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In some old course notes I'm reading to touch up on statistical forecasting methods, the book often makes reference to "non-negative definite" matrices. I know what a semi-positive definite, positive definite, and indefinite matrix are, but I've never heard this terminology before. Further, online resources don't really seem to mention it, so I'm unsure what exactly it is equivalent to as the "non-negative definite" property isn't directly applied anywhere. Any ideas?

Eric Hansen
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    My guess would be a positive semidefinite matrix, i.e. a matrix whose eigenvalues are non-negative. But we would have to look at the context to be sure. –  Dec 14 '15 at 23:33
  • @Rahul The context is mainly centered around trying to invert a really really big square matrix, and noting that "being nonnegative definite" is a great property. – Eric Hansen Dec 14 '15 at 23:37

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Non-negative definite is a synonym for positive semidefinite. One thing you should watch out for, however, is that while most authors restrict these terms to real symmetric or hermitian matrices, not all do so.

Robert Israel
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  • Are non-negative definite matrices easier to invert for some particular reason? If you don't know no worries. – Eric Hansen Dec 15 '15 at 00:06
  • @EricHansen You should check out Cholesky Decomposition and also this page: https://inst.eecs.berkeley.edu/~ee127a/book/login/l_sym_psd.html for getting started with factoring a PSD (positive semi-definite) matrix as $A = LL^T$. Such decompositions put you well on your way to solving systems that involve PSD matrices – Xoque55 Dec 15 '15 at 00:21
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An $n\times n$ matrix $A$ is non-negative definite (aka positive semi-definite) provided $x^tAx\ge 0$ for each column vector $x$ of length $n$.

John Dawkins
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An $n\times n$ real symmetric matrix A is non-negative definite (aka positive semi-definite) provided $x^TAx\geq 0$ for all $x\in\mathbb{R}^n$ where $x^T$ is the transpose of $x$.

An $n\times n$ complex symmetric matrix A is non-negative definite (aka positive semi-definite) provided $x^*Ax\geq 0$ for all $x\in\mathbb{C}^n$ where $x^*$ is the complex conjugate of $x$.

mathfun
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