Trace (Matrix)

The trace of an $n\times n$ Square Matrix A is defined by

$\begin{displaymath} \mathop{\rm Tr}\nolimits ({\hbox{\sf A}})\equiv \sum_{i=1}^n a_{ii}. \end{displaymath}$

(1)

For Square Matrices A and B, it is true that

$\displaystyle \mathop{\rm Tr}\nolimits ({\hbox{\sf A}})$	$\textstyle =$	$\displaystyle \mathop{\rm Tr}\nolimits ({\hbox{\sf A}}^{\rm T})$	(2)
$\displaystyle \mathop{\rm Tr}\nolimits ({\hbox{\sf A}}+{\hbox{\sf B}})$	$\textstyle =$	$\displaystyle \mathop{\rm Tr}\nolimits ({\hbox{\sf A}})+\mathop{\rm Tr}\nolimits ({\hbox{\sf B}})$	(3)
$\displaystyle \mathop{\rm Tr}\nolimits (\alpha{\hbox{\sf A}})$	$\textstyle =$	$\displaystyle \alpha\mathop{\rm Tr}\nolimits ({\hbox{\sf A}})$	(4)

(Lange 1987, p. 40). The trace is invariant under a Similarity Transformation

$\begin{displaymath} {\hbox{\sf A}}'\equiv{\hbox{\sf B}}{\hbox{\sf A}}{\hbox{\sf B}}^{-1} \end{displaymath}$

(5)

(Lange 1987, p. 64). Since

$\begin{displaymath} (bab^{-1})_{ij} = b_{il}a_{lk}b^{-1}_{kj}, \end{displaymath}$

(6)

$\displaystyle \mathop{\rm Tr}\nolimits ({\hbox{\sf B}}{\hbox{\sf A}}{\hbox{\sf B}}^{-1})$	$\textstyle =$	$\displaystyle b_{il}a_{lk}{b^{-1}}_{ki}$
	$\textstyle =$	$\displaystyle (b^{-1}b)_{kl}a_{lk} = \delta_{kl}a_{lk}$
	$\textstyle =$	$\displaystyle a_{kk} = \mathop{\rm Tr}\nolimits ({\hbox{\sf A}}),$	(7)

where $\delta_{ij}$ is the Kronecker Delta.

The trace of a product of square matrices is independent of the order of the multiplication since

$\displaystyle \mathop{\rm Tr}\nolimits ({\hbox{\sf A}}{}{\hbox{\sf B}})$	$\textstyle =$	$\displaystyle (ab)_{ii} = a_{ij}b_{ji} = b_{ji}a_{ij}$
	$\textstyle =$	$\displaystyle (ba)_{jj} = \mathop{\rm Tr}\nolimits ({\hbox{\sf B}}{}{\hbox{\sf A}}).$	(8)

Therefore, the trace of the Commutator of

and

is given by

$\begin{displaymath} \mathop{\rm Tr}\nolimits ([{\hbox{\sf A}},{\hbox{\sf B}}]) \... ...)-\mathop{\rm Tr}\nolimits ({\hbox{\sf B}}{\hbox{\sf A}}) = 0. \end{displaymath}$

(9)

The product of a Symmetric and an Antisymmetric Matrix has zero trace,

$\begin{displaymath} \mathop{\rm Tr}\nolimits (A_SB_A) = 0. \end{displaymath}$

(10)

The value of the trace can be found using the fact that the matrix can always be transformed to a coordinate system where the z-Axis lies along the axis of rotation. In the new coordinate system, the Matrix is

$\begin{displaymath} {\hbox{\sf A}}' = \left[{\matrix{\cos \phi & \sin \phi & 0\cr -\sin \phi & \cos \phi & 0\cr 0 & 0 & 1\cr}}\right], \end{displaymath}$

(11)

so the trace is

$\begin{displaymath} \mathop{\rm Tr}\nolimits ({\hbox{\sf A}}') = \mathop{\rm Tr}\nolimits ({\hbox{\sf A}}) \equiv a_{ii} = 1+2\cos\phi. \end{displaymath}$

(12)

References

Lang, S. Linear Algebra, 3rd ed. New York: Springer-Verlag, pp. 40 and 64, 1987.