I found a class paper that mentioned that if we have $1-1$ function $f$ that takes a vector, $(X_1, X_2, \ldots, X_n)$ and returns:

$$ f(X_1, \ldots, X_n) = \left(\sum_{i=1}^{n}X_i, X_2, \ldots, X_n\right) $$

then the inverse is defined to be:

$$ f^{-1}(y, y_2, \ldots, y_n) = \left(y-\sum_{i=2}^{n}y_i, y_2, \ldots, y_n\right) $$

where $y$ is defined as: $y = f(X_1, \ldots, X_k) = \sum_{i=1}^{n}X_i$.

I am wondering how I can find the inverse and where exactly the subtraction of the first entry of the inverse comes in. Would anyone have any ideas? Thanks.

  • 1
    $\begingroup$ "where $y$ is defined as..." No, $y$ is not defined as what follows in your post, actually $y$ is simply part of the argument for the function $f^{-1}$ hence $y$ does not need to be defined. $\endgroup$ – Did Sep 6 '16 at 7:09

We set :

$$f(X_1, \ldots, X_n) = \left(\sum_{i=1}^{n}X_i, X_2, \ldots, X_n\right)=(y, y_2, \ldots, y_n).$$

Then we have the following linear system :

$$\left\{\begin{matrix} \sum_\limits{i=1}^{n}X_i = y_1\\ X_2 = y_2 \\ \vdots\\ X_n = y_n \end{matrix}\right. \Leftrightarrow\left\{\begin{matrix} X_1+\sum_\limits{i=2}^{n} y_i = y_1\\ X_2 = y_2 \\ \vdots \\ X_n = y_n \end{matrix}\right.$$

Since $\forall i, n\geq i \geq 2, X_i=y_i $. Then the system is equivalent to : $$\left\{\begin{matrix} X_1 = y_1-\sum_\limits{i=2}^{n} y_i\\ X_2 = y_2 \\ \vdots \\ X_n = y_n \end{matrix}\right.$$

Then $$f^{-1}(y_1,\dots,y_n)=(X_1,\dots,X_n)=\left(y_1-\sum_\limits{i=2}^{n} y_i,y_2,\dots,y_n\right).$$


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