Trace (mathematics)

In mathematics, a trace is a property of a matrix and of a linear operator on a vector space. The trace plays an important role in the representation theory of groups (the collection of traces is the character of the representation) and in statistical thermodynamics (the trace of a thermodynamic observable times the density operator is the thermodynamic average of the observable).

Definition for matrices

Let A be an n × n matrix; its trace is defined by

$\mathrm {Tr} (\mathbf {A} )\;{\stackrel {\mathrm {def} }{=}}\;\sum _{i=1}^{n}A_{ii}$

where A_ii is the ith diagonal element of A.

Example

$\mathbf {A} ={\begin{pmatrix}2.1&1.3&0.0\\5.0&-0.1&8.3\\7.0&-4.7&3.0\\\end{pmatrix}}\Longrightarrow \mathrm {Tr} (\mathbf {A} )=2.1-0.1+3.0=5.0$

Theorem.

Let A and B be square finite-sized matrices, then Tr(A B) = Tr (B A).

Proof

$\mathrm {Tr} (\mathbf {AB} )=\sum _{i=1}^{n}(\mathbf {AB} )_{ii}=\sum _{i=1}^{n}\sum _{j=1}^{n}\;A_{ij}B_{ji}=\sum _{j=1}^{n}\sum _{i=1}^{n}\;B_{ji}A_{ij}=\sum _{j=1}^{n}(\mathbf {BA} )_{jj}=\mathrm {Tr} (\mathbf {BA} )$

Theorem

The trace of a matrix is invariant under a similarity transformation Tr(B⁻¹A B) = Tr(A).

Proof

$\mathrm {Tr} {\big (}\mathbf {B} ^{-1}(\mathbf {AB} ){\big )}=\mathrm {Tr} {\big (}(\mathbf {AB} )\mathbf {B} ^{-1}{\big )}=\mathrm {Tr} (\mathbf {AE} )=\mathrm {Tr} (\mathbf {A} ),$

where we used B B⁻¹ = E (the identity matrix).

Other properties of traces are (all matrices are n × n matrices):

${\begin{aligned}\mathrm {Tr} (\mathbf {A} +\mathbf {B} )&=\mathrm {Tr} (\mathbf {A} )+\mathrm {Tr} (\mathbf {B} )\\\mathrm {Tr} (\mathbf {E} )&=n\qquad {\hbox{(trace of identity matrix)}}\\\mathrm {Tr} (\mathbf {O} )&=0\qquad {\hbox{(trace of zero matrix)}}\\\mathrm {Tr} (\mathbf {ABC} )&=\mathrm {Tr} (\mathbf {CAB} )=\mathrm {Tr} (\mathbf {BCA} )\\\mathrm {Tr} (c\mathbf {A} )&=c\mathrm {Tr} (\mathbf {A} )\quad c\in \mathbb {C} \\\end{aligned}}$

Definition for a linear operator on a finite-dimensional vector space

Let V_n be an n-dimensional vector space (also known as linear space). Let ${\hat {A}}$ be a linear operator (also known as linear map) on this space,

${\hat {A}}:\quad V_{n}\rightarrow V_{n}$.

Let

$\{v_{1},v_{2},\ldots ,v_{n}\}$

be a basis for V_n, then the matrix of ${\hat {A}}$ with respect to this basis is given by

${\hat {A}}v_{i}=\sum _{j=1}^{n}\;v_{j}A_{ji}\quad {\hbox{for}}\quad i=1,\ldots ,n,\quad {\hbox{and}}\quad \mathbf {A} \equiv (A_{ij}).$

Definition: The trace of the linear operator ${\hat {A}}$ is the trace of the matrix of the operator in any basis. This definition is possible since the trace is independent of the choice of basis.

We prove that a trace of an operator does not depend on choice of basis. Consider two bases connected by the non-singular matrix B (a basis transformation matrix),

$w_{i}=\sum _{j=1}^{n}\;v_{j}B_{ji},\quad i=1,\ldots ,n.$

Above we introduced the matrix A of ${\hat {A}}$ in the basis v_i. Write A' for its matrix in the basis w_i

${\hat {A}}w_{i}=\sum _{j=1}^{n}\;w_{j}A'_{ji}\quad {\hbox{with}}\quad \mathbf {A} '=(A'_{ij}).$

It is not difficult to prove that

$\mathbf {A} '=\mathbf {B} ^{-1}\;\mathbf {A} \;\mathbf {B} \quad \Longrightarrow \quad \mathrm {Tr} (\mathbf {A} ')=\mathrm {Tr} (\mathbf {A} ),$

from which follows that the trace of ${\hat {A}}$ in both bases is equal.

