# Non symmetric matrix and its transpose and positive definiteness

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.

• positive definite $\implies P D P^T$ where $P$ is orthogonal and $D$ is diagonal $> 0$ – reuns Jan 30 '16 at 21:59
• What is your definition of "positive definite" for a matrix that is not symmetric???????????? – Will Jagy Jan 30 '16 at 22:10
• 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? – user2104150 Jan 31 '16 at 0:27
• @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. – Omnomnomnom Jan 31 '16 at 3:33

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.
• 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? – user2104150 Jan 31 '16 at 14:59
• If you use that definition of positive-definite, the antisymmetric part of $A$ makes absolutely no contribution at all: $$(A_{ij}-A_{ji})x_i x_j = 0.$$ – Chappers Jan 31 '16 at 15:20