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Trace (linear algebra)1Trace (linear algebra)In linear algebra, the trace of an n-by-n square matrix A is defined to be the sum of the elements on the maindiagonal (the diagonal from the upper left to the lower right) of A, i.e.,where aii

represents the entry on the ith row and ith column of A. Equivalently, the trace of a matrix is the sum of itseigenvalues, making it an invariant with respect to a change of basis. This characterization can be used to define thetrace for a linear operator in general. Note that the trace is only defined for a square matrix (i.e. n×n).Geometrically, the trace can be interpreted as the infinitesimal change in volume (as the derivative of thedeterminant), which is made precise in Jacobi's use of the term trace arises from the German term Spur (cognate with the English spoor), which, as a functionin mathematics, is often abbreviated to "Sp".ExamplesLet T be a linear operator represented by the matrixThen tr(T) = −2 + 1 − 1 = − trace of the identity matrix is the dimension of the space; this leads to generalizations of dimension using trace of a projection (i.e., P2 = P) is the rank of the projection. The trace of a nilpotent matrix is zero. Theproduct of a symmetric matrix and a skew-symmetric matrix has zero generally, if f(x) = (x − λ1)d1···(x − λk)dk is the characteristic polynomial of a matrix A, thenIf A and B are positive semi-definite matrices of the same order then[1]PropertiesThe trace is a linear map. That is,for all square matrices A and B, and all scalars A is an m×n matrix and B is an n×m matrix, then[2]Conversely, the above properties characterize the trace completely in the sense as follows. Let be a linearfunctional on the space of square matrices satisfying . Then and tr are proportional.[3]The trace is similarity-invariant, which means that A and P−1AP have the same trace. This is becauseA matrix and its transpose have the same trace:.

Trace (linear algebra)Let A be a symmetric matrix, and B an anti-symmetric matrix. both A and B are n by n, the trace of the (ring-theoretic) commutator of A and B vanishes: tr([A, B]) = 0; onecan state this as "the trace is a map of Lie algebras from operators to scalars", as the commutator ofscalars is trivial (it is an abelian Lie algebra). In particular, using similarity invariance, it follows that the identitymatrix is never similar to the commutator of any pair of sely, any square matrix with zero trace is the commutator of some pair of matrices.[4]

Moreover, any squarematrix with zero trace is unitarily equivalent to a square matrix with diagonal consisting of all trace of any power of a nilpotent matrix is zero. When the characteristic of the base field is zero, the conversealso holds: if for all , then is that order does matter in taking traces: in general,2In other words, we can only interchange the two halves of the expression, albeit repeatedly. This means that the traceis invariant under cyclic permutations, i.e.,However, if products of three symmetric matrices are considered, any permutation is allowed. (Proof: tr(ABC) =tr(AT

BT

CT) = tr((CBA)T) = tr(CBA).) For more than three factors this is not true. This is known as the the determinant, the trace of the product is not the product of traces. What is true is that the trace of the tensorproduct of two matrices is the product of their traces:The trace of a product can be rewritten as the sum of all elements from a Hadamard product (entry-wise product):.This should be more computationally efficient, since the matrix product of an matrix with an one(first and last dimensions must match to give a square matrix for the trace) has

multiplications followed by lications andadditions, whereas the computation of the Hadamard version (entry-wise product) requires onlyThe exponential traceExpressions like , where A is a square matrix, occur so often in some fields (e.g. multivariatestatistical theory), that a shorthand notation has become common:This is sometimes referred to as the exponential trace of a linear operatorGiven some linear map f : V → V (V is a finite-dimensional vector space) generally, we can define the trace of thismap by considering the trace of matrix representation of f, that is, choosing a basis for V and describing f as a matrixrelative to this basis, and taking the trace of this square matrix. The result will not depend on the basis chosen, sincedifferent bases will give rise to similar matrices, allowing for the possibility of a basis independent definition for thetrace of a linear a definition can be given using the canonical isomorphism between the space End(V) of linear maps on V and

V⊗V*, where V*

is the dual space of V. Let v be in V and let f be in V*. Then the trace of the decomposable element

