Talk:Determinant/Archive 3

Archive 1

Archive 2

Archive 3

Condense the lead

I propose a shorter, simpler, more concise, more user-friendly, more informative lead:

In linear algebra, the determinant is a useful value that can be computed from the elements of a square matrix. The determinant of a matrix A is denoted det(A), det A, or |A|.

In the case of a 2x2 matrix, the specific formula for the determinant is simply the upper left element times the lower right element, minus the product of the other two elements. Similarly, suppose we have a 3x3 matrix A, and we want the specific formula for its determinant |A|:

|A| = ${\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}$ = a${\begin{vmatrix}e&f\\h&i\end{vmatrix}}$-b${\begin{vmatrix}d&f\\g&i\end{vmatrix}}$+c${\begin{vmatrix}d&e\\g&h\end{vmatrix}}$ = aei+bfg+cdh-ceg-bdi-afh.

Each of the 2x2 determinants in this equation is called a "minor". The same sort of procedure can be used to find the determinant of a 4x4 matrix, and so forth.

Any matrix has a unique inverse if its determinant is nonzero. Various properties can be proved, including that the determinant of a product of matrices is always equal to the product of determinants; and, the determinant of a Hermetian matrix is always real.

Determinants occur throughout mathematics. For example, a matrix is often used to represent the coefficients in a system of linear equations, and the determinant is used to solve those equations. The use of determinants in calculus includes the Jacobian determinant in the substitution rule for integrals of functions of several variables. Determinants are also used to define the characteristic polynomial of a matrix, which is essential for eigenvalue problems in linear algebra. Sometimes, determinants are used merely as a compact notation for expressions that would otherwise be unwieldy to write down.

Of course, wikilinks would be included as appropriate.Anythingyouwant (talk) 05:16, 3 May 2015 (UTC)

I went ahead and installed this.Anythingyouwant (talk) 19:46, 3 May 2015 (UTC)

But what is it?

The lead should summarize what a determinant is, but does not: the determinant is a useful value that can be computed from the elements of a square matrix.

This is equivalent to π is a useful number in geometry.

Is there a better concise description? 15.234.216.87 (talk) 21:00, 26 June 2015 (UTC)

agreed - I came here to find out what it /represented/ conceptually, and I still don't know 92.24.197.193 (talk) 18:17, 18 October 2015 (UTC)

Part of the problem is that there isn't a simple answer: it depends on the underlying number system (field or ring). When the matrix entries are from the real numbers, the determinant is a real number that equals the scale factor by which a unit volume (or area or n-volume) in Rⁿ will be transformed by the matrix (i.e., by the transformation represented by the matrix). (In addition, a negative value indicates that the matrix reverses orientation.) That's why matrices with determinant = 0 don't have inverses: unit volumes get collapsed by (at least) one dimension, and that can't be reversed by a linear transformation. But when the numbers come from, say, the complex numbers, what the determinant means is harder to describe. (And besides, the geometric interpretation above is not necessarily important in the application at hand; e.g., to solve a system of linear equations, all you really care about is whether the determinant is non-zero, not precisely how the coefficient matrix transforms the solution space.) -- Elphion (talk) 22:24, 18 October 2015 (UTC)

Ah great! The scale factor explanation at least gives me a mental picture of something I can relate to, and make obvious the corollary about matrices of determinant 0 lacking inverses. And the "care about [...] whether the determinant is non-zero" is comparable to caring merely whether the determinant of a quadratic is positive or not. For me this is a great start to an answer - thanks! 92.25.0.45 (talk) 21:02, 20 October 2015 (UTC)

determinant expansion problem

Regarding this edit, I question the statement "The sums and the expansion of the powers of the polynomials involved only need to go up to n instead of ∞, since the determinant cannot exceed O(Aⁿ)." The expression O(Aⁿ) has no standard defined meaning that I'm aware of. The O() notation is not usually defined for matrix arguments. You can't fix the notational problem by writing instead that det(kA)=O(kⁿ) since the inner infinite sum becomes divergent if k gets too large. In fact the truncated version is true for all A but the original is not true except in some unclear formal sense. One cannot prove a true statement from a false one, so the attempt to do so is futile. McKay (talk) 05:57, 10 March 2016 (UTC)

