In linear algebra, the trace of a square matrix A, denoted tr(A), is the sum of the elements on its main diagonal, ${\displaystyle a_{11}+a_{22}+\dots +a_{nn}}$. It is only defined for a square matrix (n × n).

The trace of a matrix is the sum of its eigenvalues (counted with multiplicities). Also, tr(AB) = tr(BA) for any matrices A and B of the same size. Thus, similar matrices have the same trace. As a consequence, one can define the trace of a linear operator mapping a finite-dimensional vector space into itself, since all matrices describing such an operator with respect to a basis are similar.

The trace is related to the derivative of the determinant (see Jacobi's formula).

The trace of an n × n square matrix A is defined as $${\displaystyle \operatorname {tr} (\mathbf {A} )=\sum _{i=1}^{n}a_{ii}=a_{11}+a_{22}+\dots +a_{nn}}$$ where a_ii denotes the entry on the i th row and i th column of A. The entries of A can be real numbers, complex numbers, or more generally elements of a field F. The trace is not defined for non-square matrices.

Let A be a matrix, with $${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}={\begin{pmatrix}1&0&3\\11&5&2\\6&12&-5\end{pmatrix}}}$$

Then $${\displaystyle \operatorname {tr} (\mathbf {A} )=\sum _{i=1}^{3}a_{ii}=a_{11}+a_{22}+a_{33}=1+5+(-5)=1}$$

The trace is a linear mapping. That is, $${\displaystyle {\begin{aligned}\operatorname {tr} (\mathbf {A} +\mathbf {B} )&=\operatorname {tr} (\mathbf {A} )+\operatorname {tr} (\mathbf {B} )\\\operatorname {tr} (c\mathbf {A} )&=c\operatorname {tr} (\mathbf {A} )\end{aligned}}}$$ for all square matrices A and B, and all scalars c.

A matrix and its transpose have the same trace: $${\displaystyle \operatorname {tr} (\mathbf {A} )=\operatorname {tr} \left(\mathbf {A} ^{\mathsf {T}}\right).}$$