I know that the matrix representation of the linear transformation:
$x_1=X_1+2\lambda X_2$
$x_2=X_2 - \lambda X_1$


$ \left[ {\begin{array}{c} x_1\\ x_2\\ \end{array} } \right]= \left[ {\begin{array}{cc} 1 & 2\lambda \\ -\lambda & 1 \\ \end{array} } \right] \left[ {\begin{array}{cc} X_1\\ X_2\\ \end{array} } \right]$

but what if I have the non-linear transformation:
$x_1=X_1+2\lambda X_2X_1^2$
$x_2=X_2 - \lambda X_1$

How can it be expressed in matrix form to get it's inverse?


So, if i cannot represent it in matrix form, how can I get the inverse transformation?

  • $\begingroup$ Matrices are for expressing linear transformations. You cannot express something nonlinear in $x$ in the form $Ax$ for any matrix $A$. $\endgroup$ – Joppy Mar 15 '18 at 21:55
  • 1
    $\begingroup$ Not quite true. Quadratic forms are not linear and still one can express them using a matrix. $\endgroup$ – H. Gutsche Mar 15 '18 at 22:06
  • $\begingroup$ Thanks a lot! I edited the question, I did not know whether to comment or edit it $\endgroup$ – Joaquín Rico Mar 15 '18 at 22:07
  • $\begingroup$ What does an inverse transformation have to do with a matrix? An inverse transformation is the way to get back to the origin from an image, linear or not. $\endgroup$ – H. Gutsche Mar 15 '18 at 22:09
  • $\begingroup$ Well, I just know that method to invert the transformation. $\endgroup$ – Joaquín Rico Mar 15 '18 at 22:19

If $\lambda=0$, the transformation is clearly a bijection (geometrical interpretation : it is a symmetry with respect to the line with equation $y=x$),

For any nonzero value of $\lambda$, this transformation cannot be a bijection because point (0,2) would be the image of 3 different points by the inverse transformation. These points are :

$$\tag{1}(X_1,X_2)=\begin{cases}(0,2)&\ \text{it's a fixed point.}\\ (\frac{1}{\lambda}(-1+a),1+a)&\\ (\frac{1}{\lambda}(-1-a),1-a)& \end{cases} \ \ \ \text{with} \ \ \ a:=\tfrac{\sqrt{2}}{2}$$

Using other words, your system taken with $x_1=0$ and $x_2=2$, i.e.,:

$$\tag{2}\begin{cases}(i)&...&0&=&X_1+2\lambda X_2X_1^2\\ (ii)&...&2&=&X_2 - \lambda X_1\\ \end{cases}$$

has the 3 solutions given in (1).


  • The second and third solutions in (1) are easily found by expressing in (2)(ii) $X_2=\lambda X_1+2$ and then plugging this expression into (2)(i), getting a quadratic equation for $X_1$.

  • All points $(0,y)$ are fixed points of the transformation.

  • There are many cases where, being given $(x_1,x_2)$, there are two solutions for $(X_1,X_2)$.

  • $\begingroup$ Thanks for your answer. If I want to show mathematically that the transformation is not bijective from a certain $\lambda$ and what $\lambda$ is it, what path can I take? And what path to get the inverse transformation inside the domain where it's bijective? $\endgroup$ – Joaquín Rico Mar 15 '18 at 23:49
  • $\begingroup$ I have a definitive answer now : unless $\lambda=0$, it is never bijective (see why in my answer). $\endgroup$ – Jean Marie Mar 16 '18 at 15:21

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