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What do we mean by a matrix is positive or negative definite? Does it have any analogy with a positive real number?

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How common is it to include "symmetric" (or "hermitian" in the complex case) in the definition? And how common to omit this requirement? –  GEdgar Feb 25 '13 at 14:38

5 Answers 5

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You could view it as the parabola $Kx^2=y, K>0 $ taken up to higher dimensions. In place of the positive constant $K$, a positive definite matrix $A$ would ensure that the high dimensional parabola (visualise it as a bowl) takes all positive values for all $x\in \mathbb{R}^n$. Positive definiteness

See this question for why definiteness is needed when considering ordering among matrices.

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Answers to this and more are in the wiki article.

It is not analogous to positive and negative real numbers. That sense of "positive" has to do with the linear ordering of the real numbers.

"Positive definite" refers to a real matrix $A$ having the property that $x^TAx$ is positive for all nonzero vectors $x$. ($x^TAx$ is, of course, a real number in the case, but the whole of my discussion could have been done over any subring of $\Bbb R$.)

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An $n\times n$ matrix $A$ is said to be positive definite if $$ x^TAx>0,\quad \text{for all }x\in\mathbb{R}^n\setminus \{0\},\tag{1} $$ where $^T$ denotes the transpose. Similarly $A$ is negative definite if $(1)$ holds with $<0$ (also we have the terms non-negative definite and non-positive definite which is $\geq 0$ and $\leq 0$ respectively).

As for the analogy with real numbers, we note that a real number $a\in \mathbb{R}$ is a $1\times 1$ matrix. If $a=0$ then it can't satisfy $(1)$, but if $a\neq 0$ we have that $a$ is positive definite if and only if $$ x^2a=xax>0,\quad \text{for all } x\neq 0 $$ which is equivalent to saying that $a>0$. So positive definiteness for $1\times 1$ matrices is exactly the same as the number being strictly positive.

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To a (symmetric) square matrix $A=(a_{i,j})$ of order $n$ with real coefficients, it is associated a bilinear form $$ \phi(\vec x,\vec y)=(x_1,..x_n)A(y_1,...,y_n)^t $$ and the corresponding quadratic form $$ Q(\vec x)=\phi(\vec x,\vec x)=\sum_{i=1}^n\sum_{j=1}^na_{i,j}x_ix_j. $$ The quadratic form $Q$ (and so the matrix $A$ by extension) is said positive definite if $$ Q(\vec x)>0\qquad\text{for all $\vec x\neq0$}. $$ When $n=1$ the matrix is just a scalar $a$, and positive definiteness indeed means that $a>0$.

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To add to all the other answers, Posiitive definite matrices are the only matrices with a unique square root which is also positive definite, something reminiscent of non-negative numbers.

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