Trace (linear algebra)

Inlinear algebra,thetraceof asquare matrix $A$ ,denoted $tr(A)$ ,^[1]is defined to be the sum of elements on themain diagonal(from the upper left to the lower right) of $A$ .The trace is only defined for a square matrix ( $n \times n$ ).

In mathematical physics texts, if $tr(A)$ = 0 then the matrix is said to betraceless.This is a misnomer, but widely used, such as in thePauli Matrices.

It can be proven that the trace of a matrix is the sum of itseigenvalues(counted with multiplicities). It can also be proven that $tr(AB) = tr(BA)$ for any two matrices $A$ and $B$ of appropriate sizes. This implies thatsimilar matriceshave the same trace. As a consequence one can define the trace of alinear operatormapping a finite-dimensionalvector spaceinto itself, since all matrices describing such an operator with respect to a basis are similar.

The trace is related to the derivative of thedeterminant(seeJacobi's formula).

Definition

Thetraceof an $n \times n$ square matrix $A$ is defined as^[1]^[2]^[3]^: 34 $\operatorname {tr} (\mathbf {A} )=\sum _{i=1}^{n}a_{ii}=a_{11}+a_{22}+\dots +a_{nn}$ where $a ii$ denotes the entry on the $i$ throw and $i$ thcolumn of $A$ .The entries of $A$ can bereal numbers,complex numbers,or more generally elements of afield $F$ .The trace is not defined for non-square matrices.

Example

Let $A$ be a matrix, with $\mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}={\begin{pmatrix}1&0&3\\11&5&2\\6&12&-5\end{pmatrix}}$

Then $\operatorname {tr} (\mathbf {A} )=\sum _{i=1}^{3}a_{ii}=a_{11}+a_{22}+a_{33}=1+5+(-5)=1$

Properties

Basic properties

The trace is alinear mapping.That is,^[1]^[2] ${\begin{aligned}\operatorname {tr} (\mathbf {A} +\mathbf {B} )&=\operatorname {tr} (\mathbf {A} )+\operatorname {tr} (\mathbf {B} )\\\operatorname {tr} (c\mathbf {A} )&=c\operatorname {tr} (\mathbf {A} )\end{aligned}}$ for all square matrices $A$ and $B$ ,and allscalars $c$ .^[3]^: 34

A matrix and itstransposehave the same trace:^[1]^[2]^[3]^: 34 $\operatorname {tr} (\mathbf {A} )=\operatorname {tr} \left(\mathbf {A} ^{\mathsf {T}}\right).$

This follows immediately from the fact that transposing a square matrix does not affect elements along the main diagonal.

Trace of a product

The trace of a square matrix which is the product of two matrices can be rewritten as the sum of entry-wise products of their elements, i.e. as the sum of all elements of theirHadamard product.Phrased directly, if $A$ and $B$ are two $m \times n$ matrices, then: $\operatorname {tr} \left(\mathbf {A} ^{\mathsf {T}}\mathbf {B} \right)=\operatorname {tr} \left(\mathbf {A} \mathbf {B} ^{\mathsf {T}}\right)=\operatorname {tr} \left(\mathbf {B} ^{\mathsf {T}}\mathbf {A} \right)=\operatorname {tr} \left(\mathbf {B} \mathbf {A} ^{\mathsf {T}}\right)=\sum _{i=1}^{m}\sum _{j=1}^{n}a_{ij}b_{ij}\;.$

If one views any real $m \times n$ matrix as a vector of length $mn$ (an operation calledvectorization) then the above operation on $A$ and $B$ coincides with the standarddot product.According to the above expression, $tr(A ⊤ A)$ is a sum of squares and hence is nonnegative, equal to zero if and only if $A$ is zero.^[4]^: 7Furthermore, as noted in the above formula, $tr(A ⊤ B) = tr(B ⊤ A)$ .These demonstrate the positive-definiteness and symmetry required of aninner product;it is common to call $tr(A ⊤ B)$ theFrobenius inner productof $A$ and $B$ .This is a natural inner product on thevector spaceof all real matrices of fixed dimensions. Thenormderived from this inner product is called theFrobenius norm,and it satisfies a submultiplicative property, as can be proven with theCauchy–Schwarz inequality: $0\leq \left[\operatorname {tr} (\mathbf {A} \mathbf {B} )\right]^{2}\leq \operatorname {tr} \left(\mathbf {A} ^{2}\right)\operatorname {tr} \left(\mathbf {B} ^{2}\right)\leq \left[\operatorname {tr} (\mathbf {A} )\right]^{2}\left[\operatorname {tr} (\mathbf {B} )\right]^{2}\,$ if $A$ and $B$ are realpositive semi-definite matricesof the same size. The Frobenius inner product and norm arise frequently inmatrix calculusandstatistics.

