Talk:Symmetric matrix

message: 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)
thread_title:  Inverse Matrix 

message: 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)
replies: text: 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)

text: 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)
thread_title:  Basis, Eigenvectors 

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

text: 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)
thread_title:  Symmetric matrices are usually considered to be real valued 

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

message: ...is conspicuous by the absence of any mention of it in this article!
: 
Maybe I'll be back. Michael Hardy (talk) 02:18, 10 August 2008 (UTC)
replies: text: It's at the start of the "Properties" section. -- Jitse Niesen (talk) 10:57, 10 August 2008 (UTC)
thread_title:  The spectral theorem... 

message: Hi,
there's a mistake in the article. It's claimed that the trace of three symmetric (or hermitian) matrices is invariant under arbitrary permutations. To prove this, it's used that (CBA)^t = CBA which is simply not true because the product of symmetric (hermitian) matrices is symmetric (hermitian) if and only if they commute.  —Preceding unsigned comment added by 192.33.103.47 (talk) 09:27, 28 June 2010 (UTC)
thread_title:  trace of the product of three matrices 

message: Hello, the page currently states that for each complex symmetric matrix, there exists a unitary transformation such that the resulting diagonal matrix has real entries. The eigenvalues of complex symmetric matrices are generally themselves complex, and not all real. Somebody tell me if I'm reading this wrong, otherwise I'm going to change the wording "is a real diagonal matrix" to "is a complex diagonal matrix".  — Preceding unsigned comment added by Jeparie (talk • contribs) 18:36, 25 September 2015 (UTC)
replies: text: The entries of the diagonal matrix are the singular values not the eigenvalues of the original matrix. Best wishes, --Quartl (talk) 19:47, 25 September 2015 (UTC)
thread_title:  Complex symmetric matrix eigenvalues 

message: I changed all the math from inline math to tex. This was reverted by ‎87.254.93.231 several times. The article as it is now contains an ugly mixture of tex and inlne math. I propse changing all again to tex.
Reasons for Tex:
* You can display almost everything with Tex
* Tex is easy to distinguish from surrounding nom-math text. E.g. for a matrix A vs. for a matrix A.
Reasons for inline math
* Less work to write article
What do you think? 11:36, 15 January 2019 (UTC)
thread_title:  Math notation 

message: Note: This proof yields a short algorithm for computing the Takagi decomposition using software packages like Scipy and Matlab. The schur decomposition feature should be used for this.
Let \operatorname{uncomplexify} : M_n(\mathbb C) \to M_{2n}(\mathbb R) denote the injective ring homomorphism A + Bi \mapsto \begin{pmatrix}A & -B \\ B & A \end{pmatrix}, where we use block matrix notation. Let M = A + Bi be an arbitrary non-singular complex-symmetric matrix. (The non-singularity restriction will later be lifted.) Observe that while the matrix \operatorname{uncomplexify}(i M) = \begin{pmatrix}-B & -A \\ A & -B \end{pmatrix} is not \mathbb R-symmetric, we may define J = \begin{pmatrix}0 & I \\ I & 0 \end{pmatrix} and get that \operatorname{uncomplexify}(i M) J = \begin{pmatrix}-A & -B\\ -B & A \end{pmatrix} is \mathbb R-symmetric. By the spectral theorem it follows that \operatorname{uncomplexify}(i M) J has an orthonormal eigenbasis v_1, v_2, \dotsc, v_{2n} with real eigenvalues \lambda_1 \geq \lambda_2 \geq \dotsc \geq \lambda_{2n}. Let K = \begin{pmatrix}0 & -I \\ I & 0 \end{pmatrix}  and observe that for every eigenvector v_i  in our eigenbasis, the vector K v_i  is an eigenvector of eigenvalue -\lambda_i . We therefore improve our orthonormal eigenbasis of \operatorname{uncomplexify}(i M) J to the new orthonormal eigenbasis (v_1,v_2,\dotsc,v_n,Kv_1,\dotsc,Kv_2,Kv_n) , which we interpret as a block matrix P. Verify that each v_i is indeed orthogonal to each K v_j because they have distinct eigenvalues (which might fail if M were singular). Given our definition of P, we have that \operatorname{uncomplexify}(i M) J = P \begin{pmatrix} \Lambda & 0 \\ 0 & -\Lambda\end{pmatrix} P^T (where \Lambda = \operatorname{diag}(\lambda_1,\lambda_2,\dotsc,\lambda_n)). Observe that P  is equal to \operatorname{uncomplexify}(U)  for some matrix U ; this follows because if we write each v_i as the block matrix \begin{pmatrix}w_i \\ \hline z_i\end{pmatrix}, we have that P = \begin{pmatrix}W & -Z\\ \hline Z & W \end{pmatrix} = \operatorname{uncomplexify}(W + iZ). Observe also that U is unitary; this follows because \operatorname{uncomplexify}(I) = I = P^T P =  \operatorname{uncomplexify}(U)^T \operatorname{uncomplexify}(U) = \operatorname{uncomplexify}(U^* U), and by cancelling \operatorname{uncomplexify} due to injectivity, we get U^* U = I. We therefore have that
\begin{aligned}
\operatorname{uncomplexify}(iM) J &= P \begin{pmatrix} \Lambda & 0 \\ 0 & -\Lambda\end{pmatrix} P^T
\\&= \operatorname{uncomplexify}(U) \begin{pmatrix} \Lambda & 0 \\ 0 & -\Lambda\end{pmatrix} \operatorname{uncomplexify}(U)^T
\\&= \operatorname{uncomplexify}(U) \operatorname{uncomplexify}(i\Lambda) J \operatorname{uncomplexify}(U)^T
\\&= \operatorname{uncomplexify}(U) \operatorname{uncomplexify}(i\Lambda) \operatorname{uncomplexify}(\overline U)^T J
\\&= \operatorname{uncomplexify}(U i \Lambda \overline U^*) J
\end{aligned}
Cancelling J (as it's invertible) and then cancelling \operatorname{uncomplexify} (because it's injective) yields M = U \Lambda U^T.
The result can be extended to any singular matrix M by approximating M as a sequence of invertible matrices M_n, and then forming the pair of sequences (U_n, \Lambda_n) such that M_n = U_n \Lambda_n U_n^T. Since the components of U_n and \Lambda_n (for each n) are bounded, we may appeal to the Bolzano–Weierstrass theorem to get a pair of subsequences (U_{n_k}, \Lambda_{n_k}) that both converge. We therefore get M = \lim_{k \to \infty} M_{n_k} = \left(\lim_{k \to \infty}U_{n_k}\right)\left(\lim_{k \to \infty} \Lambda_{n_k}\right)\left(\lim_{k \to \infty} U_{n_k}\right)^T. We are done. --Svennik (talk) 14:54, 19 May 2022 (UTC)
thread_title:  Proposing adding new proof of Takagi 

