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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?

Rodrigo de Azevedo
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Ramin
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3 Answers3

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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 $$ z'Az=\frac12(z'Az+z'A'z)=z'\left(\frac{A+A'}2\right)z. $$

Martin Argerami
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  • So, This means that if $\mathbf{z'Az}\geq0$, whether $A = A'$ or not, we call $\mathbf{A}$ a Positive definite Matrix? – Ramin Dec 09 '12 at 07:48
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    I wouldn't do it, but many people does. – Martin Argerami Dec 09 '12 at 10:30
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    You mean I can find a reference that defines Positive definiteness for non-symmetric matrices? – Ramin Dec 09 '12 at 14:00
  • 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. – Martin Argerami Dec 09 '12 at 14:48
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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.

EarthwormA3aan
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Matan
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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 $z^{\mathrm{T}}Bz>0$.

Gregor Botero
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