The Frobenius inner product may be extended to ahermitian inner producton thecomplex vector spaceof all complex matrices of a fixed size, by replacing $B$ by itscomplex conjugate.

The symmetry of the Frobenius inner product may be phrased more directly as follows: the matrices in the trace of a product can be switched without changing the result. If $A$ and $B$ are $m \times n$ and $n \times m$ real or complex matrices, respectively, then^[1]^[2]^[3]^: 34^{[note 1]}

$\operatorname {tr} (\mathbf {A} \mathbf {B} )=\operatorname {tr} (\mathbf {B} \mathbf {A} )$

This is notable both for the fact that $AB$ does not usually equal $BA$ ,and also since the trace of either does not usually equal $tr(A)tr(B)$ .^{[note 2]}Thesimilarity-invarianceof the trace, meaning that $tr(A) = tr(P -1 AP)$ for any square matrix $A$ and any invertible matrix $P$ of the same dimensions, is a fundamental consequence. This is proved by $\operatorname {tr} \left(\mathbf {P} ^{-1}(\mathbf {A} \mathbf {P} )\right)=\operatorname {tr} \left((\mathbf {A} \mathbf {P} )\mathbf {P} ^{-1}\right)=\operatorname {tr} (\mathbf {A} ).$ Similarity invariance is the crucial property of the trace in order to discuss traces oflinear transformationsas below.

Additionally, for real column vectors $\mathbf {a} \in \mathbb {R} ^{n}$ and $\mathbf {b} \in \mathbb {R} ^{n}$ ,the trace of the outer product is equivalent to the inner product:

$\operatorname {tr} \left(\mathbf {b} \mathbf {a} ^{\textsf {T}}\right)=\mathbf {a} ^{\textsf {T}}\mathbf {b}$

Cyclic property

More generally, the trace isinvariant undercircular shifts,that is,

$\operatorname {tr} (\mathbf {A} \mathbf {B} \mathbf {C} \mathbf {D} )=\operatorname {tr} (\mathbf {B} \mathbf {C} \mathbf {D} \mathbf {A} )=\operatorname {tr} (\mathbf {C} \mathbf {D} \mathbf {A} \mathbf {B} )=\operatorname {tr} (\mathbf {D} \mathbf {A} \mathbf {B} \mathbf {C} ).$

This is known as thecyclic property.

Arbitrary permutations are not allowed: in general, $\operatorname {tr} (\mathbf {A} \mathbf {B} \mathbf {C} )\neq \operatorname {tr} (\mathbf {A} \mathbf {C} \mathbf {B} ).$

However, if products ofthreesymmetricmatrices are considered, any permutation is allowed, since: $\operatorname {tr} (\mathbf {A} \mathbf {B} \mathbf {C} )=\operatorname {tr} \left(\left(\mathbf {A} \mathbf {B} \mathbf {C} \right)^{\mathsf {T}}\right)=\operatorname {tr} (\mathbf {C} \mathbf {B} \mathbf {A} )=\operatorname {tr} (\mathbf {A} \mathbf {C} \mathbf {B} ),$ where the first equality is because the traces of a matrix and its transpose are equal. Note that this is not true in general for more than three factors.

