Continuity of Moore-Penrose generalized inversion

Question

Any matrix $A\in\mathbb{C}^{m\times n}$ has a unique generalized inverse $A^{\dagger}\in\mathbb{C}^{n\times m}$ with the properties $$AA^{\dagger}A=A,\qquad A^{\dagger}AA^{\dagger}=A^{\dagger},\qquad (AA^{\dagger})^*=AA^{\dagger}, \qquad (A^{\dagger}A)^*=A^{\dagger}A,$$ where $*$ is the conjugate transpose. I am mostly interested in the case of square matrices. If $A$ is invertible, its generalized inverse is really its inverse.

Under this generalized inversion $\mathbb{C}^{n\times n}$ is an inverse semigroup. It is known that, unlike the usual matrix inversion, $A\mapsto A^{\dagger}$ is not continuous. To see this, let $$ A=\begin{bmatrix} 1 & 0 \\ 1 & 0 \\ \end{bmatrix} , \qquad E=\begin{bmatrix} 0 & 0 \\ 0 & 1 \\ \end{bmatrix}.$$

Then, for any $\varepsilon\neq 0,$ $$ A^{\dagger}=\begin{bmatrix} \frac12 & \frac12 \\ 0 & 0 \\ \end{bmatrix} , \qquad (A+\varepsilon E)^{\dagger}=\begin{bmatrix} 1 & 0 \\ -\frac1\varepsilon & \frac1\varepsilon \\ \end{bmatrix}.$$

Clearly $\lim_{\varepsilon \to 0}(A+\varepsilon E)^{\dagger}\neq A^{\dagger}.$ Yet, restricted to $\operatorname{GL}_n(\mathbb{C})$ this is a continuous operation. So, my question is, what do we know about the largest inverse subsemigroup of $\mathbb{C}^{n\times n}$ on which this inversion is continuous?

Answer 1

The answer is indicated by the geometric description of the Moore-Penrose generalized inverse, which is shown here. The observation is that the restriction of a nonzero matrix to the orthogonal complement of its kernel is a bijection onto the image of the matrix. The generalized inverse first projects a vector in the codomain orthogonally onto the image, and then maps the projected vector back to the orthogonal complement of the kernel.

From this, it is clear that the map from a matrix to its generalized inverse is continuous when restricted to matrices of a given rank $k$. On the other hand, one would expect it to be discontinuous with respect to a perturbation where the rank jumps up.

I believe this can all be proved by using the normal form $A = PDQ$, where $P, Q$ are orthogonal and $D$ is diagonal.

This all works for rectangular and not just square matrices.

From all this, it appears that the largest inverse semigroup on which the generalized inverse is continuous is the set of matrices with maximal rank.