Upper bounds on the approximate inverse of a singular matrix

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.

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 $||E||\ge1$.

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 $x$
• $Px = 0 \iff Ax = 0$

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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.