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Trace (Matrix)

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

\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
{\hbox{\sf A}}'\equiv{\hbox{\sf B}}{\hbox{\sf A}}{\hbox{\sf B}}^{-1}
\end{displaymath} (5)

(Lange 1987, p. 64). Since
(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 $A$ and $B$ is given by
\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,
\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

{\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
\mathop{\rm Tr}\nolimits ({\hbox{\sf A}}') = \mathop{\rm Tr}\nolimits ({\hbox{\sf A}}) \equiv a_{ii} = 1+2\cos\phi.
\end{displaymath} (12)


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

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© 1996-9 Eric W. Weisstein