Talk:Trace (linear algebra)

message: Why is the map given by v \otimes h \mapsto (w \mapsto h(w)v)? If I'm not mistaken this would suggest that the image of any linear operator is one-dimensional. 99.240.144.168 (talk) 02:23, 31 December 2011 (UTC)
thread_title:  Coordinate-free definition 

message: I would like to suggest a change regarding the trace of the product of several (>2) matrices.  Specifically the part beginning "If A, B, and C are square...".  The phrase "all permutations" is, strictly speaking, not correct.  The last three terms in the chained equality, while looking like general permutations, are transposes of the first three with the transpose marks omitted due to symmetry.  My motivation for the change is that if more than three matrices are involved only the cyclic permutation may be used to reorder the matrices.  Transposing with symmetry assumed will make more (apparent) permutations possible, but not all posible permutations.  My quick and dirty suggestion is to delete the entire paragraph along with the chained equality.  It could be fixed but it is just a special case involving the previously discussed cyclic permutation and transpose along with the symmetry condition.  Doggit 17:19, 10 August 2006 (UTC)
thread_title:  Trace of matrix products 

message: I set the link from spur to this article. --Kajaktiger 16:47, 4 November 2006 (UTC)
thread_title:  Direct Link 

message: Don't revert back to d Tr = I. This is not true. There is a vectorisation operator present if you must present the information in this form. The peterson & pederson seem to have dropped this when they copy pasted from brookes. 
Just goes to show what sort of crap you can find on the internet. —The preceding unsigned comment was added by 150.203.45.188 (talk) 09:35, 5 December 2006 (UTC)
thread_title:  Derivatives 

message: I think that following <A,B> = tr\left(A\,^*B\right), it should be mentioned that ^*B is the conjugate transpose of B.
131.130.90.152 10:36, 22 January 2007 (UTC)
thread_title:  Inner product 

message: Trace can be used in a non-obvious way where one considers a scalar the trace of a 1×1 matrix. For example, if X is a n×1 column vector and A is a n×n square matrix, then X^TAX is a 1×1 matrix which is often regarded as a scalar, but it can be beneficial to use the trace operator instead (to use the cyclic property). This technique is used in the Estimation of covariance matrices article. Can anyone provide a simpler example to add to the main article? Roman V. Odaisky 17:23, 4 October 2007 (UTC)
thread_title:  Trace of a 1×1 matrix 

message: I’m confused by the formula 0 \leq \mathrm{tr}(AB)^n \leq \mathrm{tr}(A)^n \mathrm{tr}(B)^n. Shouldn’t it have read \mathrm{tr}((AB)^n) instead of the middle term? Or perhaps the formula should be split into two inequalities:
:\mathrm{tr} AB \leq \mathrm{tr} A \cdot \mathrm{tr} B
:\mathrm{tr} A^n \leq (\mathrm{tr} A)^n (if n is natural then this follows from the previous inequality)
Roman V. Odaisky 17:33, 4 October 2007 (UTC)
thread_title:  <math>\mathrm{tr}(AB)^n</math> 

message: Hi---just wanted to leave a note about the various edits I've been making to the article. If anyone has objections or comments, please respond here---thanks. Spireguy (talk) 18:28, 4 March 2008 (UTC)
thread_title:  Various edits 

message: I know the determinate of a transformation matrix can be thought of as the change in volume it causes. For example, if det(A) = 1, then A doesn't change the (n-dimensional) volume of an object it is applied to. Could someone provide a similar intuition for trace? What is the meaning of the sum of the eigenvalues? —Ben FrantzDale (talk) 11:59, 27 March 2008 (UTC)
thread_title:  Intuition/example 

message: Let A be a symmetric matrix, and B anti-symmetric matrix. THen
:\operatorname{tr}(AB) = 0
I was going to add this to the article, but can't see the proof on top of my head. -- Taku (talk) 02:40, 1 October 2008 (UTC)
thread_title:  Symmetric and anti-symmetric matrices 

message: Is this statement true?? 
Doesn't that only apply for hermitian matrixes which have been diagonalized?? 20:11, 17 January 2009 (UTC)
thread_title:  the trace of a matrix is the sum of its eigenvalues? 

message: ?????  —Preceding unsigned comment added by 86.20.235.138 (talk) 12:47, 14 March 2009 (UTC)
thread_title:  why is the reference section being used for footnotes? 