Trace of a Kronecker product

The trace of theKronecker productof two matrices is the product of their traces: $\operatorname {tr} (\mathbf {A} \otimes \mathbf {B} )=\operatorname {tr} (\mathbf {A} )\operatorname {tr} (\mathbf {B} ).$

Characterization of the trace

The following three properties: ${\begin{aligned}\operatorname {tr} (\mathbf {A} +\mathbf {B} )&=\operatorname {tr} (\mathbf {A} )+\operatorname {tr} (\mathbf {B} ),\\\operatorname {tr} (c\mathbf {A} )&=c\operatorname {tr} (\mathbf {A} ),\\\operatorname {tr} (\mathbf {A} \mathbf {B} )&=\operatorname {tr} (\mathbf {B} \mathbf {A} ),\end{aligned}}$ characterize the traceup toa scalar multiple in the following sense: If $f$ is alinear functionalon the space of square matrices that satisfies $f(xy)=f(yx),$ then $f$ and $\operatorname {tr}$ are proportional.^{[note 3]}

For $n\times n$ matrices, imposing the normalization $f(\mathbf {I} )=n$ makes $f$ equal to the trace.

Trace as the sum of eigenvalues

Given any $n \times n$ matrix $A$ ,there is

$\operatorname {tr} (\mathbf {A} )=\sum _{i=1}^{n}\lambda _{i}$

where $λ 1,..., λ n$ are theeigenvaluesof $A$ counted with multiplicity. This holds true even if $A$ is a real matrix and some (or all) of the eigenvalues are complex numbers. This may be regarded as a consequence of the existence of theJordan canonical form,together with the similarity-invariance of the trace discussed above.

Trace of commutator

When both $A$ and $B$ are $n \times n$ matrices, the trace of the (ring-theoretic)commutatorof $A$ and $B$ vanishes: $tr([A, B]) = 0$ ,because $tr(AB) = tr(BA)$ and $tr$ is linear. One can state this as "the trace is a map ofLie algebras $gl n \to k$ from operators to scalars ", as the commutator of scalars is trivial (it is anAbelian Lie algebra). In particular, using similarity invariance, it follows that the identity matrix is never similar to the commutator of any pair of matrices.

Conversely, any square matrix with zero trace is a linear combination of the commutators of pairs of matrices.^{[note 4]}Moreover, any square matrix with zero trace isunitarily equivalentto a square matrix with diagonal consisting of all zeros.

Traces of special kinds of matrices

The trace of the $n \times n$ identity matrixis the dimension of the space, namely $n$ .
$\operatorname {tr} \left(\mathbf {I} _{n}\right)=n$
This leads togeneralizations of dimension using trace.
The trace of aHermitian matrixis real, because the elements on the diagonal are real.
The trace of apermutation matrixis the number offixed pointsof the corresponding permutation, because the diagonal term $a ii$ is 1 if the $i$ th point is fixed and 0 otherwise.
The trace of aprojection matrixis the dimension of the target space. ${\begin{aligned}\mathbf {P} _{\mathbf {X} }&=\mathbf {X} \left(\mathbf {X} ^{\mathsf {T}}\mathbf {X} \right)^{-1}\mathbf {X} ^{\mathsf {T}}\\[3pt]\Longrightarrow \operatorname {tr} \left(\mathbf {P} _{\mathbf {X} }\right)&=\operatorname {rank} (\mathbf {X} ).\end{aligned}}$ The matrix $P X$ is idempotent.
More generally, the trace of anyidempotent matrix,i.e. one with $A 2 = A$ ,equals its ownrank.
The trace of anilpotent matrixis zero.

When the characteristic of the base field is zero, the converse also holds: if $tr(A k) = 0$ for all $k$ ,then $A$ is nilpotent.

When the characteristic $n > 0$ is positive, the identity in $n$ dimensions is a counterexample, as $\operatorname {tr} \left(\mathbf {I} _{n}^{k}\right)=\operatorname {tr} \left(\mathbf {I} _{n}\right)=n\equiv 0$ ,but the identity is not nilpotent.

Relationship to the characteristic polynomial

The trace of an $n\times n$ matrix $A$ is the coefficient of $t^{n-1}$ in thecharacteristic polynomial,possibly changed of sign, according to the convention in the definition of the characteristic polynomial.

Relationship to eigenvalues

If $A$ is a linear operator represented by a square matrix withrealorcomplexentries and if $λ 1,..., λ n$ are theeigenvaluesof $A$ (listed according to theiralgebraic multiplicities), then

$\operatorname {tr} (\mathbf {A} )=\sum _{i}\lambda _{i}$

This follows from the fact that $A$ is alwayssimilarto itsJordan form,an uppertriangular matrixhaving $λ 1,..., λ n$ on the main diagonal. In contrast, thedeterminantof $A$ is theproductof its eigenvalues; that is, $\det(\mathbf {A} )=\prod _{i}\lambda _{i}.$

Everything in the present section applies as well to any square matrix with coefficients in analgebraically closed field.

