Infer the positive definiteness of a real non-symmetric matrix from the positive definiteness of its symmetric part

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Let $A$ be a non-symmetric matrix of real numbers, $A\in\mathbb{R}^{n\times n}$.

If the quadratic form $x^\intercal Ax$ is positive definite in the sense that $x^\intercal Ax>0$ $\forall x\neq 0\in\mathbb{R}^n$, then the real part of all eigenvalues of $A$ is positive, $\Re(\lambda_i)>0$ for $i=1,\dots,n$.

However, if $\Re(\lambda_i)>0$ for $i=1,\dots,n$, it is not necessarily true that $x^\intercal Ax>0$ $\forall x\neq 0\in\mathbb{R}^n$.

Now, consider the symmetric part of $A$, $A_s=\frac{1}{2}(A+A^\intercal)$. We know by definition that $x^\intercal Ax=x^\intercal A_sx$.

What can we say about $x^\intercal Ax$ if:

  • all eigenvalues of $A_s$ are positive
  • at least one eigenvalue of $A_s$ is negative

Can we say that the $x^\intercal Ax$ is positive definite in the first case, and that it is certainly not positive definite in the second case?