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Suppose we have a non symmetric Matrix $A$. Let also we know that $A+A^T$ is positive definite. Is it the case that A is also positive definite? If A symmetric then its easy to show that $A$ is positive definite because $A=A^T$. So if it is not the case then can we find a 2x2 matrix, $B$ which is non symmetric and $B+B^T$ is positive definite but B is not positive definite?

I have another question. Suppose $A$ is a non symmetric non positive definite matrix. is it possible its inverse is positive definite?

Thanks.

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  • $\begingroup$ positive definite $\implies P D P^T$ where $P$ is orthogonal and $D$ is diagonal $> 0$ $\endgroup$ – reuns Jan 30 '16 at 21:59
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    $\begingroup$ What is your definition of "positive definite" for a matrix that is not symmetric???????????? $\endgroup$ – Will Jagy Jan 30 '16 at 22:10
  • $\begingroup$ Suppose the matrix [1 1; 0 1]. This matrix is not symmetric but positive definite as determinant is positive so are the eigenvalues. My question is suppose we know $A+A^T$ is positive definite. Is it mandatory for A to be positive definite? $\endgroup$ – user2104150 Jan 31 '16 at 0:27
  • $\begingroup$ @user2104150 note that your definition is definitely not the usual definition. For example, few people would consider the matrix [1 4; 0 1] to be positive definite. $\endgroup$ – Omnomnomnom Jan 31 '16 at 3:33
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The answer is no, at least with your definition of positive-definite; we can show this directly in the $2 \times 2$ case:

Let the symmetric part of the matrix be $$ S = \begin{pmatrix} 2a & b \\ b & 2d \end{pmatrix}, $$ and the antisymmetric part $A$ have top-right element $-c$. It is easy to compute the eigenvalues of $S$ and $S+A$. Since $S$ is symmetric, we expect it to have real eigenvalues, and indeed it does, namely $$ \lambda_{\pm}=a+d \pm \sqrt{(a-d)^2+b^2}. $$ The square root contains a sum of real squares, so is nonnegative. We also require that $$ a+d>\sqrt{(a-d)^2+b^2} $$ in order to have $\lambda_-$ positive.

On the other hand, the eigenvalues of $S+A$ are $$ \mu_{\pm}=a+d \pm \sqrt{(a-d)^2+b^2-c^2}, $$ so we require $$ (a-d)^2+b^2-c^2>0 $$ for real eigenvalues, which is certainly not necessary, given that we can choose $c$ as large as we like. Hence we can have complex eigenvalues for $S+A$, even if $S$ has positive ones. On the other hand, supposing that $\mu_{\pm} \in \mathbb{R}$, we do automatically get the other condition: $$ a+d>\sqrt{(a-d)^2+b^2} >\sqrt{(a-d)^2+b^2-c^2}, $$ since $c^2 > 0$ and the square root is increasing.

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  • $\begingroup$ So the positive definite matrix must have real eigen values? In that definition A = [1 1;-1 1] is positive definite in the sense $x^TAx > 0$ but it has complex eigenvalues? what can we say about A? is it positive definite? $\endgroup$ – user2104150 Jan 31 '16 at 14:59
  • $\begingroup$ If you use that definition of positive-definite, the antisymmetric part of $A$ makes absolutely no contribution at all: $$ (A_{ij}-A_{ji})x_i x_j = 0. $$ $\endgroup$ – Chappers Jan 31 '16 at 15:20

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