Derivative relationships

If $ΔA$ is a square matrix with small entries and $I$ denotes theidentity matrix,then we have approximately

$\det(\mathbf {I} +\mathbf {\Delta A} )\approx 1+\operatorname {tr} (\mathbf {\Delta A} ).$

Precisely this means that the trace is thederivativeof thedeterminantfunction at the identity matrix.Jacobi's formula

$d\det(\mathbf {A} )=\operatorname {tr} {\big (}\operatorname {adj} (\mathbf {A} )\cdot d\mathbf {A} {\big )}$

is more general and describes thedifferentialof the determinant at an arbitrary square matrix, in terms of the trace and theadjugateof the matrix.

From this (or from the connection between the trace and the eigenvalues), one can derive a relation between the trace function, thematrix exponentialfunction, and the determinant: $\det(\exp(\mathbf {A} ))=\exp(\operatorname {tr} (\mathbf {A} )).$

A related characterization of the trace applies to linearvector fields.Given a matrix $A$ ,define a vector field $F$ on $R n$ by $F (x) = Ax$ .The components of this vector field are linear functions (given by the rows of $A$ ). Itsdivergence $div F$ is a constant function, whose value is equal to $tr(A)$ .

By thedivergence theorem,one can interpret this in terms of flows: if $F (x)$ represents the velocity of a fluid at location $x$ and $U$ is a region in $R n$ ,thenet flowof the fluid out of $U$ is given by $tr(A) \cdot vol(U)$ ,where $vol(U)$ is thevolumeof $U$ .

The trace is a linear operator, hence it commutes with the derivative: $d\operatorname {tr} (\mathbf {X} )=\operatorname {tr} (d\mathbf {X} ).$

Trace of a linear operator

In general, given some linear map $f : V \to V$ (where $V$ is a finite-dimensional vector space), we can define the trace of this map by considering the trace of amatrix representationof $f$ ,that is, choosing abasisfor $V$ and describing $f$ as a matrix relative to this basis, and taking the trace of this square matrix. The result will not depend on the basis chosen, since different bases will give rise tosimilar matrices,allowing for the possibility of a basis-independent definition for the trace of a linear map.

Such a definition can be given using thecanonical isomorphismbetween the space $End(V)$ of linear maps on $V$ and $V \otimes V *$ ,where $V *$ is thedual spaceof $V$ .Let $v$ be in $V$ and let $g$ be in $V *$ .Then the trace of the indecomposable element $v \otimes g$ is defined to be $g (v)$ ;the trace of a general element is defined by linearity. The trace of a linear map $f : V \to V$ can then be defined as the trace, in the above sense, of the element of $V \otimes V *$ corresponding tofunder the above mentioned canonical isomorphism. Using an explicit basis for $V$ and the corresponding dual basis for $V *$ ,one can show that this gives the same definition of the trace as given above.

Numerical algorithms

Stochastic estimator

The trace can be estimated unbiasedly by "Hutchinson's trick":^[5]

Given any matrix $W\in \mathbb {R} ^{n\times n}$ ,and any random $u\in \mathbb {R} ^{n}$ with $E[uu^{T}]=I$ ,we have $E[u^{T}Wu]=tr(W)$ .(Proof: expand the expectation directly.)

Usually, the random vector is sampled from $N(0,I)$ (normal distribution) or $\{\pm n^{-1/2}\}^{n}$ (Rademacher distribution).

More sophisticated stochastic estimators of trace have been developed.^[6]

Applications

If a 2 x 2 real matrix has zero trace, its square is adiagonal matrix.

The trace of a 2 × 2complex matrixis used to classifyMöbius transformations.First, the matrix is normalized to make itsdeterminantequal to one. Then, if the square of the trace is 4, the corresponding transformation isparabolic.If the square is in the interval[0,4),it iselliptic.Finally, if the square is greater than 4, the transformation isloxodromic.Seeclassification of Möbius transformations.

