Upper bounds on the approximate inverse of a singular matrix Let $A$ be a singular matrix and $\zeta>0$ such that $A+\zeta I$ is nonsingular. 
Soft question: Is it true that if $\zeta$ is sufficiently small, then $$(A+\zeta I)^{-1}A \approx I$$ (or $A(A+\zeta I)^{-1} \approx I$)? 
More precisely: Let $E:=(A+\zeta I)^{-1}A - I$. If the answer to the above (soft) question is "yes", what are some known bounds on $E$?
Note. Using the condition number $\kappa(A)$, a well-known bound is 
$$
\frac{\Vert(A+B)^{-1} - A^{-1}\Vert}{\Vert A^{-1}\Vert}
\le \kappa(A)\frac{\Vert B\Vert}{\Vert A\Vert },
$$
but this bound requires that both $A$ and $A+B$ are nonsingular. In my question, $A$ is singular while $A+B$ is nonsingular.
 A: Try some trivial test cases, like
$$
A=\pmatrix{1&0\\0&0}\implies A(A+\zeta I)^{-1}=\pmatrix{(1+ζ)^{-1}&0\\0&0}
$$
which is nowhere close to the identity matrix.
A: There must exist some bound on $E$. But it's not true that $||E||\to0$ as $\zeta\to0$, which would seem to say that the answer to  the slightly fuzzy question of whether $(A+\zeta I)^{-1}A \approx I$ is no:
There exists $x\ne0$ with $Ax=0$. Hence $Ex=-x$, so $||E||\ge1$.
A: When $A$ is diagonalizable, we can find a change of basis $S$ such that $M = S^{-1}AS$ is diagonal, and moreover
$$
M = \pmatrix{D & 0\\0 &0}
$$
where $D$ is a diagonal and invertible matrix.  We see that
$$
M(M + \zeta I)^{-1} = \pmatrix{D(D + \zeta I)^{-1} & 0\\0 &0}.
$$
As $\zeta \to 0$, we see that $M(M + \zeta I)^{-1}$ approaches the projection onto the range of $M$ along the kernel of $M$.  Since we merely applied a change of basis, we can conclude that $A(A + \zeta I)^{-1}$ approaches the projection $P$ onto the range of $A$ along the kernel of $A$.
That is, $A(A + \zeta I)^{-1} \to P$ where $P$ satisfies:


*

*$P^2 = P$

*$PAx = Ax$ for all $x$

*$Px = 0 \iff Ax = 0$



In general, $A(A + \zeta I)^{-1}$ approaches the projection onto the image of $A^n$ along the kernel of $A^n$, where $n$ is the size of the matrix.  This holds whether or not $A$ is diagonalizable.
