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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 [[Group (mathematics)|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).  
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 [[Group (mathematics)|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==
==Definition and properties of matrix traces==
Let '''A''' be an  ''n'' × ''n'' matrix; its trace is defined by
Let '''A''' be an  ''n'' × ''n'' matrix; its trace is defined by
:<math>
:<math>
Line 16: Line 16:
\end{pmatrix} \Longrightarrow \mathrm{Tr}(\mathbf{A}) = 2.1-0.1+3.0 = 5.0
\end{pmatrix} \Longrightarrow \mathrm{Tr}(\mathbf{A}) = 2.1-0.1+3.0 = 5.0
</math>
</math>
'''Theorem.'''  
'''Theorem''' <br>
 
Let '''A''' and '''B''' be ''n''&times;''n'' matrices, then Tr('''A B''') = Tr ('''B A''').<br>
Let '''A''' and '''B''' be square finite-sized matrices, then Tr('''A B''') = Tr ('''B A''').
 
'''Proof'''
'''Proof'''
:<math>
:<math>
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</math>
</math>


'''Theorem'''
'''Theorem'''<br>
 
The trace of a matrix is invariant under a similarity transformation Tr('''B'''<sup>&minus;1</sup>'''A B''') = Tr('''A''').<br>
The trace of a matrix is invariant under a similarity transformation Tr('''B'''<sup>&minus;1</sup>'''A B''') = Tr('''A''').
 
'''Proof'''
'''Proof'''
:<math>
:<math>
Line 44: Line 40:
\mathrm{Tr}( \mathbf{ABC}) &= \mathrm{Tr}( \mathbf{CAB})=\mathrm{Tr}( \mathbf{BCA}) \\
\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} \\
\mathrm{Tr}(c\mathbf{A}) & = c \mathrm{Tr}(\mathbf{A}) \quad c\in\mathbb{C} \\
\mathrm{Tr}(\mathbf{A}^\mathrm{T}) & = \mathrm{Tr}(\mathbf{A}) \\
\end{align}
\end{align}
</math>
</math>
'''Theorem'''<br>
Let '''S''' be a symmetric matrix, '''S'''<sup>T</sup> = '''S''',  and '''A''' be an antisymmetric matrix, '''A'''<sup>T</sup> = &minus;'''A'''. Then
:<math>
\mathrm{Tr}(\mathbf{S}\mathbf{A}) = \mathrm{Tr}(\mathbf{A}\mathbf{S}) =0 .
</math><br>
'''Proof'''
:<math>
\mathrm{Tr}(\mathbf{SA}) = \mathrm{Tr}\big((\mathbf{SA})^\mathrm{T}\big) = \mathrm{Tr}(\mathbf{A}^\mathrm{T}\mathbf{S}^\mathrm{T}) = -\mathrm{Tr}(\mathbf{AS}) = -\mathrm{Tr}(\mathbf{SA})
</math>
A number equal to minus itself can only be zero.
===Relation to eigenvalues===
===Relation to eigenvalues===
We will show that  ''the trace of an n&times;n  matrix  is equal to the sum of its n eigenvalues (the n roots of its secular equation)''.
We will show that  ''the trace of an n&times;n  matrix  is equal to the sum of its n eigenvalues (the n roots of its secular equation)''.
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\Delta(\lambda) = (-\lambda)^n + P_1(-\lambda)^{n-1} + P_2(-\lambda)^{n-2}+ \cdots +P_{n-1}(-\lambda) + P_n = 0.
\Delta(\lambda) = (-\lambda)^n + P_1(-\lambda)^{n-1} + P_2(-\lambda)^{n-2}+ \cdots +P_{n-1}(-\lambda) + P_n = 0.
</math>
</math>
The coefficient ''P''<sub>1</sub> of &lambda;<sup>''n''&minus;1</sup> is equal to the trace of '''A''' (and incidentally ''P''<sub>n</sub> is the determinant of '''A''').  If the field ''F'' is algebraically closed (such as the field of complex numbers) then the [[fundamental theorem of algebra]] states that the secular equation has exactly ''n'' roots (zeros) &lambda;<sub>''i''</sub>, ''i'' =1, ..., ''n'', the [[eigenvalue]]s of '''A''' and the following factorization holds
The coefficient ''P''<sub>1</sub> of (&minus;&lambda;)<sup>''n''&minus;1</sup> is equal to the trace of '''A''' (and incidentally ''P''<sub>n</sub> is the determinant of '''A''').  If the field ''F'' is algebraically closed (such as the field of complex numbers) then the [[fundamental theorem of algebra]] states that the secular equation has exactly ''n'' roots (zeros) &lambda;<sub>''i''</sub>, ''i'' =1, ..., ''n'', the [[eigenvalue]]s of '''A''' and the following factorization holds
:<math>
:<math>
\Delta(\lambda) = (\lambda_1-\lambda)(\lambda_2-\lambda)\cdots(\lambda_n-\lambda).
\Delta(\lambda) = (\lambda_1-\lambda)(\lambda_2-\lambda)\cdots(\lambda_n-\lambda).
</math>
</math>
Expansion shows that the coefficient ''P''<sub>1</sub> of &lambda;<sup>''n''&minus;1</sup> is equal to
Expansion shows that the coefficient ''P''<sub>1</sub> of (&minus;&lambda;)<sup>''n''&minus;1</sup> is equal to
:<math>
:<math>
  \sum_{i=1}^n \lambda_i = P_1 =\mathrm{Tr}(\mathbf{A}).
  \sum_{i=1}^n \lambda_i = P_1 =\mathrm{Tr}(\mathbf{A}).
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==Infinite-dimensional space==
==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).
The trace of a linear operator on an infinite-dimensional linear space is not always defined.  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., the trace is undefined. In fact, the property of having a finite trace is a severe restriction on a linear operator.


