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I posted this question on math.stackexchange but got no answer, so I decided to post it here instead. Sorry about the impreciness, not professional mathematician here.

Let's assume we have a set of 2D-points, and their coordinates are in a $n \times 2$ matrix. My claim is that if that set has at least one valid symmetry axis, then at least one of those axes is equivalent to one of the two principal component vectors, ie. eigenvectors of the covariance matrix.

For example, let's assume we have points (2,0), (-2,0), (0,1) and (0,-1). The eigenvectors are (1,0) and (0,1) (times some non-zero real number). In this case, those vectors are also the symmetry axes of the set of points.

Then again, take points (0,1), (0,-1), (5,2), (5,-2). Now vector (1,0) is a symmetry axis but (0,1) isn't.

In both cases at least one eigenvector is also a symmetry axis. Is this always the case, and if it is, how could I prove it?

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  • $\begingroup$ The m.se reference is math.stackexchange.com/questions/869684/… $\endgroup$
    – Gerry Myerson
    Commented Jul 21, 2014 at 0:11
  • $\begingroup$ The plural of "axis" is "axes". There is no such word in the English language as "axises". That's why I edited your question. You edited it back. Why? $\endgroup$
    – Gerry Myerson
    Commented Jul 22, 2014 at 0:32
  • $\begingroup$ By accident, I thought I'd left them there. I changed them back. $\endgroup$
    – tziki
    Commented Jul 22, 2014 at 5:23

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It is sufficient to consider the situation where the axis in question is the first coordinate axis (if it is not, apply rotation). In this situation, the matrix of covariances has zero off-diagonal entries. To see this, reflect all your data with respect to the first axis. On the one hand, by doing so you will change the signs of those covariances, on the other hand, the data and covariances won't change.

Once you know that the off-diagonal entries are zero, it is clear that the eigenvectors coincide with the coordinate vectors.

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  • $\begingroup$ Thanks. Correct me if I'm wrong, but this also implies that the existance of symmetry axes for a set of 2D points can be determined in a linear time by taking the two principal component vectors and checking if either of them is a symmetry axis, no? $\endgroup$
    – tziki
    Commented Jul 22, 2014 at 5:25
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    $\begingroup$ In practice this will usually work, but if the covariance matrix is scalar (or numerically indistinguishable from scalar) then every line through the origin is still a candidate symmetry axis. $\endgroup$
    – Noam D. Elkies
    Commented Jul 22, 2014 at 5:30
  • $\begingroup$ @tziki: Yes, but beware of finite precision and the situation described in Noam D. Elkies's comment. $\endgroup$
    – Yuri Bakhtin
    Commented Jul 22, 2014 at 15:10

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