4
$\begingroup$

Is symmetry a necessary condition for positive (or negative) definiteness?

If not:

It can be proved that if $\mathbf{A} \in \mathbb{R}^{m\times m}$ is a square (non-symmetric) matrix, then $ \mathbf{z'Az=z'Bz},~~\mathbf{B=B'= \frac{A+A'}{2}} $

On the other hand, a positive definite matrix is a symmetric matrix for which:

$\mathbf{z'Bz}>0,~~ \mathbf{z\ne 0}$

Can we imply that $\mathbf{A}$ which is a non-symmetric matrix, is positive definite?

3 Answers 3

5

It looks like you are working with real matrices.

Most often the definition of positive definite includes symmetric, but sometimes this is not required.

In any case, if $z'Az\geq0$ then $z'A'z=(z'Az)'=z'Az\geq0$. So $ z'Az=\frac12(z'Az+z'A'z)=z'\left(\frac{A+A'}2\right)z. $

  • 0
    The thing is that when you work with a complex vector space, you get Hermitian for free; but not in the real case. This issue with the definition is mentioned in the wikipedia article: http://en.wikipedia.org/wiki/Positive-definite_matrix#On_the_definition but I'm not familiar with references dealing with positive definite matrices.2012-12-09
5

Not sure if that was your question, but if it is

does there exist a nonsymmetric matrix s.t. $v^T A v >0$ for all $v\neq 0$

then the answer is yes. Take for example the matrix $ A = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}. $

You can check that for any $v=(x,y)$ you get $v^T A v = x^2+y^2 > 0$, if $x$ or $y$ is not $0$, or equivalently $v$ is not the zero vector.

  • 0
    See also: https://en.wikipedia.org/wiki/Positive-definite_matrix#Extension_for_non-symmetric_matrices2017-09-22
1

Yes, it is possible. In fact, it follows easily from the properties of taking the transpose:

$0 < z^{\mathrm{T}}\left(A+A^{\mathrm{T}}\right) z = z^{\mathrm{T}}Az + z^{\mathrm{T}}A^{\mathrm{T}}z = z^{\mathrm{T}}Az + (z^{\mathrm{T}}Az)^{\mathrm{T}} = 2z^{\mathrm{T}}Az$

Taking the transpose of a real number doesn't change anything. Therefore $z^{\mathrm{T}}Az> 0$ if and only if $z^{\mathrm{T}}Bz>0$.

  • 0
    Would you please see my comment at Martin Argerami answer. Thanks2012-12-09