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I'm asked to find ALL affine transformations from $\mathbb{R^3}$ to itself which satisfy that the point $(-1,2,2)$ is fixed and that the lines $$\textbf{a}: y-z=y+z-2=0$$ and $$\textbf{b}: z-1=x-z=0$$ are invariant.

So, I think I've found successfully ONE affine transformation satisfying all of them at the same time, which is $$T(x,y,z)=(-x-4z+6,-y+2z,z)$$ I did it sending $(-1,2,2)$ to itself, sending the intersection point of both lines $(1,1,1)$ to itself and choosing two random points $(0,1,1)$ and $(1,0,1)$ from the lines and associating them to another random point $(2,1,1)$ and $(1,2,1)$ on their respective lines, respectively. Of course, point $(-1,2,2)$ is fixed and I've checked several points contained in the lines above and seen that their images are still on the same lines. The problem is that, supposedly, I think that there could be more than one affine transformation satisfying those conditions at the same time. Maybe, a family of affine transformations depending on a parameter. How can I get all possible affine maps satisfying the conditions given or prove that an affine map is unique under a certain set of circumstances (although this might not be the case in this exercise)?

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You’ve made a good start by identifying a second fixed point of the transformation. The required family of affine transformation has two degrees of freedom remaining. To see this, consider a nonsingular affine transformation on a line: it consists of a scaling followed by a translation. For each of the two invariant lines, then, if we choose an arbitrary nonzero scale factor, the translation part of the restricted affine transformation is determined by the fixed intersection point, so we can take those two arbitrary scale factors as parameters of the transformation.

I find it convenient to work in homogeneous coordinates for problems like these. An affine transformation can be represented by a single matrix, and its invariant flats can be described in terms of eigenvectors of this matrix. So, we seek a matrix of the form $$M = \begin{bmatrix}m_{11}&m_{12}&m_{13}&m_{14}\\m_{21}&m_{22}&m_{23}&m_{24}\\m_{31}&m_{32}&m_{33}&m_{34}\\0&0&0&1\end{bmatrix}.$$ The two fixed points are eigenvectors of $M$ with eigenvalue $1$. The directions of the two invariant lines must also be preserved, so the homogeneous vectors that represent their directions are also eigenvectors of $M$, but with arbitrary nonzero eigenvalues (these are the scale factors from above). The direction vectors of the two lines can be found by computing the cross products of the normals of the planes in their defining equations.

Using the “bulk” eigenvector equation $MX=X\Lambda$, we find that $$M = \begin{bmatrix}-1 & 1 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0\end{bmatrix} \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&\lambda&0\\0&0&0&\mu\end{bmatrix} \begin{bmatrix}-1 & 1 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0\end{bmatrix}^{-1} = \begin{bmatrix}\lambda & 0 & 2(\lambda-1) & 3(1-\lambda) \\ 0 & \mu & 1-\mu & 0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix},$$ i.e., $$T:\begin{bmatrix}x\\y\\z\end{bmatrix} \mapsto \begin{bmatrix} \lambda x+2(\lambda-1)z+3(1-\lambda) \\ \mu y+(1-\mu)z \\ z\end{bmatrix}$$ for $\lambda\mu\ne0$. The transformation you found is obtained by setting $\lambda=\mu=-1$.

By construction, the two fixed points are mapped to themselves, but it’s worth checking anyway to ensure that you haven’t made any algebraic errors along the way. In addition, $T(1,0,0)=(\lambda,0,0)$ and so $(1,1,1)+t(1,0,0)$ is mapped to $(1,1,1)+\lambda t(1,0,0)$, which is obviously another parameterization of the same line. A similar simple calculation verifies that the other line is also mapped onto itself.

You can also make a slight modification to your solution method to get the general solution: Instead of mapping $(0,1,1)$ and $(1,0,1)$ to specific points on the respective lines, map them to the arbitrary points $(1,1,1)+\lambda(1,0,0)$ and $(1,1,1)+\mu(0,1,0)$, respectively, and then carry out the same calculations that you made to obtain the one transformation that you did find.

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  • $\begingroup$ Your answer was really helpful !!! but I didn't understand how you built the last row of the X matrix. I mean, for that matrix, you put the two fixed points and the vectors of the lines in columns but in the last row you put [1 1 0 0]. Why is it like that? (it may be elementary but I really appreciate your help if you can help with that last thing) $\endgroup$ – Mary Clark Dec 31 '18 at 2:34
  • $\begingroup$ @MaryClark Each column is the homogeneous coordinate vector of a point. For a finite point, this is obtained by appending a $1$ to the inhomogeneous coordinate vector; for a point at infinity (direction vector), you append $0$ instead. $\endgroup$ – amd Dec 31 '18 at 8:34
  • $\begingroup$ Everything clear. Answer accepted. Thanks! $\endgroup$ – Mary Clark Dec 31 '18 at 15:54

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