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I'm trying to find a expression for the matrix derivative with respect to the pseudo-inverse of a matrix. So, i have some function $f(A)$ of a matrix $A$, which is singular. If it weren't I could use that $$ \frac{df(A)}{dA^{-1}} = -A^{-1}\frac{df(A)}{dA}A^{-1}, $$ but I can't right? So does anyone know where I could find a pseudo-inverse version of this? So basically I want an expression for $\frac{df(A)}{dA^+}$, and yes I reckon it won't be as cleas and simple as the one above. Also, does anyone know where I could find pseudo-inverse generalizations of all those classic matrix inversion lemmas?

Thanks in advance for any comments!

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2 Answers 2

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To address your second question, here are two useful references:

  1. An Extension of the Matrix Inversion Lemma by Nariyasu Minamide in SIAM J. Alg. and Disc. Methods, 6, pp. 371-377 (1985).
  2. The Moore-Penrose generalized inverse for sums of matrices by J. A. Fill and D. E. Fishkind. (1998)
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  • $\begingroup$ Cool, thanks for the links. This seems to answer the second question pretty well! $\endgroup$
    – alexsuse
    Commented Feb 7, 2012 at 13:55
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@ alexsuse , your formula is false. To see that, choose $u(A)=A^{-1}$.

You want the derivative of the function $g:A^{-1}\rightarrow A\rightarrow f(A)$, that is $g=f\circ u$ with $Du_{A^{-1}}:H\rightarrow -AHA$. Thus $Dg_{A^{-1}}:H\in M_n\rightarrow Df_A(-AHA)$. In other words, if $A^{-1}$ is a function of $t$, then $\dfrac{\partial{g}}{\partial{t}}=Df_A(-A\dfrac{\partial{A^{-1}}}{\partial{t}} A)$.

EDIT: let $t\rightarrow A(t)$ be a real analytic function (I don't know if this is a necessary condition) s.t. $A(0)$ is singular and, for every $t\not=0$ (in a neighborhood of $0$), $A(t)$ is not singular. Then it seems to me that my last formula works in the form $\dfrac{\partial{g}}{\partial{t}}(0)=\lim_{t\rightarrow 0}Df_A(-A\dfrac{\partial{A^{-1}(t)}}{\partial{t}} A)$. See this example: $f(A)=A(t)=\begin{pmatrix}-3t^2+t&5t^2-t\\-6t^2+4t&-t^2+6t\end{pmatrix}$. Yet, if $A(t)$ is singular, then my last formula (replacing $A^{-1}$ with $A^+$) does not work; for instance: $f(A)=A(t)=\begin{pmatrix}2&3\\4+2t&6+3t\end{pmatrix}$. Instead of $\dfrac{\partial{g}}{\partial{t}}(0)=\begin{pmatrix}0&0\\2&3\end{pmatrix}$ we obtain $\dfrac{\partial{g}}{\partial{t}}(0)=\begin{pmatrix}4/5&6/5\\8/5&12/5\end{pmatrix}$.

EDIT: the inverse function $u$ was mixed up with $f$ originally.

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  • $\begingroup$ Not sure what you meant. What is $u$ in your notation, and what do you mean by the $Du_{A^{-1}}:H$ notation? Note that I'm considering the derivative of a matrix with respect to another matrix. In that sense, $df/dA$ is a matrix ($(df/dA)_{ij} = \frac{\partial f}{\partial A_{ij}}$), and so is $df/d(A^{-1})$. In your example, I think you are considering a matrix which is a function of a parameter and looking at the derivative with respect to that parameter. $\endgroup$
    – alexsuse
    Commented Sep 12, 2014 at 10:57
  • $\begingroup$ @ alexsuse, $Du_{A^{-1}}$ is the derivative of $u:X\rightarrow X^{-1}$ in $A^{-1}$; it is a linear application that is defined on $M_n(\mathbb{R})$ with values in $M_n(\mathbb{R})$. The derivative of the function composition is $Dg_{A^{-1}}=Df_A\circ Du_{A^{-1}}$. My last formula is obtained when we compose with any function $t\rightarrow A^{-1}(t)$. Instead of reading the Matrix Cookbook, read a good book about differential calculus. $\endgroup$
    – loup blanc
    Commented Sep 12, 2014 at 11:36
  • $\begingroup$ Such snark, much calculus :D. FWIW, I think you meant $u(A) = A^{-1}$ in the first line. At least it makes more sense that way. And technically speaking, you didn't show that my formula is false. My formula is written in terms of $A^{-1}$, so it is only natural that it would not hold if $A$ is not invertible (that is actually part of my question). That being said, thanks for the comments. $\endgroup$
    – alexsuse
    Commented Feb 3, 2015 at 17:49

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