We consider an infinite-dimensional space with an inner product (a [[Hilbert space]]). Let  <font style="vertical-align: top"><math>\hat{T}</math></font> be a linear operator on this space with the property
We consider an infinite-dimensional space with an inner product (a [[Hilbert space]]). Let  <font style="vertical-align: top"><math>\hat{T}</math></font> be a linear operator on this space with the property
:<math>
:<math>
(\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},
(\hat{T}^\dagger\hat{T})\; v_i = \alpha_i^2 \; v_i,\quad i=1,2,\ldots,\infty \quad \hbox{and} \quad \alpha_i^2\in \mathbb{R},
</math>  
</math>  
where {''v''<sub>''i''</sub>} is an [[orthonormal]] basis of the space. Note that the operator
where {''v''<sub>''i''</sub>} is an [[orthonormal]] basis of the space.  
Note that the operator
<font style="vertical-align: top"><math>\hat{T}^\dagger\hat{T}</math></font> is [[self-adjoint operator|self-adjoint]] and [[positive definite]], i.e.,
<font style="vertical-align: top"><math>\hat{T}^\dagger\hat{T}</math></font> is [[self-adjoint operator|self-adjoint]] and [[positive definite]], i.e.,
:<math>
:<math>
  \langle (T^\dagger T) w |  w \rangle = \langle  w | (T^\dagger T) w \rangle = \langle T w | T w \rangle  \ge 0 \quad\hbox{for any}\quad w.
  \langle (T^\dagger T) w |  w \rangle = \langle  w | (T^\dagger T) w \rangle = \langle T w | T w \rangle  \ge 0 \quad\hbox{for any}\quad w.
</math>
</math>
When the following sum converges,  
From this follows that the eigenvalues of  <font style="vertical-align: top"><math>\hat{T}^\dagger\hat{T}</math></font> are positive&mdash;so that they may be written as squares&mdash;and its eigenvectors  ''v''<sub>''i''</sub> are orthonormal.
 
If the following sum of square roots of eigenvalues converges,  
:<math>
:<math>
\sum_{i=1}^\infty \alpha_i < \infty
\sum_{i=1}^\infty \alpha_i < \infty,
</math>
</math>
one may define  the trace of  <font style="vertical-align: top"><math>\hat{T}</math></font> by
then the trace of  <font style="vertical-align: top"><math>\hat{T}</math></font> can be defined by
:<math>
:<math>
\mathrm{Tr}(\hat{T}) \equiv \sum_{i=1}^\infty \langle v_i |T| v_i \rangle,
\mathrm{Tr}(\hat{T}) \equiv \sum_{i=1}^\infty \langle v_i |T| v_i \rangle,
</math>
</math>
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,
i.e., it can be proved that this summation converges as well.  Operators that have  a well-defined trace are called "trace class operators" or sometimes "nuclear operators".
 