I fear you may be overthinking it... All the formula does is write down Jacobi's formula det (I+A) = exp(tr log(I+A)) in terms of formal series expansions for matrix functions, which one always does! ... unless you had a better way, maybe using matrix norms, etc... But, in the lore of the Cayley-Hamilton theorem, one always counts matrix elements, powers of matrices, traces, etc... as a given power, just to keep track, using your scaling variable k, and few mistakes are made. Fine, take the O(n) notation to be on all matrix elements: that includes powers of A! The crucial point is that this is an algorithm for getting the determinant quickly, with the instruction to cut off the calculation at order n, as anything past that automatically vanishes, by the Cayley-Hamilton theorem, "as it should". Check the third order for 2x2 matrices to see exactly how it works. If you had a slicker way of putting it, fine, but just cutting off the sums at finite order won't do, since they mix orders. You have to cut everything consistently at O(n). Perhaps you could add a footnote, with caveats on formal expansions, etc... which however would not really tell a pierful of fishermen that properly speaking there is a proof fish don't exist, a cultural glitch applications-minded people often have to endure! The "as it should" part reminds the reader he need not check the Cayley-Hamilton magic of suppressing higher orders--it must work. But, yes, it is only an algorithm, as most of its users have always understood. Cuzkatzimhut (talk) 11:58, 10 March 2016 (UTC)

physical meaning of determinant

Let suppose if we have a vector which make 2x2 matrix and if we take its determinant then this value will represent its magnitude or something else? What are the meaning of determinant? Please explain it using vectors Muzammalsafdar (talk) 15:55, 19 April 2016 (UTC)

Product of scaling factors (eigenvalues) of its eigenvectors. May wish to go to eigenvalue and eigenvector article. This is the wrong place. Here, the physical connection to areas and volumes is expounded and illustrated in sections 1.1, 1.2, 8.3. Cuzkatzimhut (talk) 19:31, 19 April 2016 (UTC)

Simple proof of inequality $\operatorname {tr} (I-A^{-1})\leq \ln \det(A)\leq \operatorname {tr} (A-I)$

Let λ_i are positive eigenvalues of A. Inequality 1 - 1/λ_i ≤ ln(λ_i) ≤ λ_i - 1 is known from standard courses in math. By taking the sum over i we obtain Σ_i(1 - 1/λ_i) ≤ ln(Π_iλ_i) ≤ Σ_i(λ_i - 1). In terms of trace and determinant functions, tr(I - Λ^-1) ≤ ln(det(Λ)) ≤ tr(Λ - I), where Λ = diag(λ₁,λ₂,...,λ_n). Substituting Λ = UAU^-1 and eliminating U, we obtain the inequality tr(I - A^-1) ≤ ln(det(A)) ≤ tr(A - I). Trompedo (talk) 12:00, 17 July 2016 (UTC)

You may wish to insist on strictly positive-definite matrix A. Cuzkatzimhut (talk) 14:45, 17 July 2016 (UTC)

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Replacing false claim

I am replacing this:

An important arbitrary dimension n identity can be obtained from the Mercator series expansion of the logarithm when $A\in B(0,1)$

${\begin{aligned}\det(I+A)=\sum _{k=0}^{\infty }{\frac {1}{k!}}\left(-\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}\operatorname {tr} (A^{j})\right)^{k}\,,\end{aligned}}$

where I is the identity matrix. The sums and the expansion of the powers of the polynomials involved only need to go up to n instead of ∞, since the determinant cannot exceed O(Aⁿ).

The main problem is the last sentence, which is simply wrong. It is only necessary to try a random 2x2 matrix to see that it does not hold. The reason given is meaningless as far as I can tell. In case the reasoning is that convergent power series in an nxn matrix are equal to a polynomial of degree n in that matrix: yes, but you get the polynomial by factoring by the minimal polynomial of the matrix, not by truncating the power series. McKay (talk) 01:57, 20 January 2017 (UTC)

You are right that the the last sentence is clumsy, but you threw out the baby with the bathwater. Insert a parameter s in front of A: the left-hand side is a polynomial of order sⁿ, and so must be the right hand side. By the Cayley–Hamilton theorem all orders > n on the right-hand side must vanish, and so can be discarded. Set s=1. The argument is presented properly on the C-H thm page, but was badly mangled here. Perhaps you may restore an echo of it. Cuzkatzimhut (talk) 14:47, 20 January 2017 (UTC)

Decomposition Methods

The article in general seems to be packed full of overly complicated ways of doing simple things. For instance the Decomposition Methods section fails to mention Gaussian Elimination (yes, it is mentioned in the Further Methods section). It makes no sense to me (so either I don't understand or the section is simply wrong) that det(A) = e*det(L)*det(U) given that det(A) = det(L') = det(U') for the upper triangular matrix U' and the lower triangular matrix L' that are EASILY arrived at (order O ~ n^3) by Gaussian Elimination. In other words, given how easy and efficient Gaussian Elimination is, LU Decomposition, etc. should be clearly justified here, and it is (they are) not. I also note the Wiki-article on Gaussian Elimination claims right up front that it operates on Rows of a matrix. This is just WRONG, it can operate on either Rows or Columns and if memory serves me (it may not) it can operate on both. Restricting discussion to row reduction is nonsense. Anyway, my main point is why should LU decomposition be mentioned here if it is MORE EXPENSIVE than Gaussian Elimination? Before discussing its details, that needs to be addressed - since as far as I can see it is only useful for reasons other than calculation of the determinant of a matrix.98.21.212.196 (talk) 22:55, 27 May 2017 (UTC)