message: Note: This proof does not treat singular matrices as a special case.
Let \operatorname{uncomplexify} : M_n(\mathbb C) \to M_{2n}(\mathbb R) denote the injective ring homomorphism A + Bi \mapsto \begin{pmatrix}A & -B \\ B & A \end{pmatrix}, where we use block matrix notation. Let M = A + Bi be an arbitrary complex-symmetric matrix. Observe that while the matrix \operatorname{uncomplexify}(i M) = \begin{pmatrix}-B & -A \\ A & -B \end{pmatrix} is not \mathbb R-symmetric, we may define J = \begin{pmatrix}0 & I \\ I & 0 \end{pmatrix} and get that \operatorname{uncomplexify}(i M) J = \begin{pmatrix}-A & -B\\ -B & A \end{pmatrix} is indeed \mathbb R-symmetric. Let K = \operatorname{uncomplexify}(iI) = \begin{pmatrix}0 & -I \\ I & 0 \end{pmatrix}  and observe that for every eigenvector v of eigenvalue \lambda of \operatorname{uncomplexify}(i M) J, the vector K v  is an eigenvector of eigenvalue -\lambda. We therefore build an orthonormal eigenbasis of \operatorname{uncomplexify}(i M) J in the following way:
: We start with the empty basis B and the linear map L : \operatorname{span}(B)^\bot \to \operatorname{span}(B)^\bot, v \mapsto \operatorname{uncomplexify}(i M) J v. 
: While \operatorname{span}(B)^\bot is not 0-dimensional, we let v be a unit eigenvector of L. Observe that as well as v being orthogonal to  \operatorname{span}(B), so is Kv. We replace B with B \cup \{v,Kv\}, and replace L with its restriction to \operatorname{span}(B)^\bot.
We arrange our resulting eigenbasis into the block matrix P = (v_1,v_2,\dotsc,v_n,Kv_1,Kv_2,\dotsc,Kv_n). Given our definition of P, we have that \operatorname{uncomplexify}(i M) J = P \begin{pmatrix} \Lambda & 0 \\ 0 & -\Lambda\end{pmatrix} P^T (where \Lambda consists of the eigenvalues corresponding to the eigenvectors v_1 to v_n). Observe that P is equal to \operatorname{uncomplexify}(U)  for some matrix U ; this follows because if we write each v_i as the block matrix \begin{pmatrix}w_i \\ \hline z_i\end{pmatrix}, we have that P = \begin{pmatrix}W & -Z\\ \hline Z & W \end{pmatrix} = \operatorname{uncomplexify}(W + iZ), so that U = W + iZ. Observe also that U is unitary; this follows because \operatorname{uncomplexify}(I) = I = P^T P =  \operatorname{uncomplexify}(U)^T \operatorname{uncomplexify}(U) = \operatorname{uncomplexify}(U^* U), and by cancelling \operatorname{uncomplexify} (due to injectivity) we get I = U^* U. We therefore have that
\begin{aligned}
\operatorname{uncomplexify}(iM) J &= P \begin{pmatrix} \Lambda & 0 \\ 0 & -\Lambda\end{pmatrix} P^T
\\&= \operatorname{uncomplexify}(U) \begin{pmatrix} \Lambda & 0 \\ 0 & -\Lambda\end{pmatrix} \operatorname{uncomplexify}(U)^T
\\&= \operatorname{uncomplexify}(U) \operatorname{uncomplexify}(i\Lambda) J \operatorname{uncomplexify}(U)^T
\\&= \operatorname{uncomplexify}(U) \operatorname{uncomplexify}(i\Lambda) \operatorname{uncomplexify}(\overline U)^T J
\\&= \operatorname{uncomplexify}(U i \Lambda \overline U^*) J
\end{aligned}
Cancelling J (as it's invertible) and then cancelling \operatorname{uncomplexify} (because it's injective) yields M = U \Lambda U^T. --Svennik (talk) 19:46, 19 May 2022 (UTC)
thread_title:  Proof 3 of Takagi