What non-symmetric matrices satisfy $x^TAx>0,\forall x\neq 0$

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.

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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. – Sangchul Lee Apr 13 '11 at 15:18
why not post it as an answer? – shamovic Apr 13 '11 at 15:23
Any $B+C$ with $B$ symmetric and positive definite, and $C$ anti-symmetric... – Did Apr 13 '11 at 15:49
@sos440: It's the job of the voters to decide which answer is the most enlightening! :) – Hans Lundmark Apr 13 '11 at 17:17
Something related... – J. M. Apr 14 '11 at 1:26

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.
As long as $A$ is diagonalizable and have positive eigenvalues then $A$ can be written as $P^T D P$ where $P$ is an orthogonal matrix and $D$ is a diagonal matrix with positive entries along the diagonal. Then clearly given any $x \in R$, $x^T A x = x^T P^T D P x=y^TDy>0$ where $y=Px$. As long as $x\neq0$, $y$ will not be the zero vector.
If $A$ can be written as $P^TDP$, of course $x^TAx>0, \forall x\in R^n$. You may not get my question well. I think sos440 has given a right answer. Thanks anyway. – Shiyu Apr 23 '11 at 5:57