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

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