Limit of regularized inverse matrix consider a (quadratic, but otherwise arbitrary) matrix $A$. I am interested in the long-time behaviour of
$$ r(t) = (I + t\cdot A)^{-1} r_0,$$
i.e. what is $\lim_{t\to\infty} r(t)$?
An interesting arrangement (by inserting $(I+t\cdot A - t\cdot A)$) is the following:
$$r(t) = r_0 - A(t^{-1}I + A)^{-1}r_0$$
The second part looks fairly similar to the limit definition of the Moore-Penrose pseudoinverse: Given a matrix $B$, the M-P pseudoinverse is given by $B^+ = \lim_{t\to\infty} B^T(\frac1 t I + BB^T)$. But as you can see, this doesn't exactly match this form (the problem being that A is not symmetric so it cannot be interpreted as a term of the form $BB^T$).
My hypothesis is something along the lines of "the components of $r$ in the kernel of $A$ stay constant, while the components in the orthogonal complement converge to 0".
 A: We assume that $A$ is a matrix with complex entries. Then there exists a matrix $T$ such that  $A=T^{-1}JT$, where the matrix $J$ has Jordan normal form. Then $(I+tA)^{-1}= T^{-1}(I+tJ)^{-1}T$, provided $t$ is not an eigenvalue of $A$.
Let $f(x)=x^{-1}$. Let $J_{\mu, n}$ be an $n\times n$ Jourdan cell corresponding to an eigenvalue $\mu$, $t\ne 0$, and $\lambda=\tfrac 1t +\mu \ne 0$. According to Wikipedia
$$\displaystyle f(J_{\lambda, n})={\begin{bmatrix}f(\lambda )&f'(\lambda )&{\tfrac {f''(\lambda )}{2}}&\dots &{\tfrac {f^{(n-1)}(\lambda )}{(n-1)!}}\\ 0&f(\lambda )&f'(\lambda )&\dots&{\tfrac {f^{(n-2)}(\lambda )}{(n-2)!}}\\\vdots &\vdots &\ddots &\ddots &\vdots \\ 0&0&0&f(\lambda )&f'(\lambda )\\ 0&0&0&0&f(\lambda )\end{bmatrix}}.$$
Since $f^{(k)}(\lambda)=(-1)^{k}k!\lambda^{-k-1}$ for each non-negative integer $k$, $$\lim_{t\to\infty\, t\ne 0\, 1+t\mu\ne 0} f(1+tJ_{\mu, n})=$$ $$\lim_{t\to\infty\, t\ne 0\, 1+t\mu\ne 0} f(t(\frac 1t+J_{\mu, n}))=$$ $$\lim_{t\to\infty\, t\ne 0\, 1+t\mu\ne 0} \frac 1t f(J_{\lambda, n}).$$
If $\mu\ne 0$ then the latter limit is $0$, because all entries of $J_{\lambda, n}$ are continuous functions with respect to $\lambda\ne 0$, and when $t>0$ tends to infinity then $\lambda$ tends to $\mu$.
If $\mu=0$ then $\lambda=1/t$ and so the entry of  $\tfrac 1t f(J_{\lambda, n})$ corresponding to $k$, that is $\tfrac {f^{(k)}(\lambda)}{k!}$ equals $\tfrac 1t (-1)^{k}\lambda^{-k-1}=(-t)^k$, that is $\tfrac 1t f(J_{\lambda, n})$ diverges if $n>1$. If $n=1$ then $\tfrac 1t f(J_{\lambda, n})=\tfrac 1t f(1/t)=1=\operatorname{const}$.
The above observations determine the convergence behaviors of $(I+tA)^{-1}=T^{-1}(I+tJ)^{-1}T $ (for sufficiently large $t$) and the product  $r(t)=(I+tA)^{-1}r_0$.