The trace is used to definecharactersofgroup representations.Two representations $A, B : G \to GL (V)$ of a group $G$ are equivalent (up to change of basis on $V$ ) if $tr(A (g)) = tr(B (g))$ for all $g \in G$ .

The trace also plays a central role in the distribution ofquadratic forms.

Lie algebra

The trace is a map of Lie algebras $\operatorname {tr}:{\mathfrak {gl}}_{n}\to K$ from the Lie algebra ${\mathfrak {gl}}_{n}$ of linear operators on an $n$ -dimensional space ( $n \times n$ matrices with entries in $K$ ) to the Lie algebra $K$ of scalars; as $K$ is Abelian (the Lie bracket vanishes), the fact that this is a map of Lie algebras is exactly the statement that the trace of a bracket vanishes: $\operatorname {tr} ([\mathbf {A},\mathbf {B} ])=0{\text{ for each }}\mathbf {A},\mathbf {B} \in {\mathfrak {gl}}_{n}.$

The kernel of this map, a matrix whose trace iszero,is often said to betracelessortrace free,and these matrices form thesimple Lie algebra ${\mathfrak {sl}}_{n}$ ,which is theLie algebraof thespecial linear groupof matrices with determinant 1. The special linear group consists of the matrices which do not change volume, while thespecial linear Lie algebrais the matrices which do not alter volume ofinfinitesimalsets.

In fact, there is an internaldirect sumdecomposition ${\mathfrak {gl}}_{n}={\mathfrak {sl}}_{n}\oplus K$ of operators/matrices into traceless operators/matrices and scalars operators/matrices. The projection map onto scalar operators can be expressed in terms of the trace, concretely as: $\mathbf {A} \mapsto {\frac {1}{n}}\operatorname {tr} (\mathbf {A} )\mathbf {I}.$

Formally, one can compose the trace (thecounitmap) with the unit map $K\to {\mathfrak {gl}}_{n}$ of "inclusion ofscalars"to obtain a map ${\mathfrak {gl}}_{n}\to {\mathfrak {gl}}_{n}$ mapping onto scalars, and multiplying by $n$ .Dividing by $n$ makes this a projection, yielding the formula above.

In terms ofshort exact sequences,one has $0\to {\mathfrak {sl}}_{n}\to {\mathfrak {gl}}_{n}{\overset {\operatorname {tr} }{\to }}K\to 0$ which is analogous to $1\to \operatorname {SL} _{n}\to \operatorname {GL} _{n}{\overset {\det }{\to }}K^{*}\to 1$ (where $K^{*}=K\setminus \{0\}$ ) forLie groups.However, the trace splits naturally (via $1/n$ times scalars) so ${\mathfrak {gl}}_{n}={\mathfrak {sl}}_{n}\oplus K$ ,but the splitting of the determinant would be as the $n$ th root times scalars, and this does not in general define a function, so the determinant does not split and the general linear group does not decompose: $\operatorname {GL} _{n}\neq \operatorname {SL} _{n}\times K^{*}.$

Bilinear forms

Thebilinear form(where $X$ , $Y$ are square matrices) $B(\mathbf {X},\mathbf {Y} )=\operatorname {tr} (\operatorname {ad} (\mathbf {X} )\operatorname {ad} (\mathbf {Y} ))\quad {\text{where }}\operatorname {ad} (\mathbf {X} )\mathbf {Y} =[\mathbf {X},\mathbf {Y} ]=\mathbf {X} \mathbf {Y} -\mathbf {Y} \mathbf {X}$ is called theKilling form,which is used for the classification of Lie algebras.

The trace defines a bilinear form: $(\mathbf {X},\mathbf {Y} )\mapsto \operatorname {tr} (\mathbf {X} \mathbf {Y} ).$

The form is symmetric, non-degenerate^{[note 5]}and associative in the sense that: $\operatorname {tr} (\mathbf {X} [\mathbf {Y},\mathbf {Z} ])=\operatorname {tr} ([\mathbf {X},\mathbf {Y} ]\mathbf {Z} ).$

For a complex simple Lie algebra (such as ${\mathfrak {sl}}$ ), every such bilinear form is proportional to each other; in particular, to the Killing form^{[citation needed]}.

