# Talk:Determinant

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## Right handed coordinante

the following sentence is not clear. "The determinant of a set of vectors is positive if the vectors form a right-handed coordinate system, and negative if left-handed." what does "right-handed coordinate system" means? the "coordinate system" article does not mention it. amit man

### types of matrices

special linear group, special orthogonal group, special unitary group, indefinite special orthogonal group, modular group, unimodular matrix, matrices with multidimensional indices

### geometry, analysis

conformal map?, Gauss curvature, orientability, Integration by substitution, Wronskian, invariant theory, Monge–Ampère equation, Brascamp–Lieb_inequality, Liouville's formula, absolute value of cx numbers and quaternions (see 3-sphere), distance geometry (Cayley–Menger determinant), Delaunay_triangulation

### examples

reflection matrix, Rotation matrix, Vandermonde matrix, Circulant matrix, Hessian matrix (Blob_detection#The_determinant_of_the_Hessian), block matrix, Gram determinant, Elementary_matrix, Orr–Sommerfeld_equation, det of Cartan matrix

### generalizations

Hyperdeterminant, Quasideterminant, Continuant (mathematics), Immanant of a matrix, permanent, Pseudo-determinant, det's of infinite matrices / regularized det / functional determinant (see also operator theory), Fredholm determinant, superdeterminant

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## 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 ${\displaystyle \operatorname {tr} (I-A^{-1})\leq \ln \det(A)\leq \operatorname {tr} (A-I)}$

Let λi are positive eigenvalues of A. Inequality 1 - 1/λi ≤ ln(λi) ≤ λi - 1 is known from standard courses in math. By taking the sum over i we obtain Σi(1 - 1/λi) ≤ ln(Πiλi) ≤ Σii - 1). In terms of trace and determinant functions, tr(I - Λ-1) ≤ ln(det(Λ)) ≤ tr(Λ - I), where Λ = diag(λ12,...,λn). 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)