For notational convenience I want to introduce a short-hand notation: Let $A\in\mathbb R^{n\times(k+1)}$ and $x\in\mathbb R^k$. My "product" should represent the affine-transformation $A\bullet x = Sx + a_0$ (where $a_0$ is the first column of $A$ and $S$ is the submatrix of $A$ when the first column $a_0$ is delted). Does this make sense? Are there any problems with this definition? Is there maybe already a similar concept I can build on?

edit: some background: My function looks like this: $$f(x) = A\left[\begin{array}{c} g\left(B\left[\begin{array}{c} h(x) \\ 1\end{array}\right]\right) \\ 1 \end{array}\right]$$ Using the notation I could simply write $f(x) = A\bullet g(B\bullet h(x))$.

  • $\begingroup$ The usual way of performing affine transformations via matrix multiplication is to append an extra (1) to all of your vectors and then let the last column of the matrix represent your constant $a_0$. See en.wikipedia.org/wiki/… $\endgroup$ – Jair Taylor Mar 8 at 0:30
  • $\begingroup$ Unfortunately, my $x$ is a function (which again depends on another function and so on. If I used the idea suggested there, I would end up with $$f(x) = A\left[\begin{array}{c} g\left(B\left[\begin{array}{c} h(x) \\ 1\end{array}\right]\right) \\ 1 \end{array}\right]$$ which looks horrible. $\endgroup$ – Syd Amerikaner Mar 8 at 0:33
  • $\begingroup$ You can define your $h: \mathbb{R}^{k+1} \rightarrow \mathbb{R}^{k+1}$ so that they do not change the last coordinate. Then it just looks like $f(𝑥)=A(g(B(h(𝑥))))$. $\endgroup$ – Jair Taylor Mar 8 at 1:36

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