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 $<n$. $\square$


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.