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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.2011-04-13
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    why not post it as an answer?2011-04-13
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    Well, it's because I think someone may find a more enlightening answer...2011-04-13
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    @sos440: Thanks for the good answer. Then $x^TAx>0, \forall x\neq 0$ if and only if $A+A^T$ is positive definite. Then a further question appears: what kind of non-symmetric $A$ has $A+A^T$ as a positive definite matrix?2011-04-13
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    Any $B+C$ with $B$ symmetric and positive definite, and $C$ anti-symmetric...2011-04-13
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    @sos440: It's the job of the voters to decide which answer is the most enlightening! :)2011-04-13
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    @Didier: I see. Thanks. @sos440: by the way, please post an answer so I can accept it. Thanks2011-04-14
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    [Something related...](http://dx.doi.org/10.1016/0024-3795(79)90122-8)2011-04-14
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    @J.M.: Thanks. That seems a good reference for this problem.2011-04-14

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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.

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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.

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    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.2011-04-23