Talk:Symmetric matrix

From Wikipedia, the free encyclopedia

Mathematics Portal This article is within the scope of WikiProject Mathematics, which collaborates on articles related to mathematics.
Mathematics rating:	Start Class	Mid Priority	Field: Algebra
Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments. Please add to or update the comments to suggest improvements to the article. Geometry guy 22:00, 17 June 2007 (UTC)

Shouldn't be better to create a distinct entry for 'skew-symmetric matrix' ?

Contents 1 Inverse Matrix 2 Basis, Eigenvectors 3 Symmetric matrices are usually considered to be real valued 4 Products of Symmetric Matrices: Eigenspaces Closed Under Transformation 5 eigenvalues

[edit] Inverse Matrix

Does the inverse of a square symmetrical matrix have any special properties? Does being symmetrical provide any shortcut to finding an inverse? 58.107.136.85 (talk) 03:56, 11 April 2008 (UTC)

[edit] Basis, Eigenvectors

It's easy to identify a symmetric matrix when it's written in terms of an orthogonal basis, but what about when it's not? Is a real-valued matrix symmetrix iff its eigenvectors are orthogonal? —Ben FrantzDale 00:31, 11 September 2006 (UTC)

Reading more carefully answers my question: "Every symmetric matrix is thus, up to choice of an orthonormal basis, a diagonal matrix." So apparently the answer is yes. —Ben FrantzDale 15:27, 11 September 2006 (UTC)

I believe you're confusing a couple of concepts here. A matrix is a rectangular array of numbers, and it's symmetric if it's, well, symmetric. Of course, a linear map can be represented as a matrix when a choice of basis has been fixed. On the other hand, the concept of symmetry for a linear operator is basis independent. Greg Woodhouse 01:34, 30 November 2006 (UTC)

being symmetric with real entries implies unitarily diagonalizable; the converse need not be true. anti-symmetric matrices with real entries are normal therefore unitarily diagonalizable. but the eigenvalues are no longer real, so one must speak of unitary matrices, rather than orthogonal. Mct mht 04:07, 12 September 2006 (UTC)

[edit] Symmetric matrices are usually considered to be real valued

I've made several changes to indicate that symmetric matrices are generally assumed to be real valued. With this, the real spectral theorem can be stated properly. VectorPosse 05:03, 12 September 2006 (UTC)

Would it be better to have a little more detailed discussion of Hermitian? --TedPavlic 16:21, 19 February 2007 (UTC)

It may be worthwhile to add a section on complex symmetric matrices, or matrices that are (complex) symmetric w/r/t an orthonormal basis. They are not as useful as self-adjoint operators, but the category includes toeplitz matrices, hankel matrices and any normal matrix. 140.247.23.104 04:43, 12 January 2007 (UTC)

I agree. We just need to make sure it's in a different section so that it doesn't get mixed up with the stuff about the spectral theorem. VectorPosse 19:28, 19 February 2007 (UTC)

[edit] Products of Symmetric Matrices: Eigenspaces Closed Under Transformation

As the article states, products of symmetric matrices are symmetric if and only if the matrices commute. However, it also says, "Two real symmetric matrices commute if and only if they have the same eigenspaces." This makes no sense. Consider arbitrary matrix A and the identity matrix I. Certainly, AI = IA, so these matrices commute. However, in general A and I will not have the same eigenspaces! I think this statement was supposed to be, "Two real symmetric matrices commute if and only if they are simultaneously diagonalizable," or, "Two real symmetric matrices commute if and only if the eigenspace for one matrix is closed under the other matrix." Both of these statements sound complicated compared to the original statement. I'm not sure if it's worthwhile to even mention it. However, I'm going to make a change. I'm okay with someone removing the statement entirely. --TedPavlic 17:34, 19 February 2007 (UTC)

the previous version was correct. two real symmetric matrix commute iff they can be simultaneously diagoanlized iff they have the same eigenspaces. please undo your change. Mct mht 10:24, 21 February 2007 (UTC)

As far as I can see, Ted's counterexample (identity matrix and arbitrary symmetric matrix) shows that two symmetric matrices can commute without having the same eigenspaces. Please tell me where we go wrong. -- Jitse Niesen (talk) 11:25, 21 February 2007 (UTC)

hm, that depends on what's meant by "having the same eigenspaces", no? if that means "the collection of eigenspaces coincide", then you would be right. (however, seems to me the wording of the comment, which i removed, about the "closure" of eigenspaces can be improved.) perhaps it's more precise to say two real symmetric matrix commute iff there exists a basis consisting of common eigenvectors. Mct mht 12:17, 21 February 2007 (UTC)

also, the identity matrix is really a degenerate case. since it and its multiples are the only matrices that's diagonal irrespective of the basis chosen. excluding such cases (if A restricted to a subspace V is a · I, remove V), seems to me that the general claim is true: real symmetric matrices {A_i} commute pairwise iff the family of eigenspaces of A_i and the family of eigenspaces of A_j are the same for all i and j. Mct mht 15:42, 21 February 2007 (UTC)

I agree with "two real symmetric matrices commute iff there exists a basis consisting of common eigenvectors". I think the more common formulation is "two real symmetric matrices commute iff they are simultaneously diagonalizable", so I'd prefer that. I agree that the formulation "the eigenspace for one matrix is closed under the other matrix" is rather unfortunate as I had to read that sentence a couple of times before I understood what is meant.

I don't understand what you mean with "if A restricted to a subspace V is a · I, remove V". Every matrix is a multiple of the identity when restricted to an eigenspace, and after removing the eigenspaces of a symmetric matrix there's nothing left. -- Jitse Niesen (talk) 04:04, 22 February 2007 (UTC)

shoot, you're right. well, remove V if dimension V is > 1. that better? Mct mht 04:10, 22 February 2007 (UTC)

hm, forget it Jitse, that did not make it better. you're right there. Mct mht 12:20, 22 February 2007 (UTC)

[edit] eigenvalues

are the eigenvalues of A:n×n, A=A^T always {0,...,0,tr(A)} ?
applies for matrix B^TB with B=[1,2,3,4]
--Saippuakauppias ⇄ 10:48, 31 December 2007 (UTC)

No. For instance, the identity matrix is symmetric, but has eigenvalues {1,1,…,1}. However, every matrix of the form A = B^TB does have {0,…,0,tr(A)) as its eigenvalues. Such matrices are called rank-one matrices, because their rank is one. -- Jitse Niesen (talk) 15:24, 31 December 2007 (UTC)