trace of a matrix

Definition
Let $A=(a_{i,j})$ be a square matrix of order $n$ . The trace of the matrix is the sum of the main diagonal:

$\operatorname{trace}(A)=\sum\limits_{i=1}^{n}a_{i,i}$

Notation:
The trace of a matrix $A$ is also commonly denoted as $\operatorname{Tr}(A)$ or $\operatorname{Tr}A$ .

Properties:

1.

The trace is a linear transformation from the space of square matrices to the real numbers. In other words, if $A$ and $B$ are square matrices with real (or complex) entries, of same order and $c$ is a scalar, then

$\displaystyle\operatorname{trace}(A+B)$ $\displaystyle=$ $\displaystyle\operatorname{trace}(A)+\operatorname{trace}(B),$

$\displaystyle\operatorname{trace}(cA)$ $\displaystyle=$ $\displaystyle c\cdot\operatorname{trace}(A).$
2.

For the transpose and conjugate transpose, we have for any square matrix $A$ with real (or complex) entries,

$\displaystyle\operatorname{trace}(A^{t})$ $\displaystyle=$ $\displaystyle\operatorname{trace}(A),$

$\displaystyle\operatorname{trace}(A^{\ast})$ $\displaystyle=$ $\displaystyle\overline{\operatorname{trace}(A)}.$
3.

If $A$ and $B$ are matrices such that $A B$ is a square matrix, then

$\operatorname{trace}(AB)=\operatorname{trace}(BA).$

For this reason it is possible to define the trace of a linear transformation, as the choice of basis does not affect the trace. Thus, if $A, B, C$ are matrices such that $A B C$ is a square matrix, then

$\operatorname{trace}(ABC)=\operatorname{trace}(CAB)=\operatorname{trace}(BCA).$
4.

If $B$ is in invertible square matrix of same order as $A$ , then

$\operatorname{trace}(A)=\operatorname{trace}(B^{-1}AB).$

In other words, the trace of similar matrices are equal.
5.

Let $A$ be a square matrix of order $n$ with real (or complex) entries $a_{ij}$ . Then

$\displaystyle\operatorname{trace}A^{\ast}A$ $\displaystyle=$ $\displaystyle\operatorname{trace}AA^{\ast}$

$\displaystyle=$ $\displaystyle\sum_{i,j=1}^{n}|a_{ij}|^{2}.$

Here ${}^{\ast}$ is the complex conjugate, and $|\cdot|$ is the complex modulus. In particular, $\operatorname{trace}A^{\ast}A\geq 0$ with equality if and only if $A=0$ . (See the Frobenius matrix norm.)
6.

Various inequalities for $\operatorname{trace}$ are given in [2].

See the proof of properties of trace of a matrix.

References

1 The Trace of a Square Matrix. Paul Ehrlich, [online] http://www.math.ufl.edu/ ehrlich/trace.htmlhttp://www.math.ufl.edu/ ehrlich/trace.html
2 Z.P. Yang, X.X. Feng, A note on the trace inequality for products of Hermitian matrix power, Journal of Inequalities in Pure and Applied Mathematics, Volume 3, Issue 5, 2002, Article 78, http://www.emis.de/journals/JIPAM/v3n5/082_02.htmlonline.

Title	trace of a matrix
Canonical name	TraceOfAMatrix
Date of creation	2013-03-22 11:59:56
Last modified on	2013-03-22 11:59:56
Owner	Daume (40)
Last modified by	Daume (40)
Numerical id	20
Author	Daume (40)
Entry type	Definition
Classification	msc 15A99
Related topic	ShursInequality

	$\displaystyle\operatorname{trace}(A+B)$	$\displaystyle=$	$\displaystyle\operatorname{trace}(A)+\operatorname{trace}(B),$
	$\displaystyle\operatorname{trace}(cA)$	$\displaystyle=$	$\displaystyle c\cdot\operatorname{trace}(A).$

	$\displaystyle\operatorname{trace}(A^{t})$	$\displaystyle=$	$\displaystyle\operatorname{trace}(A),$
	$\displaystyle\operatorname{trace}(A^{\ast})$	$\displaystyle=$	$\displaystyle\overline{\operatorname{trace}(A)}.$

	$\displaystyle\operatorname{trace}A^{\ast}A$	$\displaystyle=$	$\displaystyle\operatorname{trace}AA^{\ast}$
		$\displaystyle=$	$\displaystyle\sum_{i,j=1}^{n}\|a_{ij}\|^{2}.$