As in the finite-dimensional case the trace is independent of the choice of (orthonormal) basis,
:<math>
:<math>
\mathrm{Tr}(\hat{T}) = \sum_{i=1}^\infty \langle w_i |T| w_i \rangle < \infty,
\mathrm{Tr}(\hat{T}) = \sum_{i=1}^\infty \langle w_i |T| w_i \rangle < \infty,
</math>
</math>
for any orthonormal basis {''w''<sub>''i''</sub>}. Operators that have  a well-defined trace are called "trace class operators" or "nuclear operators".
for any orthonormal basis {''w''<sub>''i''</sub>}.  


An important example is the exponential of the self-adjoint operator ''H'',
An important example of a trace class operator is the exponential of the self-adjoint operator ''H'',
:<math>
:<math>
e^{-\beta\hat{H}},\quad \beta \in \mathbb{R},\quad 0< \beta < \infty.
e^{-\beta\hat{H}},\quad \beta \in \mathbb{R},\quad 0< \beta < \infty.
</math>
</math>
The operator ''H'' being self-adjoint has only real eigenvalues &epsilon;<sub>''i''</sub>.  When ''H'' is bounded from below (its lowest eigenvalue is finite) then the sum
The operator ''H'', being self-adjoint, has only real eigenvalues &epsilon;<sub>''i''</sub>.  When ''H'' is bounded from below (its lowest eigenvalue is finite) then the sum
:<math>
:<math>
\mathrm{Tr}e^{-\beta H} = \sum_{i=1}^\infty e^{-\beta \epsilon_i} < \infty
\mathrm{Tr}e^{-\beta H} = \sum_{i=1}^\infty e^{-\beta \epsilon_i} < \infty
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*F. R. Gantmacher, ''Matrizentheorie'', Translated from the Russian by H. Boseck, D. Soyka, and K. Stengert, Springer Verlag, Berlin (1986). ISBN 3540165827
*F. R. Gantmacher, ''Matrizentheorie'', Translated from the Russian by H. Boseck, D. Soyka, and K. Stengert, Springer Verlag, Berlin (1986). ISBN 3540165827
*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
*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
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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 and properties of matrix traces

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{Tr}(\mathbf{A})\; \stackrel{\mathrm{def}}{=} \; \sum_{i=1}^n A_{ii} }

where Aii is the ith diagonal element of A.

Example

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 n×n matrices, then Tr(A B) = Tr (B A).
Proof

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{Tr}\big(\mathbf{B}^{-1}(\mathbf{AB})\big) = \mathrm{Tr}\big((\mathbf{AB})\mathbf{B}^{-1}\big) = \mathrm{Tr}(\mathbf{A E}) = \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):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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} \\ \mathrm{Tr}(\mathbf{A}^\mathrm{T}) & = \mathrm{Tr}(\mathbf{A}) \\ \end{align} }

Theorem
Let S be a symmetric matrix, ST = S, and A be an antisymmetric matrix, AT = −A. Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{Tr}(\mathbf{S}\mathbf{A}) = \mathrm{Tr}(\mathbf{A}\mathbf{S}) =0 . }

Proof

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{Tr}(\mathbf{SA}) = \mathrm{Tr}\big((\mathbf{SA})^\mathrm{T}\big) = \mathrm{Tr}(\mathbf{A}^\mathrm{T}\mathbf{S}^\mathrm{T}) = -\mathrm{Tr}(\mathbf{AS}) = -\mathrm{Tr}(\mathbf{SA}) }

A number equal to minus itself can only be zero.

Relation to eigenvalues

We will show that the trace of an n×n matrix is equal to the sum of its n eigenvalues (the n roots of its secular equation).