-- umm, dude, the usual LU decomposition *is* gaussian elimination - U is the resulting row echelon form, L contains the sequence of pivot multipliers used during the elimination procedure. Which, btw, is generally done on rows because the RHS of a set of simultaneous equations - used to augment a matrix to solve those equations, which was the whole *point* of gaussian elimination - adds one or more additional columns (not rows), so the corresponding elimination procedure must act on rows. — Preceding unsigned comment added by 174.24.230.124 (talk) 06:26, 1 June 2017 (UTC)

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Determinant inequalities

There is a problem with the determinant inequalities in the properties with determinants section. If we take the determinant of a commutator

$XY-YX\in {\mathfrak {gl}}(2)$

then

$\det(XY-YX)\geq \det(XY)+(-1)^{2}\det(YX)$

But if $Y=Id$ and $X$ has a non-zero determinant, there is a contradiction. — Preceding unsigned comment added by Username6330 (talk • contribs) 21:56, 10 November 2017 (UTC)

Geometric explanation for 2x2 matrices

I would like to precise only the following part of the explanation:

"yet it may be expressed more conveniently using the cosine of the complementary angle to a perpendicular vector, e.g. a^⊥ = (-b, a), such that |a^⊥||b|cosθ' , which can be determined by the pattern of the scalar product to be equal to ad − bc"

As far as I understand it, for the scalar product to be ad − bc the vector a^⊥ must be (-b, a). So I would change "to a perpendicular vector, e.g." by "to the perpendicular vector". Actually, would that be the case, the module of a and a^⊥ are the same, so the final expression could be better written in terms of a. Happy to do it, I would like just to confirm. Conjugado (talk) 00:52, 31 January 2018 (UTC)

Determinant identities

I suggest merging Determinant identities into this article. There's not enough there to justify a separate article. The original article creator was blocked a few times for creating inappropriate article fragments in the mainspace. -Apocheir (talk) 21:26, 8 April 2018 (UTC)

I would be in mild, perichiral, agreement. Cuzkatzimhut (talk) 21:30, 8 April 2018 (UTC)

• Support. Most there is already here, but the remainder could easily be accommodated. –Deacon Vorbis (carbon • videos) 20:06, 22 April 2018 (UTC)
• Yes, for now. At the moment Determinant identities has very little, but it could be expanded. There are lots of determinant identities out there which could make for a more useful page. McKay (talk) 05:05, 23 April 2018 (UTC)
• Yes, why not?. It seems odd that Determinant identities even exists as a separate page. LudwikSzymonJaniuk (talk) 18:43, 7 May 2018 (UTC)
• Support. P.Shiladitya^✍talk 19:03, 19 May 2018 (UTC)

Computationality

The page says; although other methods of solution are much more computationally efficient.

No example is given HumbleBeauty (talk) 09:23, 3 June 2019 (UTC)

Permutations

Is this already in it? - I couldn't find it.

Let S_n be the set of all permutations of {1,..,n}

Then the elements of a permutation k, when applied to the rows of a determinant (e.g.2314 means 2nd element in row 1, 3rd in row 2. 1st row 3, 4th row 4.

Multiply elements together and by odd/even -> -/+1, and add them all together to get the determinant

Darcourse (talk) 15:03, 21 May 2020 (UTC)

Your description is obscure, but seems to refer to the formula given at the beginning of section Determinant#n × n matrices. D.Lazard (talk) 15:50, 21 May 2020 (UTC)

Found it1 Thanks. @dl

Darcourse (talk) 02:03, 22 May 2020 (UTC)

3x3 column vectors

It would be very useful to simply state in the 3x3 section that when a 3x3 matrix is given by its column vectors A = [ a | b | c] then det(A) = a^T (b x c). — Preceding unsigned comment added by 2A00:23C6:5486:4A00:7DEB:F569:CA90:8032 (talk) 11:52, 21 June 2020 (UTC)

Determinants in "The Nine Chapters"

I have read the russian translation by Berezkina and found nothing that can be considered as finding the determinant. In the 8th "book" gaussian elimination is considered, but the question of existance of solution is not discussed. In case there is such topic in the book (maybe I missed it or maybe the translation is flawed), it would be helpful to have a citation and direct the reader to particluar problem in the book. Medvednikita (talk) 13:21, 16 October 2020 (UTC)

Interesting. I don't have access to this, but maybe someone else can weigh in? I'd also question the leap from "system is (in)consistent" to full-blown "determinant" for anyone, as this section seems to be doing. –Deacon Vorbis (carbon • videos) 13:34, 16 October 2020 (UTC)

Illustration of "Sarrus' scheme" (for 3x3 matrices) - why is it missing a "d"?

