-1}\Lambda V$$, where $V$ is the matrix of normalized eigenvectors and $\Lambda = \operatorname{diag} (\lambda_1,\cdots,\lambda_n)$ is a diagonal matrix of the eigenvalues.