Trace (linear algebra)v⊗f is defined to be f(v); the trace of a general element is defined by linearity. Using an explicit basis for V and thecorresponding dual basis for V*, one can show that this gives the same definition of the trace as given above.3Eigenvalue relationshipsIf A is a square n-by-n matrix with real or complex entries and if λ1,...,λn

are the (complex and distinct) eigenvaluesof A (listed according to their algebraic multiplicities), thenThis follows from the fact that A is always similar to its Jordan form, an upper triangular matrix having λ1,...,λn

onthe main diagonal. In contrast, the determinant of is the product of its eigenvalues; i.e.,More generally,DerivativesThe trace is the derivative of the determinant: it is the Lie algebra analog of the (Lie group) map of the is made precise in Jacobi's formula for the derivative of the determinant (see under determinant). As a particularcase, : the trace is the derivative of the determinant at the identity. From this (or from the connectionbetween the trace and the eigenvalues), one can derive a connection between the trace function, the exponential mapbetween a Lie algebra and its Lie group (or concretely, the matrix exponential function), and the determinant:det(exp(A)) = exp(tr(A)).For example, consider the one-parameter family of linear transformations given by rotation through angle θ,These transformations all have determinant 1, so they preserve area. The derivative of this family at θ = 0 is theantisymmetric matrixwhich clearly has trace zero, indicating that this matrix represents an infinitesimal transformation which preservesarea.A related characterization of the trace applies to linear vector fields. Given a matrix A, define a vector field F on Rnby F(x) = Ax. The components of this vector field are linear functions (given by the rows of A). The divergence divF is a constant function, whose value is equal to tr(A). By the divergence theorem, one can interpret this in terms offlows: if F(x) represents the velocity of a fluid at the location x, and U is a region in Rn, the net flow of the fluid outof U is given by tr(A)· vol(U), where vol(U) is the volume of trace is a linear operator, hence its derivative is constant:

Trace (linear algebra)4ApplicationsThe trace is used to define characters of group representations. Two representations

group are equivalent (up to change of basis on ) if

The trace also plays a central role in the distribution of quadratic all .of aLie algebraThe trace is a map of Lie algebras from the Lie algebra gln

of operators on a n-dimensional space (matrices) to the Lie algebra k of scalars; as k is abelian (the Lie bracket vanishes), the fact that this is a mapof Lie algebras is exactly the statement that the trace of a bracket vanishes:

The kernel of this map, a matrix whose trace is zero, is said to be traceless or tracefree, and these matrices form thesimple Lie algebra sln, which is the Lie algebra of the special linear group of matrices with determinant 1. Thespecial linear group consists of the matrices which do not change volume, while the special linear algebra is thematrices which infinitesimally do not change fact, there is a internal direct sum decomposition of operators/matrices into tracelessoperators/matrices and scalars operators/matrices. The projection map onto scalar operators can be expressed interms of the trace, concretely as:Formally, one can compose the trace (the counit map) with the unit map

obtain a map

yielding the formula terms of short exact sequences, one hasof "inclusion of scalars" tomapping onto scalars, and multiplying by n. Dividing by n makes this a projection,which is analogous tofor Lie groups. However, the trace splits naturally (via times scalars) so but the splitting of thedeterminant would be as the nth root times scalars, and this does not in general define a function, so the determinantdoes not split and the general linear group does not decompose:

Bilinear formsThe bilinear formis called the Killing form, which is used for the classification of Lie trace defines a bilinear form:(x, y square matrices).The form is symmetric, non-degenerate[5]

and associative in the sense that:In a simple Lie algebra (e.g.,

Killing form.), every such bilinear form is proportional to each other; in particular, to theTwo matrices x and y are said to be trace orthogonal if

Trace (linear algebra)5Inner productFor an m-by-n matrix A with complex (or real) entries and

