Suppose a 3D configuration of points is given, $X\in\mathbb{R}^{n\times 3}$, and a matrix $Q\in\mathbb{3\times 2}$, with orthonormal columns. Now, suppose a mapping to 2D is obtained as $$Y=XQ.$$ Could it be safely assumed that $Y$ simply corresponds to a rotated or reflected configuration $X$ restricted to first 2 axes? I read that the application of orthogonal matrices, as in the above, preserves isometries in Euclidean space, meaning that it should correspond a rotation, reflection and translation.

However, in


(p.4), it is stated that "Orthogonal matrices with determinant $-1$ are not rotations, but most of them are not reflections either". On the other hand, the statement from


(p.1) makes a general statement on orthogonal matrices corresponding to a rotation and reflection. A claim that a $3\times 3$ orthogonal matrix $Q$ with determinant $-1$ corresponds to a rotation + reflection is given in


(p. 3). As you may observe, the claims are not the same, i.e., there is no certainty that the application of any orthogonal $Q$ corresponds to a rigid transformation of on a configuration (rotation and/or reflection). So, I'm interested if a general statement that the application of an orthogonal matrix simply corresponds to a rigid transformation.


Another concept that adds to confusion is the orthogonal projector matrix, $P\in\mathbb{R}^{n\times n}$. What does $$Z=PX$$ imply, and how does it differ from the above $Y=XQ$?


1 Answer 1


(i) Think of $Q$ to be the first two columns of an orthogonal matrix $U$. The rows of $XU$ are the points viewed in a rotated and/or reflected coordinate system, (i.e., a rotation or a rotation followed by a reflection) and distances are preserved in this transformation. (As 0 is preserved, there is no translation involved.)

Discarding the last column of the transformed data means that you look at a 2-dimensional projection of the rotated/reflected point set. The projection generally changes distances. Thus multiplication with rectangular orthogonal matrices need not be an isometry, and in your case it isn't. Thus your transformation is not rigid. (You can check this easily by comuting a few random distences before and after the transformation.)

But as you can change the sign of the third column of $U$, you can always make the determinant 1, you may interpret your transformation as the result of a rotation only.

(ii) $Z=PX$ is a completely unrelated transformation, as it mixes different points to create a new set of points.

  • $\begingroup$ For (i): I actually meant the rigid transformation in the full dimensional space, i.e., $Q\in\mathbb{R}^{3\times 3}$. So, isometry is preserved in 3D, and clearly not in 2D. I suppose it's safe to assume that the mapping $Y=XQ$ corresponds to a axonometric parallel projection ? Just an additional note: what would happen if the columns of $Q$ were not orthonormal? What kind of transformation on the original configuration would that than correspond to? I guess a rotation/reflection with some shearing (subset of an affine transformation). For (ii): clear $\endgroup$
    – usero
    May 18, 2012 at 14:00
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    $\begingroup$ If the columns are not orthonormal, you just get an affine transformation. You can perform a SVD of $Q$ and get the transformation represented as a product of a 3D rotation, a scale transformed projection, and a rotation in the 2D image space. $\endgroup$ May 18, 2012 at 14:17
  • $\begingroup$ A very good approach to think about it. I suppose that by "scale transformed projection" you mean the effect of a diagonal matrix with singular values, hence a different scale transformation applied to different axes? $\endgroup$
    – usero
    May 18, 2012 at 19:22

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