Suppose
$\vec{n_{1}} = \begin{bmatrix} n_{1x} & n_{1y} & n_{1z}\\ \end{bmatrix}$
$\vec{n_{2}} = \begin{bmatrix} n_{2x} & n_{2y} & n_{2z}\\ \end{bmatrix}$
$\vec{\sigma} = \begin{bmatrix} \sigma_{x}\\ \sigma_{y}\\ \sigma_{z}\end{bmatrix} $ where each sigma is a Pauli - operator matrix.
How can I convince myself that
$Tr[(\vec{n_{1}}\cdot\vec{\sigma})(\vec{n_{2}}\cdot\vec{\sigma})] = \sum_{k,l}(n_1)_{k}(n_{2})_l2\delta_{kl}$
The pauli operators are defined below: https://en.wikipedia.org/wiki/Pauli_matrices
My attempt: \begin{align} (\vec{n}_1\cdot\vec{\sigma})(\vec{n}_2\cdot\vec{\sigma})&=(\vec{n}_1\cdot\vec{n}_2)I+i(\vec{n}_1\times\vec{n}_2)\cdot\vec{\sigma}\\ &=(\vec{n}_1\cdot\vec{n}_2)I+i(\vec{n}_1\times\vec{n}_2)_x\sigma_x+i(\vec{n}_1\times\vec{n}_2)_y\sigma_y+i(\vec{n}_1\times\vec{n}_2)_z\sigma_z \end{align} Thus, \begin{align} \text{Tr}[(\vec{n}_1\cdot\vec{\sigma})(\vec{n}_2\cdot\vec{\sigma})]&=(\vec{n}_1\cdot\vec{n}_2)\text{Tr}(I)+i(\vec{n}_1\times\vec{n}_2)_x\text{Tr}(\sigma_x)+i(\vec{n}_1\times\vec{n}_2)_y\text{Tr}(\sigma_y)+i(\vec{n}_1\times\vec{n}_2)_z\text{Tr}(\sigma_z)\\&=2(\vec{n}_1\cdot\vec{n}_2). \end{align} Of course, \begin{align} 2(\vec{n}_1\cdot\vec{n}_2)=2\sum_{k=1}^3(n_1)_k(n_2)_k=\sum_{k=1}^3\sum_{l=1}^3(n_1)_k(n_2)_l2\delta_{kl}. \end{align}