# Need help understanding an equation: composition, addition and inverse

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.

• 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. – Arturo Magidin Apr 11 at 16:58
• 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. – Arturo Magidin Apr 11 at 16:59
• 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? – David Apr 11 at 17:58
• 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 $x$, $(f+g)\circ h = (f\circ h) + (g\circ h)$. – Arturo Magidin Apr 11 at 19:41
• @ArturoMagidin Thank you! Now I understand. I was thinking about the whole thing wrong. – David Apr 11 at 22:22