Talk:Cholesky decomposition

From Wikipedia, the free encyclopedia

Contents

[edit] Hermitian property

Shouldn't A be symmetric or Hermitian ?

A is positive definite, and all positive-definite matrices are Hermitian (see positive-definite matrix). I hope this concise explanation suffices. -- Jitse Niesen 14:13, 26 Apr 2005 (UTC)
Is the "all PD matrices are Hermitian" fact available in Wikipedia? I don't know who wrote the original question, but to be fair, the Wikipedia definition: positive-definite matrix gives only the properties that a Hermitian matrix (M) must have to be PD, but doesn't explicitly say than a non-symmetric or non-hermitian matrix can't be PD. Ideally, this would be stated or derived among the 'properties' listed on the positive-definite matrix page, rather than here on a talk page.

Wragge 15:17, 2005 Apr 26 (UTC)

(In case there is any confusion, not all Hermitians are PD, so this page, Cholesky decomposition, is fine as it is.)
As I feared, my explanation was not well formulated. What I meant to say is that a positive-definite matrix is by definition a Hermitian matrix A such that v*Av > 0 for all nonzero vectors v. So, it is built into the definition that Hermitian matrices are positive definite. You (Wragge) are right that this is not stated explicitly in the article positive-definite matrix.
The question which Wragge probably wants to see answered is: if you know that v*Av > 0 for all nonzero vectors v, does it follow that A is Hermitian? If we are restricting ourselves to the reals, the answer is no: there are real matrices A which are not symmetric, but which still have vTAv > 0 for all nonzero real vectors v. I need to check what happens for the complex case; one complication is that v*Av is not necessarily real if A is not Hermitian.
I'll check some books when I have time to make sure I've got the details right and rewrite positive-definite matrix if necessary. -- Jitse Niesen 16:57, 26 Apr 2005 (UTC)


[edit] PD MUST BE HERMITIAN IN DISPUTE

Thanks for the simple explanation Jitse, I wasn't sure whether all matrices, A, where v*A.v > 0 were Hermitian or not.

Isn't that the general definition of PD though? Is it specified that A must also be Hermitian to qualify as PD? Based on my hazy education, and what I see on Wolfram Research, and Computation Methods in Linear Algebra I am forced to dispute this definition.

Everything I've seen specifies simply the first condition (the multiplication property) and not the Hermitian property (Planet Math fudges the issue by saying that "If A is a square Hermitian then ... is it a positive definite matrix", without saying that this is the non-Hermitians with the same property are not. I have seen no definitions explicitly specifying this condition.) For instance, the definition Google uses, from University of Waterloo simply says ([1]):

Positive definite matrix (A). x'Ax > 0 for all nonzero x.

Importantly, Computation Methods in Linear Algebra Proposition 6.2.2 clearly implies:

  1. Being Hermitian is not a condition for a matrix to be PD (because it states that both conditions must be fulfilled for Cholesky factorization to succeed they must be logically independent).
  2. Therefore, the Cholesky decomposition page must make this condition (symmetry/Hermitian) explicit, as the original questioner requested

I conclude that the original questioner raised a useful & correct gap in the definition, but if I have made a mistake, It would be interesting to see where. Can you tell me why you (Jitse Niesen) think that Matrices, A, where all non-zero v vectors give: v*A.v > 0 must also be Hermitian to qualify as a positive-definite matrix?

I only work with symmetric matrices, and most programs and references refer to "Positive Definite Symmetric" as a subset of "Positive Definite", implying that there is no such thing as "Asymmetric Positive Definite Matrix". If such a thing does exist, then some people (me anyway) would be interested in a topic or heading with that name, and an example. It would be one of the few on the Internet.

Also, I am not a pure mathematician, so please forgive any basic errors in understanding that I may have made. I don't know if this is relevant, but the structure of "Definite bilinear form" has the same "if Hermitian then PD" form, without saying "iff Hermitian then PD". Even if this is right, I think it should be cleared up.

- Wragge 22:19, 2005 Apr 28 (UTC)

I pulled some books from the shelf and they disagree on this issue. Most use a formulation like "A Hermitian matrix A is positive definite if v*Av > 0," which is ambiguous. Some imply further on that they consider being Hermitian to be a condition for a matrix to be positive definite and some imply that Hermitian and positive definite are independent properties. So, I think it is safest to stress in this article that we need A to be both Hermitian and positive definite. I wrote a small discussion over the relation between positive definite and Hermitian at Positive-definite matrix#Non-Hermitian matrices. -- Jitse Niesen 21:33, 29 Apr 2005 (UTC)

Thanks, Jitse. The Positive-definite matrix#Non-Hermitian matrices is a good summary, with a nice example. I think the reason for the ambiguity is that non-Hermitian matrices have so few applications that the literature concentrates on SPD matrices. Anyway, I don't know enough to take this any further, perhaps it's something to watch out for, if you are looking through textbooks - Wragge 21:47, 2005 Apr 29 (UTC)


[edit] The semidefinite case

Question:

The page says that only a positive-definite matrix may be factorized as seen. Are positive-SEMIdefinite matrices excluded from this factorization?

I added a paragraph to address this question. -- Jitse Niesen (talk) 04:37, 10 May 2006 (UTC)


[edit] The real matrix vs. complex matrix decomposition

I belive the formulas given for the decomposition relate only to real (real SPD) case. I think they should be generalised to the complex Hermitian positive-definite case. Than they will look like that:

l_{ii} = \sqrt{a_{ii} - \sum_{k = 1}^{i - 1}{|l_{ik}|^2}}

l_{ij} = \frac{1}{l_{jj}} \left( a_{ij} - \sum_{k = 1}^{i - 1}{l_{ik}\bar{l}_{jk}} \right)


One more question came to my mind...Why do we consider only positive-definite matrices in complex case??? There exists square root for any complex number (not unique however). So we should be able to use above formulas for any Hermitian matrix (not necessarily PD). Of course the diagonal elements of L would not be positive reals anymore.

What is more, we should be able to use the original formulas for symmetric real case in symmetric complex case. So why do we care about positive-definitness???

--A-lbi 09:49, 22 August 2007 (UTC)

[edit] Possible mistake?

l_{i,i} = \sqrt{ a_{i,i} - \sum_{k=1}^{i-1} l_{i,k}^2 }.

Are we sure \sum_{k=1}^{i-1} is right? For the first value of the matrix we have i=1, so:

l_{1,1} = \sqrt{ a_{1,1} - \sum_{k=1}^{0} l_{1,k}^2 }.

So k goes from 1 to 0? Also, the next value is (i = 1, j = 2):

l_{i,j} = \frac{1}{l_{j,j}} \left( a_{i,j} - \sum_{k=1}^{j-1} l_{i,k} l_{j,k} \right), \qquad\mbox{for } i>j.
l_{1,2} = \frac{1}{l_{2,2}} \left( a_{1,2} - \sum_{k=1}^{0} l_{1,k} l_{2,k} \right)

Again the sum seems strange. —Preceding unsigned comment added by 84.223.150.164 (talk) 21:54, 27 March 2008 (UTC)

By convention,
\sum_{k=1}^{0} l_{1,k}^2 = 0,
see empty sum. So, for the (1,1) entry, we get
l_{1,1} = \sqrt{ a_{1,1} - \sum_{k=1}^{0} l_{1,k}^2 } = \sqrt{a_{1,1}}.
Indeed, if we look at a 1x1 matrix [a], then its Cholesky decomposition is [a] = [l]* [l], so l is the square root of a. So at first sight everything seems fine. -- Jitse Niesen (talk) 22:23, 27 March 2008 (UTC)