# Prove a formula related to the Moore-Penrose pseudo-inverse of operators

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 $$\lambda_i\not=0$$ 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 $$. $$\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.

• 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. – Student Dec 24 '18 at 8:32
• 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 $a=min\{|\lambda|; \lambda\in spectrum(A)\setminus\{0\}\}$ – loup blanc Dec 24 '18 at 10:37