Is there a reason why this illustration is missing the letter "d" - i.e. why the matrix elements are listed as "a b c e f g h i j", rather than "a b c d e f g h i" (as in the preceding two illustrations for the Laplace and Leibniz formulas)? (If the "d" is included, then the resulting formula would come out to be identical to the Leibniz formula, as one would expect.) — Preceding unsigned comment added by PatricKiwi (talk • contribs) 09:45, 26 February 2021 (UTC)

Using a single style for the design of formulas

In this article, three styles of design of mathematical expressions intersect at once ({{math|...}}, {{math|''...''}}, ''...'' and <math>...</math>; e.g., for 'n': n, n, n and $n$). This is very noticeable when styles are mixed within a paragraph, sentence. The suggestion is to bring all formulas to any one style, for example, <math>...</math>. — Preceding unsigned comment added by Alexey Ismagilov (talk • contribs) 12:18, 17 October 2021 (UTC)

Please, do not forget to sign your contributions in talk page with four tildes (~~~~).

This has been discuted many times and the consensus is summarized in MOS:FORMULA. Since the last version of this manual of style, the rendering of <math> </math> has been improved, and the consensus has slightly evolved. Namely, it is recommended to replace raw html (''n'' and n) by {{math|''n''}} or {{mvar|n}} (these are equivalent for isolated variables), under the condition of doing this in a whole article, or at least in a whole section. The change from {{math|}} to <math></math> is recommended for formulas that contain special characters (see MOS:BBB, for example). Otherwise, the preference of preceding editors must be retain per MOS:VAR. D.Lazard (talk) 11:00, 17 October 2021 (UTC)

According to the article you are referring to, it is said that the community has not come to a single agreement on the style of formulas. However, display formulas have to be written in LaTeX notation. Therefore, for inline formulas, it seems to me, it is also worth using LaTeX in order to adhere to the uniformity of the design. I agree that the edits that I made to the article and which were later canceled by the user D.Lazard contain several bad changes. I apologize for these changes. Expressions do not contain special characters (MOS:BBB), so I think it's better not to touch anything, according to MOS:VAR. Alexey Ismagilov (talk) 12:26, 17 October 2021 (UTC)

Multiplication in determinants is not commutative

Am I wrong in believing many of the equations presented in this article are incorrect? Ex. the very first equation

${\begin{aligned}|A|={\begin{vmatrix}a&b\\c&d\end{vmatrix}}=ad-bc.\end{aligned}}$

should be written as

${\begin{aligned}|A|={\begin{vmatrix}a&b\\c&d\end{vmatrix}}=ad-cb.\end{aligned}}$

The original equation works fine if all of the terms are just single numbers but does not work in some cases when they are vectors. Below is an identity seen in semidefinite programming, where $x$ is a vector

${\begin{aligned}{\begin{bmatrix}1&x^{T}\\x&xx^{T}\end{bmatrix}}\succeq 0.\end{aligned}}$

Since the matrix is positive semidefinite, the determinant of the matrix must be nonnegative. Using the original equation presented,

${\begin{aligned}{\begin{vmatrix}1&x^{T}\\x&xx^{T}\end{vmatrix}}=1*xx^{T}-x^{T}*x.\end{aligned}}$

This is impossible to evaluate since the dimensions on $x^{T}*x$ do not match. instead using the second equation,

${\begin{aligned}{\begin{vmatrix}1&x^{T}\\x&xx^{T}\end{vmatrix}}=1*xx^{T}-x*x^{T}=xx^{T}-xx^{T}=0.\end{aligned}}$

Which is possible to evaluate

--Karsonkevin2 (talk) 00:16, 7 December 2021 (UTC)

This looks like an anomalous cancellation. In any case, the last formula is not well formed, as expressing the equality of a scalar (a determinant) and a non-scalar matrix.

Moreover, it is explicitly said in § Definition that commutativity is required for multiplication of matrix entries. So, in the 2x2 case, $cb=bc.$ Therefore, there is no reason to not follow the alphabetical order. D.Lazard (talk) 07:50, 7 December 2021 (UTC)