The secular determinant of an n × n matrix A is the determinant of A −λ E, where λ is a number (an element of a field F). If we put the secular determinant equal to zero we obtain the secular equation of A (also known as the characteristic equation),

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(\lambda) \equiv \begin{vmatrix} A_{11}-\lambda & A_{12} & \cdots & \cdots & A_{1n} \\ A_{21} & A_{22}-\lambda & \cdots & \cdots & A_{2n} \\ \cdots & \cdots & \ddots \\ A_{n1} & A_{n2} & &\cdots & A_{nn}-\lambda \\ \end{vmatrix} = 0 }

The secular determinant is a polynomial in λ:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(\lambda) = (-\lambda)^n + P_1(-\lambda)^{n-1} + P_2(-\lambda)^{n-2}+ \cdots +P_{n-1}(-\lambda) + P_n = 0. }

The coefficient P1 of (−λ)n−1 is equal to the trace of A (and incidentally Pn is the determinant of A). If the field F is algebraically closed (such as the field of complex numbers) then the fundamental theorem of algebra states that the secular equation has exactly n roots (zeros) λi, i =1, ..., n, the eigenvalues of A and the following factorization holds

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta(\lambda) = (\lambda_1-\lambda)(\lambda_2-\lambda)\cdots(\lambda_n-\lambda). }

Expansion shows that the coefficient P1 of (−λ)n−1 is equal to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^n \lambda_i = P_1 =\mathrm{Tr}(\mathbf{A}). }

Note: It is not necessary that A has n linearly independent eigenvectors, although any A has n eigenvalues in an algebraically closed field.

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

Let Vn be an n-dimensional vector space (also known as linear space). Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} be a linear operator (also known as linear map) on this space,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}:\quad V_n \rightarrow V_n } .

Let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{v_1, v_2, \ldots, v_n\} }

be a basis for Vn, then the matrix of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} with respect to this basis is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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),

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_i = \sum_{j=1}^n\; v_j B_{ji}, \quad i=1,\ldots, n. }

Above we introduced the matrix A of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} in the basis vi. Write A' for its matrix in the basis wi

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} in both bases is equal.

Theorem

Let a linear operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} on Vn have n linearly independent eigenvectors,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{Tr}(\hat{A}) = \sum_{i=1}^n \alpha_i. }

Proof

The matrix of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} in basis of its eigenvectors is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 δji is the Kronecker delta.

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

Finite-dimensional inner product space

When the n-dimensional linear space Vn 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{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 Vn. The symbol δij stands for the Kronecker delta. The matrix of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A}} with respect to this basis is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{A} v_i = \sum_{j=1}^n\; v_j A_{ji} . }

Project with vk:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 a linear operator on an infinite-dimensional linear space is not always defined. 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., the trace is undefined. In fact, the property of having a finite trace is a severe restriction on a linear operator.

We consider an infinite-dimensional space with an inner product (a Hilbert space). Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}} be a linear operator on this space with the property

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\hat{T}^\dagger\hat{T})\; v_i = \alpha_i^2 \; v_i,\quad i=1,2,\ldots,\infty \quad \hbox{and} \quad \alpha_i^2\in \mathbb{R}, }

where {vi} is an orthonormal basis of the space. Note that the operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}^\dagger\hat{T}} is self-adjoint and positive definite, i.e.,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle (T^\dagger T) w | w \rangle = \langle w | (T^\dagger T) w \rangle = \langle T w | T w \rangle \ge 0 \quad\hbox{for any}\quad w. }

From this follows that the eigenvalues of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}^\dagger\hat{T}} are positive—so that they may be written as squares—and its eigenvectors vi are orthonormal.

If the following sum of square roots of eigenvalues converges,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^\infty \alpha_i < \infty, }

then the trace of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{T}} can be defined by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{Tr}(\hat{T}) \equiv \sum_{i=1}^\infty \langle v_i |T| v_i \rangle, }

i.e., it can be proved that this summation converges as well. Operators that have a well-defined trace are called "trace class operators" or sometimes "nuclear operators".

As in the finite-dimensional case the trace is independent of the choice of (orthonormal) basis,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{Tr}(\hat{T}) = \sum_{i=1}^\infty \langle w_i |T| w_i \rangle < \infty, }

for any orthonormal basis {wi}.

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-\beta\hat{H}},\quad \beta \in \mathbb{R},\quad 0< \beta < \infty. }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

  • F. R. Gantmacher, Matrizentheorie, Translated from the Russian by H. Boseck, D. Soyka, and K. Stengert, Springer Verlag, Berlin (1986). ISBN 3540165827
  • 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