# Moore-Penrose Pseudoinverse identity

I came across the following:

Let $$n \leq m$$, $$L \in \mathbb{R}^{m \times n}$$, with $$\operatorname{Ker} L=\{0\}$$ and $$A \in \mathbb{R}^{n \times m}$$.

Denote by $$L^{\dagger}$$ the Moore-Penrose pseudoinverse of $$L$$, and $$\Pi_F$$ is the orthogonal projector onto $$F$$. We know that $$L^{\dagger}=\left(L^{\top} L\right)^{-1} L^{\top}$$.

I read $$D L^{\dagger}(A)=-L^{\dagger} A L^{\dagger}+\left(L^{\top} L\right)^{-1} A^{\top} \Pi_{(\mathrm{Im} L)^{\perp}}$$, but there is no mention of $$D$$. I suspect that $$D$$ might be a diagonal matrix.

Does this identity (or a corrected version of it) ring a bell for anyone?

• So $n\le m$?.... And do you mean $A\in\mathbb R^{m\times n}$? – amsmath Sep 6 '19 at 21:41
• If you assume that $m=n$ and $A = I$, then the equality reads $DL^{-1} = -(L^{-1})^2+(L^TL)^{-1}P_{(\text{im}L)^\perp}$ and so $D = -L^{-1}$. Just saying... – amsmath Sep 6 '19 at 21:50
• @amsmath yes $n \leq m$, I edited. – user384617 Sep 6 '19 at 21:52
• "I came across the following" Came across where? – darij grinberg Sep 6 '19 at 22:07
• Is $D$ some kind of derivative? As in the exterior differential or something like that? – Malkoun Sep 6 '19 at 22:16

The symbol $$D$$ signifies derivative. Let $$M$$ be an invertible matrix. Since $$M^{-1}=\operatorname{adj}(M)/\det(M)$$ is a rational function in the entries of $$M$$, $$DM^{-1}$$ exists when $$M^{-1}$$ exists. As \begin{aligned} (M+H)(M+H)^{-1}-I &=(M+H)\left(M^{-1}+DM^{-1}(H)+O(\|H\|^2)\right)-I\\ &=MDM^{-1}(H) + HM^{-1} + O(\|H\|^2) \end{aligned} we get $$DM^{-1}(H)=-M^{-1}HM^{-1}$$ and hence $$(M+H)^{-1}=M^{-1}-M^{-1}HM^{-1}+O(\|H\|^2).$$
Similarly, $$L^+=(L^TL)^{-1}L^T$$ is a rational function in the entries of $$L$$. So, it is differentiable whenever it exists. Now, suppose $$A$$ is a small perturbation to $$L$$. Then $$(L+A)^T(L+A)=\underbrace{L^TL}_M+\underbrace{L^TA+A^TL+O(\|A\|^2)}_H=M+H.$$ It follows that \begin{aligned} &\left[(L+A)^T(L+A)\right]^{-1}(L+A)^T-(L^TL)^{-1}L^T\\ =&(M+H)^{-1}(L+A)^T-M^{-1}L^T\\ =&\left(M^{-1}-M^{-1}HM^{-1}+O(\|H\|^2)\right)(L+A)^T-M^{-1}L^T\\ =&\left(M^{-1}-M^{-1}HM^{-1}+O(\|A\|^2)\right)(L+A)^T-M^{-1}L^T\\ =&M^{-1}A^T-M^{-1}HM^{-1}L^T+O(\|A\|^2)\\ =&M^{-1}A^T-M^{-1}\left(L^TA+A^TL+O(\|A\|^2)\right)M^{-1}L^T+O(\|A\|^2)\\ =&-M^{-1}L^TAM^{-1}L^T+M^{-1}A^T(I-LM^{-1}L^T)+O(\|A\|^2)\\ =&-L^+AL^+ + (L^TL)^{-1}A^T(I-LL^+)+O(\|A\|^2). \end{aligned} Therefore $$DL^+(A)=-L^+AL^+ + (L^TL)^{-1}A^T\Pi_{(\operatorname{Im} L)^{\perp}}$$.