Talk:Positive-definite matrix

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[edit] Old remarks

The statements about the complex/real case were incorrect. Here's a simple counterexample:

\begin{bmatrix} 1 & 2i \\ -2i & 1 \end{bmatrix}

This Hermitian matrix has $x^T A x > 0$ for all real $x$, but not for $x = [1 , i]$.

Would it make any difference if x was in Cn?

shd: In that case it should be something like x*Mx where * refers to complex conjugate transpose

I added that to the article. AxelBoldt

Does this encompass all positive-definite matrices? The Mathworld page leads me to believe that there are positive-definite matrices that are not Hermitian. From the page: 'A necessary and sufficient condition for a complex matrix to be positive definite is that the Hermitian part ... be positive definite.' inferno

If you think about xtMx as a quadratic form in the vector x, and write the square matric M as a sum of a symmetric part M1 and an antisymmetric part M2, you see that the quadratic form is independent of M2. So if you look at a symmetric positive definite M1, you can add any M2 you want, and still be positive definite. If that's useful ... And the same thing if you split up into a hermitian and anti-hermitian part, for complex x and introducing complex conjugation in the form.

Charles Matthews 20:11, 16 Jun 2004 (UTC)


Actually, from the point of view of applications, the restriction of the definition to symmetric matrices (in the real case) is unfortunate, since for instance discretizations of advection problems yield non-symmetric positive definite matrices. I'll change this unless I receive serious complaints. Unfortunately, the article Normal matrix becomes wrong then. -- Guido Kanschat 20:15, 3 November 2005 (UTC)

This page is linked to from Niemeier lattices. The usage there is a "positive definite unimodular lattice". So we should probably explain here what "positive definite" means in the context of a lattice. I don't know the answer. A5 06:02, 20 Jun 2005 (UTC)

I'm afraid I can't find the link. The first sentence in Niemeier lattices links to positive definite, which redirect to definite bilinear form. However, the best explanantion of "positive definite" in this context can be found on Lattice (group). -- Jitse Niesen 12:18, 20 Jun 2005 (UTC)

[edit] English

This is all very nice and I'm sure a "rigorous definition" is great for some people. However, for someone who just got redirected from positive-semidefinite and doesn't have a degree in pure math, this is not very usefull.

Agreed - can someone with a better understanding of the subject please try to give a more intuitive description somewhere? 70.93.249.46

I may be a mathematician, but I have no idea what could be made clearer. Of course, you'll likely encounter only real matrices for which the definition is simply that B is symmetric and x^T B x >= 0 for all vectors x. Geometrically, the set of n x n positive semidefinite matrices is a cone (you can add two together and multiply them with a nonnegative number), but that's rather trivial, and the definition of the barrier function that makes them useful in optimization is not helpful either. What sort of "useful" facts would you like to learn? Wandrer2 16:35, 2 March 2006 (UTC)
I've just expanded property 2 in the definition to give a spectral intuition about how positive definite M are special within the set of Hermitian operators. For me, this intuition is fundamental and was not quite there before. Hope it helps. Eclecticos 08:02, 20 September 2006 (UTC)

[edit] Proofs of Positive Semidefiniteness

I would like to prove that the difference between two general matrices (each of a certain class) is a positive semidefinite matrix. I am not up to the task without some examples; would anybody mind posting examples of positive semidefinite (or definite) proofs?

[edit] Positive Eigenvalue?

What is ment by "A positive definite if and only if all eigenvalues are positive"?. Is all eigenvalues >0 or is all eigenvalues \geq0?

I have managed to prove the following: Let $A$ be a positive definite $n\times n$-matrix with eigenvalues \lambda_1,\lambda_2,\ldots,\lambda_n

then \lambda_i\geq0, \quad i=1,2,3\ldots,n and there exist a k\in\{1,2,3\ldots,n\} such that λk > 0.

But I havn't managed to prove the following:

Let $A$ be a positive definite $n\times n$-matrix with eigenvalues \lambda_1,\lambda_2,\ldots,\lambda_n then \lambda_i>0,\quad i=1,2,3,\ldots,n.

To be clearer A= \begin{pmatrix} 3&0&0\\ 0&2&0\\ 0&0&0\\ \end{pmatrix} has the eigenvalues λ1 = 3,λ2 = 2,λ3 = 0.

The matix is positive definite since x^tAx=\sum_{i=1}^3\lambda_ix_i^2>0 for all vectors x=(x_1,x_2,x_3)\in\mathbf{R^3},x\neq0.

Therefore A symmetric and positive definite doesn't imply that all eigenvalues of $A$ is positive (in the sence >0).

However maybe this might just be the case when the matrix contain a row $j$ and column $j$ that are both zerovectors.

Can anybody help me??? I don't get it.

/Tobias mathstudent

"A positive definite if and only if all eigenvalues are positive" means that all eigenvalues have to be > 0.
Your matrix A is not positive definite, because xTAx = 0 for x = (0,0,1), which is not the zero vector.
I hope this clarifies the matter. If not, feel free to ask. -- Jitse Niesen (talk) 10:53, 16 August 2006 (UTC)
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