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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 $T^+$?

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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 $\lambda_i\not=0$ and $p+q=n$.

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

On the other hand, if $t\not= 0$ and $t\not=-\lambda_i^2$, 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 $a=\min\{|λ|;λ\in spectrum(A) \setminus{\{0\}}\}$.

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

Note also (cf. below) that $(*)$ $\Delta\sim 1/\lambda_i$ when $\lambda_i\rightarrow 0$ 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 $N=\infty$, $\lambda_i\rightarrow 0$.

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

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.

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  • $\begingroup$ Thank you for your answer, please is the formula $(1)$ is well known in the literature? If yes do you know a reference ? because I want to cite it. $\endgroup$ – Student Dec 24 '18 at 8:32
  • $\begingroup$ No, I did not know this formula. Practically, you obtain an approximation of $A^+$, giving a small value to $t$. The error is $\approx t/a^3$ where $a=min\{|\lambda|; \lambda\in spectrum(A)\setminus\{0\}\}$ $\endgroup$ – loup blanc Dec 24 '18 at 10:37

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