Theorem

Let a linear operator ${\hat {A}}$ on V_n have n linearly independent eigenvectors,

${\hat {A}}\;v_{i}=\alpha _{i}v_{i}\quad {\hbox{with}}\quad \alpha _{i}\in \mathbb {C} \quad {\hbox{and}}\quad i=1,\ldots ,n.$

Then its trace is the sum of the eigenvalues

$\mathrm {Tr} ({\hat {A}})=\sum _{i=1}^{n}\alpha _{i}.$

Proof

The matrix of ${\hat {A}}$ in basis of its eigenvectors is

${\hat {A}}\;v_{i}=\sum _{j=1}^{n}\;v_{j}(\alpha _{j}\delta _{ji})\quad \Longrightarrow \quad \mathbf {A} ={\begin{pmatrix}\alpha _{1}&0&\cdots &0\\0&\alpha _{2}&\cdots \\\cdots &&\ddots \\0&&&\alpha _{n}\\\end{pmatrix}},$

where δ_ji is the Kronecker delta.

Note. To avoid misunderstanding: not all linear operators on V_n possess n linearly independent eigenvectors.

Finite-dimensional inner product space

When the n-dimensional linear space V_n is equipped with a positive definite inner product, an expression for the matrix of a linear operator and its trace can be given. These expressions can be generalized to inner product spaces of infinite dimension and are of great importance in quantum mechanics.

Let

$\{v_{1},v_{2},\ldots ,v_{n}\}\quad {\hbox{with}}\quad \langle v_{i}|v_{j}\rangle =\delta _{ij},\quad i,j=1,\ldots ,n,$

be an orthonormal basis for V_n. The symbol δ_ij stands for the Kronecker delta. The matrix of ${\hat {A}}$ with respect to this basis is given by

${\hat {A}}v_{i}=\sum _{j=1}^{n}\;v_{j}A_{ji}.$

Project with v_k:

$\langle v_{k}|{\hat {A}}|v_{i}\rangle =\sum _{j=1}^{n}\;\langle v_{k}|v_{j}\rangle \;A_{ji}=\sum _{j=1}^{n}\;\delta _{kj}\;A_{ji}=A_{ki}.$

Hence

$A_{ij}=\langle v_{i}|{\hat {A}}|v_{j}\rangle \quad \Longrightarrow \quad \mathrm {Tr} ({\hat {A}})=\sum _{i=1}^{n}\langle v_{i}|{\hat {A}}|v_{i}\rangle .$

Infinite-dimensional space

The trace of an operator on an infinite-dimensional linear space is not well-defined for all operators on all infinite-dimensional spaces. Even if we restrict our attention to infinite-dimensional spaces with countable bases, the generalization of the definition is not always possible. For instance, we saw above that the trace of the identity operator on a finite-dimensional space is equal to the dimension of the space, so that a simple extension of the definition leads to a trace of the identity operator that is infinite (i.e., not defined).

We consider an infinite-dimensional space with an inner product (a Hilbert space). Let ${\hat {T}}$ be a linear operator on this space with the property

$({\hat {T}}^{\dagger }{\hat {T}})\;v_{i}=\alpha _{i}^{2}\;v_{i},\quad i=1,2,\ldots ,\infty \quad {\hbox{and}}\quad \alpha _{i}\in \mathbb {R} ,$

where {v_i} is an orthonormal basis of the space. Note that the operator ${\hat {T}}^{\dagger }{\hat {T}}$ is self-adjoint and positive definite, i.e.,

$\langle (T^{\dagger }T)w|w\rangle =\langle w|(T^{\dagger }T)w\rangle =\langle Tw|Tw\rangle \geq 0\quad {\hbox{for any}}\quad w.$

When the following sum converges,

$\sum _{i=1}^{\infty }\alpha _{i}<\infty$

one may define the trace of ${\hat {T}}$ by

$\mathrm {Tr} ({\hat {T}})\equiv \sum _{i=1}^{\infty }\langle v_{i}|T|v_{i}\rangle ,$

i.e., it can be shown that this summation converges as well. As in the finite-dimensional case it can be proved that the trace is independent of the choice of (orthonormal) basis,

$\mathrm {Tr} ({\hat {T}})=\sum _{i=1}^{\infty }\langle w_{i}|T|w_{i}\rangle <\infty ,$

for any orthonormal basis {w_i}. Operators that have a well-defined trace are called "trace class operators" or "nuclear operators".

An important example is the exponential of the self-adjoint operator H,

$e^{-\beta {\hat {H}}},\quad \beta \in \mathbb {R} ,\quad 0<\beta <\infty .$

The operator H being self-adjoint has only real eigenvalues ε_i. When H is bounded from below (its lowest eigenvalue is finite) then the sum

$\mathrm {Tr} e^{-\beta H}=\sum _{i=1}^{\infty }e^{-\beta \epsilon _{i}}<\infty$

converges. This trace is the canonical partition function of statistical physics.

Reference

N. I Achieser and I. M. Glasmann, Theorie der linearen Operatoren im Hilbert Raum, Translated from the Russian by H. Baumgärtel, Verlag Harri Deutsch, Thun (1977). ISBN 3871443263