message: The article contains the text "When both A and B are n by n, the trace of the commutator of A and B vanishes: tr([A, B]) = 0. Conversely, any square matrix with zero trace is the commutator of some two matrices.[3] In particular, the identity matrix is never similar to the commutator of any matrices.". Surely this can't be true! I mean, if you take two nxn matrices which commute, then their trace will equal the trace of the identity matrix, which is never zero. Is this meant to say something else? SetaLyas (talk) 12:53, 25 May 2009 (UTC)
thread_title:  Trace of a commutator 

message: "If A is a square n-by-n matrix with real or complex entries and if λ1,...,λn are the (complex and distinct) eigenvalues of A (listed according to their algebraic multiplicities)"
The line says that A is n-by-n and that the eigenvalues \lambda_1...\lambda_n should be distinct. Then it requires the eigenvalues to be listed accordingly to 
their algebraic multiplicity. But if the eigenvalues are all distinct, of course their algebraic multiplicities are 1. So in what the (listed according ...) helps?  —Preceding unsigned comment added by 79.138.132.212 (talk) 08:41, 4 September 2009 (UTC)
thread_title:  Algebraic Multiplicities 

message: Can somebody point me to a proof of the statement: 
0 \leq \mathrm{tr}(AB)^n \leq \mathrm{tr}(A)^n \mathrm{tr}(B)^n?
:No, since it's totally false. Algebraist 22:06, 24 February 2010 (UTC)
replies: text:  I've found a proof for the case of n=1 for a real positive semidefinit matrix, so for general n it's also proven. wpoely86 (talk) 11:24, 25 February 2010 (UTC)
thread_title:  <math>\mathrm{tr}(AB)^n</math> 

message: Perhaps it should be mentioned that the matrix P must have the same dimensions as A. --Tomtheebomb (talk) 22:46, 3 April 2010 (UTC)
thread_title:  Similarity Invariance 

message: The first line of the article says "the trace of an n-by-n diagonal matrix A is defined to be the sum of the elements ...", whereas, the first example shows calculation of trace for a non-diagonal matrix. As far as I know, trace is defined for square matrix, and does not require diagonal matrix. -- Samikrc (talk) 05:30, 7 February 2011 (UTC)
thread_title:  Trace only for diagonal matrix? 

message: How can it be zero always? It is mentioned under the section example. — Preceding unsigned comment added by Tagib (talk • contribs) 07:03, 14 December 2011 (UTC)
replies: text: Because it is false. Any simple example can prove that. However, the trace of a product of a symmetric and skew-symmetric matrix is always zero. wpoely86 (talk) 13:22, 3 October 2012 (UTC)
thread_title:  Trace of a symmetric matrix 0? 

message: The trace of a projection matrix is the dimension of the target space. 
P_{X}=X\left(X^{T}X\right)^{-1}X, we have \text{Tr}\left(P_{X}\right)=\text{rank}\left(X\right). Jackzhp (talk) 03:08, 9 February 2011 (UTC)
thread_title: Trace of a projection matrix

message: This doesn't make any sense.  Can we clarify this point?  —Preceding unsigned comment added by 152.3.68.8 (talk) 23:12, 23 March 2011 (UTC)
thread_title: "Trace is cognate to 'spoor'"

message: Is a sum of the elements on a diagonal parallel to the primary diagonal given a name?
At least one use of \operatorname{tr_p}() is familiar.
If D is a distance matrix, a permutation of indices of D which minimizes \operatorname{tr_1}(D) is a solution of the traveling salesman problem.  Regards, ... PeterEasthope (talk) 02:52, 7 October 2019 (UTC)
thread_title:  Paratrace? 

message: The claim: "the space of linear operators on a finite-dimensional vector space  (defined over the field ) is isomorphic to the space  is not correct: the space of linear operators on   has dimension n^2 while the space  has dimension 2n.
Clearly that map between the two spaces is not surjective.
: You are incorrect. The space  has dimension n^2. Gumshoe2 (talk) 02:36, 27 January 2022 (UTC)
thread_title:  Coordinate free definition (2) 

message: In the article, cyclicity of trace is shown by the equation
Why are 4 matrices used here as opposed to 3 (say just A, B, C)? Using three matrices would demonstrate perfectly well that trace is invariant under cycling. Zephyr the west wind (talk) 10:43, 11 July 2022 (UTC)
thread_title:  Cyclic property of trace: why 4 matrices instead of 3? 

message: The statement "It can also be proven that tr(AB) = tr(BA) for any two matrices A and B." is incorrect. This equality holds only for matrices with compatible structure so that both AB and BA are defined. If A is an m x n matrix, B is an n x p matrix and m /= p, the BA product is not defined, and the trace is not defined for AB either.
The statement can be rectified by either
a) stating that A and B are m x n and n x m matrices respectively (as below, but this is confusing, it took me a while to notice this special structure in the text)
b) equivalently, stating that AB product is a square matrix (then BA is also square)
c) stating that "tr(AB) = tr(BA) for any two matrices A and B for which the trace is defined." (implying (b)).
d) Using the "Frobenius product" article notation: "tr(ATB) = tr(BAT) for any two m x n matrices A and B"
Which would be the most clear? Pterodaktilis (talk) 07:01, 22 July 2023 (UTC)
thread_title:  Trace of AB and BA matrices