Two matrices $X$ and $Y$ are said to betrace orthogonalif $\operatorname {tr} (\mathbf {X} \mathbf {Y} )=0.$

There is a generalization to a general representation $(\rho,{\mathfrak {g}},V)$ of a Lie algebra ${\mathfrak {g}}$ ,such that $\rho$ is a homomorphism of Lie algebras $\rho:{\mathfrak {g}}\rightarrow {\text{End}}(V).$ The trace form ${\text{tr}}_{V}$ on ${\text{End}}(V)$ is defined as above. The bilinear form $\phi (\mathbf {X},\mathbf {Y} )={\text{tr}}_{V}(\rho (\mathbf {X} )\rho (\mathbf {Y} ))$ is symmetric and invariant due to cyclicity.

Generalizations

The concept of trace of a matrix is generalized to thetrace classofcompact operatorsonHilbert spaces,and the analog of theFrobenius normis called theHilbert–Schmidtnorm.

If $K$ is a trace-class operator, then for anyorthonormal basis $(e_{n})_{n}$ ,the trace is given by $\operatorname {tr} (K)=\sum _{n}\left\langle e_{n},Ke_{n}\right\rangle,$ and is finite and independent of the orthonormal basis.^[7]

Thepartial traceis another generalization of the trace that is operator-valued. The trace of a linear operator $Z$ which lives on a product space $A \otimes B$ is equal to the partial traces over $A$ and $B$ : $\operatorname {tr} (Z)=\operatorname {tr} _{A}\left(\operatorname {tr} _{B}(Z)\right)=\operatorname {tr} _{B}\left(\operatorname {tr} _{A}(Z)\right).$

For more properties and a generalization of the partial trace, seetraced monoidal categories.

If $A$ is a generalassociative algebraover a field $k$ ,then a trace on $A$ is often defined to be any map $tr: A \mapsto k$ which vanishes on commutators; $tr([a, b]) = 0$ for all $a, b \in A$ .Such a trace is not uniquely defined; it can always at least be modified by multiplication by a nonzero scalar.

Asupertraceis the generalization of a trace to the setting ofsuperalgebras.

The operation oftensor contractiongeneralizes the trace to arbitrary tensors.

Traces in the language of tensor products

Given a vector space $V$ ,there is a natural bilinear map $V \times V * \to F$ given by sending $(v,φ)$ to the scalar $φ(v)$ .Theuniversal propertyof thetensor product $V \otimes V *$ automatically implies that this bilinear map is induced by a linear functional on $V \otimes V *$ .^[8]

Similarly, there is a natural bilinear map $V \times V * \to Hom(V, V)$ given by sending $(v,φ)$ to the linear map $w \mapsto φ(w) v$ .The universal property of the tensor product, just as used previously, says that this bilinear map is induced by a linear map $V \otimes V * \to Hom(V, V)$ .If $V$ is finite-dimensional, then this linear map is alinear isomorphism.^[8]This fundamental fact is a straightforward consequence of the existence of a (finite) basis of $V$ ,and can also be phrased as saying that any linear map $V \to V$ can be written as the sum of (finitely many) rank-one linear maps. Composing the inverse of the isomorphism with the linear functional obtained above results in a linear functional on $Hom(V, V)$ .This linear functional is exactly the same as the trace.

Using the definition of trace as the sum of diagonal elements, the matrix formula $tr(AB) = tr(BA)$ is straightforward to prove, and was given above. In the present perspective, one is considering linear maps $S$ and $T$ ,and viewing them as sums of rank-one maps, so that there are linear functionals $φ i$ and $ψ j$ and nonzero vectors $v i$ and $w j$ such that $S (u) = Σ φ i (u) v i$ and $T (u) = Σ ψ j (u) w j$ for any $u$ in $V$ .Then

(S\circ T)(u)=\sum _{i}\varphi _{i}\left(\sum _{j}\psi _{j}(u)w_{j}\right)v_{i}=\sum _{i}\sum _{j}\psi _{j}(u)\varphi _{i}(w_{j})v_{i}

for any $u$ in $V$ .The rank-one linear map $u \mapsto ψ j (u) φ i (w j) v i$ has trace $ψ j (v i) φ i (w j)$ and so

\operatorname {tr} (S\circ T)=\sum _{i}\sum _{j}\psi _{j}(v_{i})\varphi _{i}(w_{j})=\sum _{j}\sum _{i}\varphi _{i}(w_{j})\psi _{j}(v_{i}).

