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Let $E$ be complex Hilbert space and $mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.

Definition: Let $T in mathcal{L}(E)$. The Moore-Penrose inverse of $T$, denoted by $T^{+}$, is defined as the unique linear extension of $(bar{T})^{-1}$ in
$$D(T^{+}) = mathcal{R}(T)+mathcal{R}(T)^{perp},$$
with $mathcal{N}(T^{+}) = mathcal{R}(T)^{perp}$ and $bar{T}$ is the isomorphism
$$bar{T}:=T|_{{mathcal{N}(T)}^{perp}}: {mathcal{N}(T)}^{perp} longrightarrow mathcal{R}(T).$$
Moreover, $T^{+}$ is the unique solution of the four ''Moore-Penrose equations'':
$$TXT = T,quad XTX = X,quad XT = P_{N{(T)^{bot}}},,mbox{and},,quad TX = P_{overline{mathcal{R}(T)}}{{|}_{D(T^{+})}}.$$

Here $mathcal{R}(T)$ and $mathcal{N}(T)$ denote respectively the range and the nullspace of $T$. Also $P_{F}$ denote the orthogonal projection onto $F$.

It is well known that $T^{+}$ is bounded if and only if $T$ has a closed range.

According to this answer, we have

• If $A$ is selfadjoint matrix, then
  $$ A^{+}= lim_{t to 0}(A^2+tI)^{-1} A.;;(1).$$
  Note that this limit is with respect to the strong topology.




• If $A$ is not selfadjoint, then $A^+=A^*(AA^*)^+$ (which is equal to $(A^*A)^+A^*$).

How we prove the formula $(1)$?

In a general situation .i.e. when $T$ is an operator and not a matrix, it is possible to make sense of (1) as a strong limit on the domain of $T^+$?

Yes, (1) is true for a hermitian matrix $A$ but not for any matrix.

Proof. Up to a change of orthonormal basis, we may assume that $A=diag(0_p,lambda_1,cdots,lambda_q)$ where $lambda_inot=0$ and $p+q=n$.

Then $A^+=diag(0_p,1/lambda_1,cdots,1/lambda_q)$.

On the other hand, if $tnot= 0$ and $tnot=-lambda_i^2$, then $(A^2+tI)^{-1}A=diag(0_p,lambda_1/(lambda_1^2+t),cdots,lambda_q/(lambda_q^2+t))$ and, since there is only a finite number of non-zero eigenvalues, the limit is clearly $A^+$.

To obtain a counter-example for any $A$, choose a random $A$ with rank $<n$. $square$

EDIT and CORRECTION.

i) Practically, we obtain an approximation $B$ of $A^+$, giving a small value $t_0$ to $t$.

Proposition. The error $||B-A^+||$ is $approx t_0/a^3$ where $a=min{|λ|;λin spectrum(A) setminus{{0}}}$.

Proof. Indeed $Delta=1/lambda-lambda/(lambda^2+t)=dfrac{t}{lambda_i(lambda_i^2+t)}sim t/lambda^3$ when $trightarrow 0^+$.

Note also (cf. below) that $(*)$ $Deltasim 1/lambda_i$ when $lambda_irightarrow 0$ and $t$ is fixed.

ii) Now, in a Hilbert, we can consider a bounded, compact, self-adjoint operator $H$ and use the Hilbert-Schmidt theorem in order to diagonalize it. Unfortunately, there is a sequence of non-zero real eigenvalues $lambda_i, i = 1, ..., N$, with $N=rank(A)$, such that $||lambda_i||$ is monotonically non-increasing and, when $N=infty$, $lambda_irightarrow 0$.

In this last case, according to $(*)$, the considered (strong) limit does not exist, even if $trightarrow 0^+$.

That works only if $H$ has a finite number of distinct eigenvalues.

iii) Clearly, when $H$ is a bounded self-adjoint operator, it works even worse.