We rewrite this relation as

$$\langle \mathbf{M}(t) \rangle = \mathbf{\hat{w}}(t)\mathbf{m}=\sum_i\lambda_i^t\alpha_i\mathbf{\hat{v}}_i\mathbf{m}_i.$$

If we define $m_i=\mathbf{\hat{v}}_i\mathbf{m}_i$ as the expectation value of $\mathbf{M}$ in the $i^{\mathrm{th}}$ eigenstate we can rewrite the last equation as

$$\langle \mathbf{M}(t) \rangle = \sum_i\lambda_i^t\alpha_im_i.$$

Since we have that in the limit $t\rightarrow \infty$ the mean value is dominated by the the largest eigenvalue $\lambda_0$, we can rewrite the last equation as

$$\langle \mathbf{M}(t) \rangle = \langle \mathbf{M}(\infty) \rangle+\sum_{i\ne 0}\lambda_i^t\alpha_im_i.$$

We define the quantity

$$\tau_i=-\frac{1}{log\lambda_i},$$

and rewrite the last expectation value as

$$\langle \mathbf{M}(t) \rangle = \langle \mathbf{M}(\infty) \rangle+\sum_{i\ne 0}\alpha_im_ie^{-t/\tau_i}. \tag{21}$$