*

being the conjugate transpose, we havewith equality if and only if A = 0. The assignmentyields an inner product on the space of all complex (or real) m-by-n norm induced by the above inner product is called the Frobenius norm. Indeed it is simply the Euclidean norm ifthe matrix is considered as a vector of length lizationThe concept of trace of a matrix is generalised to the trace class of compact operators on Hilbert spaces, and theanalog of the Frobenius norm is called the Hilbert-Schmidt partial trace is another generalization of the trace that is A is a general associative algebra over a field k, then a trace on A is often defined to be any map tr: A → k whichvanishes on commutators: tr([a, b]) = 0 for all a, b in A. Such a trace is not uniquely defined; it can always at least bemodified by multiplication by a nonzero scalar.A supertrace is the generalization of a trace to the setting of operation of tensor contraction generalizes the trace to arbitrary nate-free definitionWe can identify the space of linear operators on a vector space V with the space

. We also have a canonical bilinear function

applying an element of to an element of to get an element of

, wherethat consists ofin symbolsThis induces a linear function on the tensor product (by its universal property)which, as it turns out, when that tensor product is viewed as the space of operators, is equal tothe also clarifies why

may interpret the composition map

coming from the pairing

and why

ason the middle terms. Taking the trace of the product then comes fromas composition of operatorsone(multiplication of matrices) and trace can be interpreted as the same pairing. Viewing

pairing on the outer terms, while taking the product in the opposite order and then taking the trace just switcheswhich pairing is applied first. On the other hand, taking the trace of A and the trace of B corresponds to applying thepairing on the left terms and on the right terms (rather than on inner and outer), and is thus coordinates, this corresponds to indexes: multiplication is given by

and

which is V finite-dimensional, with basis and dual basis

operator with respect to that basis. Any operator

defined as above,

otherwise. This shows that

which is the same, while, then is the entry of the matrix of the. With

sois therefore a sum of the form

. The latter, however, is just the Kronecker delta, being 1 if i=j and 0is simply the sum of the coefficients along the diagonal. This method, however,makes coordinate invariance an immediate consequence of the definition.

Trace (linear algebra)6DualFurther, one may dualize this map, obtaining a map

inclusion of scalars, sending

One can then compose these,

dimension of the vector s are the unit, while trace is the yields multiplication by n, as the trace of the identity is theThis map is precisely theto the identity matrix: "trace is dual to scalars". In the language of bialgebras,Notes[1]Can be proven with the Cauchy-Schwarz inequality.[2]This is immediate from the definition of matrix multiplication.[3]Proof:if and only if

and thusand (with the standard basis ),.More abstractly, this corresponds to the decomposition as tr(AB)=tr(BA) (equivalently,) defines the trace on sln, which has complement the scalar matrices, and leaves one degree offreedom: any such map is determined by its value on scalars, which is one scalar parameter and hence all aremultiple of the trace, a non-zero such map.[4]Proof: is a semisimple Lie algebra and thus every element in it is the commutator of some pair of elements, otherwise the derivedif and only if

algebra would be a proper ideal.[5]This follows from the fact that

Article Sources and Contributors7Article Sources and ContributorsTrace (linear algebra) Source: /w/?oldid=402560193 Contributors: Achab, Adiel, Aetheling, Algebraist, Archelon, AxelBoldt, BenFrantzDale, Berland,Btyner, CYD, Calc rulz, CattleGirl, Charles Matthews, Chochopk, Dima373, Dysprosia, Edinborgarstefan, Egriffin, Email4mobile, Eranb, Eric Olson, Fropuff, Ged.R, Giftlite, Haseldon,JabberWok, Japanese Searobin, Jewbacca, Jshadias, Kaarebrandt, Kan8eDie, Katzmik, Keyi, Kruusamägi, Ksyrie, Laurentius, Lethe, Lukpank, MER-C, MarSch, Mct mht, Melchoir, MichaelHardy, Mon4, Nathanielvirgo, Nbarth, Nineteen O'Clock, Octahedron80, Oleg Alexandrov, Oyz, Phe, Phys, Pokipsy76, Pt, Robinh, Salgueiro, Saretakis, Sciyoshi, Spiel496, Spireguy,StradivariusTV, j, TakuyaMurata, Tarquin, Tercer, Tffff, Thehotelambush, Tsirel, V1adis1av, Whaa?, WhiteHatLurker, Wpoely86, Wshun, 84 anonymous editsLicenseCreative Commons Attribution-Share Alike 3.0 Unported/licenses/by-sa/3.0/