Trace

Definition

$\mathrm{tr}(A)={\sum }_{i=1}^n{a}_{ii}=a_{11}+a_{22}+\cdots +a_{nn}$

The sum of elements on the main diagonal

Facts

$\mathrm{tr}(A+B)=\mathrm{tr}(A)+\mathrm{tr}(B)$

$\mathrm{tr}(cA)=c\mathrm{tr}(A)$

$\mathrm{tr}(A^{⊺})=\mathrm{tr}(A)$

$\mathrm{tr}(AB)=\mathrm{tr}(BA)$

$\mathrm{tr}(A{B}^{⊺})=\mathrm{tr}(B{A}^{⊺})$

$\mathrm{tr}(ABCD)=\mathrm{tr}(BCDA)=\mathrm{tr}(CDAB)=\mathrm{tr}(DABC)$^[Cyclic property]

$\mathrm{tr}(\mathbf{A})={\sum }_{i=1}^n{\lambda }_i$ The Trace of the matrix is equal to the sum of the eigenvalues of the matrix.

Proof Since the matrix $\mathbf{A}-\lambda \mathbf{I}$ only has $\lambda$ term on diagonal, and the calculation of cofactor deletes a row and a column, the coefficient of ${\lambda }^{n-1}$ of $\text{det}(\mathbf{A}-\lambda \mathbf{I})$ is always $-(a_{11}+a_{22}+\cdots +a_{nn})$. Also, since ${\lambda }_i$ are the solution of the Characteristic Polynomial, the expression is factorized as $\text{det}(\mathbf{A}-\lambda \mathbf{I})=(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\dots (\lambda -{\lambda }_n)$ and the coefficient of ${\lambda }^{n-1}$ become a $-({\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n)$ Therefore, $-(a_{11}+a_{22}+\cdots +a_{nn})=-({\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n)$ and $\text{Trace}(\mathbf{A})={\sum }_{i=1}^n{\lambda }_i$.

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