Matrix Inversion of $f(X) = (X^{-1}-A)$

Question

We have a function $$ f(X) = X^{-1} - A, $$ and I have to show with the help of Newtons's method: $$ X_{k+1} = X_k - Df(X_k)^{-1}f(X_k) $$ that the iteration takes the following form for $f(X_k)$: $$ X_{k+1} = X_k + X_k(I-AX_k). $$ I tried to solve it with: $$ Df(X_k)^{-1}f(X_k) = -f^{-1}(X_k)f(X_k)f^{-1}(X_k), $$ but didn't succeed. Note: $f$ is defined in $\mathbb{R}^{n\times n}$

Answer 1

You could approach this problem using differentials $$\eqalign{ \def\X{X^{-1}} F &= \X - A \\ dF &= -\X\;dX\;\X \\ }$$ Since you want the updated value of the function $F_+$ to equal zero $$dF = \Big(F_+-F\Big) = -F$$ Substitute into the differential expression and solve $$\eqalign{ -F &= -\X\,dX\;\X \\ dX &= XFX \\ (X_+-X) &= X\Big(\X-A\Big)X \\ X_+ &= X \;+\; \Big(X\X X-XAX\Big) \\ X_+ &= X \;+\; X\Big(I-AX\Big) \\ }$$ which has the desired form.

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NB: For ease of typing, I use a streamlined notation which suppresses the iteration counter, i.e. $$\eqalign{ X_+ = X_{k+1},\qquad X = X_k,\qquad dX=dX_k=\big(X_{k+1}-X_k\big) \\ }$$ The initial matrix $X_0$ is very important for this iteration to converge. The following should work $$X_0 = \frac{A^T}{\|A\|^2_F}$$