Suppose $M$ is a $n$-dimensional matrix and $v$ and $w$ are two vectors in $n-$dimensions. I want to be able to say that $\vert \langle e^{-M \cdot t} v, w \rangle \vert$ or $e^{-M \cdot t} v$ itself, is function monotonically decreasing to $0$ with $t$.
What is a necessary and/or sufficient condition for the above conclusion to hold?
Likewise, is $M$ being a positive definite matrix enough?
For $e^{-Mt}v$ to diminish to zero, what is required is that $v\in E^+(M)$, the subspace generated by eigenspaces of eigenvalues having strictly positive real part. If $M$ is positive definite self-adjoint, then $E^+(M)$ is the whole vector space.
If $v\in E^+(M)$, then $v=\sum_i\alpha_ie_i$ with $Me_i=\lambda_ie_i$, $\mathrm{Re}(\lambda_i)>0$. $$e^{-Mt}v=\sum_i\alpha_ie^{-\mathrm{Re}(\lambda_i)t}e^{\mathrm{i Im}(\lambda_i)t}e_i\to0$$ This is not true when $v$ has a component in $E^0(M)$ or $E^-(M)$.
For $\langle w,e^{-Mt}v\rangle$ to go to zero, the condition can be relaxed to $$w\in(E^0(M)+E^-(M))^\perp$$ because then any component of $v$ in those spaces will be dotted to zero.