Following the same procedure with $S$ and $T$ reversed, one finds exactly the same formula, proving that $tr(S \circ T)$ equals $tr(T \circ S)$ .

The above proof can be regarded as being based upon tensor products, given that the fundamental identity of $End(V)$ with $V \otimes V *$ is equivalent to the expressibility of any linear map as the sum of rank-one linear maps. As such, the proof may be written in the notation of tensor products. Then one may consider the multilinear map $V \times V * \times V \times V * \to V \otimes V *$ given by sending $(v, φ, w, ψ)$ to $φ (w) v \otimes ψ$ .Further composition with the trace map then results in $φ (w) ψ (v)$ ,and this is unchanged if one were to have started with $(w, ψ, v, φ)$ instead. One may also consider the bilinear map $End(V) \times End(V) \to End(V)$ given by sending $(f, g)$ to the composition $f \circ g$ ,which is then induced by a linear map $End(V) \otimes End(V) \to End(V)$ .It can be seen that this coincides with the linear map $V \otimes V * \otimes V \otimes V * \to V \otimes V *$ .The established symmetry upon composition with the trace map then establishes the equality of the two traces.^[8]

For any finite dimensional vector space $V$ ,there is a natural linear map $F \to V \otimes V'$ ;in the language of linear maps, it assigns to a scalar $c$ the linear map $c \cdotid V$ .Sometimes this is calledcoevaluation map,and the trace $V \otimes V' \to F$ is calledevaluation map.^[8]These structures can be axiomatized to definecategorical tracesin the abstract setting ofcategory theory.

Notes

^This is immediate from the definition of thematrix product: $\operatorname {tr} (\mathbf {A} \mathbf {B} )=\sum _{i=1}^{m}\left(\mathbf {A} \mathbf {B} \right)_{ii}=\sum _{i=1}^{m}\sum _{j=1}^{n}a_{ij}b_{ji}=\sum _{j=1}^{n}\sum _{i=1}^{m}b_{ji}a_{ij}=\sum _{j=1}^{n}\left(\mathbf {B} \mathbf {A} \right)_{jj}=\operatorname {tr} (\mathbf {B} \mathbf {A} ).$
^For example, if $\mathbf {A} ={\begin{pmatrix}0&1\\0&0\end{pmatrix}},\quad \mathbf {B} ={\begin{pmatrix}0&0\\1&0\end{pmatrix}},$ then the product is $\mathbf {AB} ={\begin{pmatrix}1&0\\0&0\end{pmatrix}},$ and the traces are $tr(AB) = 1 \neq 0 \cdot 0 = tr(A)tr(B)$ .
^Proof: Let $e_{ij}$ the standard basis and note that $f\left(e_{ij}\right)=0$ if and only if $i\neq j$ and $f\left(e_{jj}\right)=f\left(e_{11}\right)$ $f(\mathbf {A} )=\sum _{i,j}[\mathbf {A} ]_{ij}f\left(e_{ij}\right)=\sum _{i}[\mathbf {A} ]_{ii}f\left(e_{11}\right)=f\left(e_{11}\right)\operatorname {tr} (\mathbf {A} ).$ More abstractly, this corresponds to the decomposition ${\mathfrak {gl}}_{n}={\mathfrak {sl}}_{n}\oplus k,$ as $\operatorname {tr} (AB)=\operatorname {tr} (BA)$ (equivalently, $\operatorname {tr} ([A,B])=0$ ) defines the trace on ${\mathfrak {sl}}_{n},$ which has complement the scalar matrices, and leaves one degree of freedom: any such map is determined by its value on scalars, which is one scalar parameter and hence all are multiple of the trace, a nonzero such map.
^Proof: ${\mathfrak {sl}}_{n}$ is asemisimple Lie algebraand thus every element in it is a linear combination of commutators of some pairs of elements, otherwise thederived algebrawould be a proper ideal.
^This follows from the fact that $tr(A * A) = 0$ if and only if $A = 0$ .