5
$\begingroup$

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.

  • 0
    @J.$M$.: Thanks. That seems a good reference for this problem.2011-04-14

2 Answers 2

3

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.

1

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.

  • 0
    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