$A \in \mathbb{R}^{d \times d}$ is a real non-symmetrix square matrix and we know there exists a real vector $x \in \mathbb{R}^d$ such that $x^T A x > 0$. Can we conclude that $A$ has at least one eigenvalue with a positive real part? If not, what is a sufficient condition to show $A$ has at least one eigenvalue with a positive real part?
An easy $d=2$ case is $A = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}$ and $x = [1,0]^T$, where the two eigenvalues of $A$ are 2 and 1.
Best Answer
If one off diagonal entry of the matrix is positive and dominating we can achieve $x^TAx>0$ for $x=(1,1,\ldots,1)^T$ no matter what eigenvalues are.
Let $$A = \begin{bmatrix} -2 & 10 \\ 0 &- 1 \end{bmatrix}$$ For $x=(1,1)^T$ we have $$x^TAx=-2+10-1=7 $$ The eigenvalues are negative and the matrix is non symmetric.
If we replace the assumption by $x^TAx>0$ for every nonzero vector $x$ then the conclusion holds. Indeed for every vector $e_k$ from the standard basis we have $$a_{kk}=e_k^TAe_k>0$$ Hence ${\rm Tr}A>0.$ on the other hand the trace is equal the sum of all the eigenvalues. Therefore at least one eigenvalue has positive real part.