Editing Trace (linear algebra) (section)

===Characterization of the trace===
The following three properties:
<math display="block">\begin{align}
\operatorname{tr}(\mathbf{A} + \mathbf{B}) &= \operatorname{tr}(\mathbf{A}) + \operatorname{tr}(\mathbf{B}), \\
\operatorname{tr}(c\mathbf{A}) &= c \operatorname{tr}(\mathbf{A}), \\
\operatorname{tr}(\mathbf{A}\mathbf{B}) &= \operatorname{tr}(\mathbf{B}\mathbf{A}),
\end{align}</math>
characterize the trace [[up to]] a scalar multiple in the following sense: If <math>f</math> is a [[linear functional]] on the space of square matrices that satisfies <math>f(xy) = f(yx),</math> then <math>f</math> and <math>\operatorname{tr}</math> are proportional.<ref group="note">Proof: Let <math>e_{ij}</math> the standard basis and note that <math>f\left(e_{ij}\right) = f\left(e_{i} e_{j}^\top\right) = f\left(e_i e_1^\top e_1 e_j^\top\right) = f\left(e_1 e_j^\top e_i e_1^\top\right) = f\left(0\right) = 0</math> if  <math>i \neq j</math> and <math>f\left(e_{jj}\right) = f\left(e_{11}\right)</math>
<math display="block">f(\mathbf{A}) = \sum_{i, j} [\mathbf{A}]_{ij} f\left(e_{ij}\right) = \sum_i [\mathbf{A}]_{ii} f\left(e_{11}\right) = f\left(e_{11}\right) \operatorname{tr}(\mathbf{A}).</math>

More abstractly, this corresponds to the decomposition
<math display="block">\mathfrak{gl}_n = \mathfrak{sl}_n \oplus k,</math>
as <math>\operatorname{tr}(AB) = \operatorname{tr}(BA)</math> (equivalently, <math>\operatorname{tr}([A, B]) = 0</math>) defines the trace on <math>\mathfrak{sl}_n,</math> which has complement the scalar matrices, and leaves one degree of freedom: any such map is determined by its value on scalars, which is one scalar parameter and hence all are multiple of the trace, a nonzero such map.</ref>

For <math>n\times n</math> matrices, imposing the normalization <math>f(\mathbf{I}) = n</math> makes <math>f</math> equal to the trace.