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I have found an interesting paper on a digital image registration algorithm. There are many equations in the paper that I only understand partially, but there is a particular one I would like to understand better (at the top of page 4).

$(T^{(-1)}+w)^{(-1)}=T \circ(Id + w \circ T)^{(-1)}$

Where $T$ is a non-parametric transformations that map every point $x$ in image $I$ to $T(x)$ in image $J$. We can find $T^{(-1)}$ iteratively: $T^{(-1)}_{k+1}(x)=(T^{(-1)}_k\circ w_k)(x)$ where $w$ is also a transformation called the adjustment field.

My questions are:

1: What are the steps needed to arrive from the LHS of the equation to the RHS?

2: What are the necessary assumptions for this equation to hold?

Just by looking at the equation, my guess is that something like this might have happened:

$(T^{(-1)}+w)^{(-1)}=T\circ T^{(-1)} \circ (T^{(-1)}+w)^{(-1)}$

$T\circ T^{(-1)} \circ (T^{(-1)}+w)^{(-1)} =T\circ (T^{(-1)}\circ T+w \circ T)^{(-1)}=T \circ(Id + w \circ T)^{(-1)}$

But I don't know why would it be possible to pass $T^{(-1)}$ through the parenthesis or why would the function composition be distributive with addition, or even if this is really the way this equation was done.

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    $\begingroup$ If $A$ and $B$ are invertible, then $A\circ B$ is invertible, and $(A\circ B)^{-1} = B^{-1}\circ A^{-1}$. So you can go from $T^{-1}\circ(T^{-1}+wI)^{-1}$ to $\bigl( (T^{-1}+wI)\circ T^{-1}\bigr)^{-1}$, and then use the fact that composition of linear transformations distributes over the sum of linear transformations to get the result. $\endgroup$ – Arturo Magidin Apr 11 at 16:58
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    $\begingroup$ If $T$, $U$, and $R$ are linear transformations, then $(U+R)T = UT+RT$. You can verify this by simply noting that they evaluate to the same thing at every vector. $\endgroup$ – Arturo Magidin Apr 11 at 16:59
  • $\begingroup$ Thank you for the answer @ArturoMagidin . I am pretty sure however, that these transformations in the paper are non-linear. Do you think the equation can be solved for non-linear transformations as well? $\endgroup$ – David Apr 11 at 17:58
  • $\begingroup$ The first comment (on inverses) is valid in general. The second is valid in all contexts where the expression makes sense. By definition the function $(f+g)$ is defined by $(f+g)(x) = f(x)+g(x)$ (evaluate each function, add the results). So $(f+g)\circ h$, evaluated at $x$, gives $(f+g)\circ h(x) = (f+g)(h(x)) = f(h(x)) + g(h(x)) = (f\circ h)(x) + (g\circ h)(x) = ( (f\circ h) + (g\circ h) )(x)$; as this hold for any $x$, $(f+g)\circ h = (f\circ h) + (g\circ h)$. $\endgroup$ – Arturo Magidin Apr 11 at 19:41
  • $\begingroup$ @ArturoMagidin Thank you! Now I understand. I was thinking about the whole thing wrong. $\endgroup$ – David Apr 11 at 22:22

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