TITLE: What non-symmetric matrices satisfy $x^TAx>0,\forall x\neq 0$
QUESTION [6 upvotes]: Let $A\in R^{n\times n}$ be a matrix. It is positive definite if and only if $A$ is symmetric and $x^TAx>0,\forall x\in R^n$.
My question is: if $x^TAx>0,\forall x\in R^n$ but $A$ is not symmetric, what does $A$ look like?
I have an example. For a rotation matrix $A$ whose rotation angle is less than 90 degrees, $x^TAx>0,\forall x\in R^n$ but $A$ is not symmetric. Is this the only type of non-symmetric matrices that satisfy $x^TAx>0,\forall x\in R^n$? Can you give any other examples of this kind of matrices? Many thanks.

REPLY [3 votes]: For the sake of having an answer, here is sos440's answer from the comments.
There is a unique way of decomposing $A$ into the sum of a symmetric matrix $A_{+}$ and an antisymmetric matrix $A_{-}$, namely $A = (A + A^{T})/2 + (A - A^{T})/2$. Then note that $x^{T} A x = x^{T} A_{+} x$. That is, $x^{T} A x$ does not depend on the antisymmetric part $A_{-}$, so the matrix $A$ satisfying $x^{T} A x > 0$ for all $x \neq 0$ is characterized as the sum of a positive definite matrix and an antisymmetric matrix.