I'm reading Szeliski's Computer Vision:Algorithms and Applications. On page 212-213, corner detector is discussed. I can understand upto equation 4.8. From then on, I don't understand how come eigenvectors of A specified the direction of slowest and fastest changes, and why does the shorter eigenvector specified the direction of fastest change?

The book can be found here http://szeliski.org/Book/drafts/SzeliskiBook_20100903_draft.pdf

Any help would be appreciated.

  • $\begingroup$ related: dsp.stackexchange.com/questions/3336/… $\endgroup$ Commented Dec 1, 2014 at 13:56
  • $\begingroup$ thanks nikie. It does not mention eigenvalue and eigenvectors though. It's more for the intuition $\endgroup$
    – nglinh
    Commented Dec 2, 2014 at 19:47
  • 2
    $\begingroup$ The intuitive idea behind using an eigendecomposition is that you "rotate" the image so that the diagonal components of the autocorrelation matrix vanish. Intuitively speaking: you rotate it so that the "canyon" mentioned in the other answer is aligned with the x-Axis. And since $A=Q^{\mathsf{T}}.\Lambda .Q$, the eigenvector matrix $Q$ is exactly the rotation to do that. And the eigenvalues tell you how "steep" the canyon is in the rotated X/Y-Directions. $\endgroup$ Commented Dec 2, 2014 at 20:33
  • $\begingroup$ So what are the points in the diagram (in Szeliski's book). Are they the partial derivatives? In the linked answer, Matrix A were mentioned as covariance matrix. But how come partial derivatives can be come covariance? $\endgroup$
    – nglinh
    Commented Dec 2, 2014 at 20:41
  • $\begingroup$ The partial derivatives come from the first order Taylor series (as explained on page 212, right at the top). $\endgroup$ Commented Dec 3, 2014 at 7:32

1 Answer 1


The eigenvector corresponding to the largest eigenvalue of the autocorrelation matrix indicates the direction of fastest change, while the eigenvector of the smallest eigenvalue indicates the direction of slowest change. Nothing strange here.

What's confusing is the Uncertainty Ellipse (Figure 4.6, page 213). The direction of slowest change is longer because it's scaled by the square root of the inverse eigenvalue of A. Really what it represents is the direction of the highest uncertainty.

Uncertainty and speed of change are different sides of the same coin:

  • high uncertainty <--> slow change
  • low uncertainty <--> fast change

The purpose of all this is to help us find good match points in an image. An example of a good match point is shown in Figure 4.5b. The reason it's "good" is because it has a well defined optimum. This occurs when the image has fast change in more than one direction (i.e. both eigenvalues of the autocorrelation matrix are large).

Figure 4.5c shows a match point that doesn't have a well defined optimum (it's more like a ridge). The fact that we cannot precisely point to the optimum means that there is uncertainty in the direction of the ridge. The autocorrelation function of this patch will have one large eigenvalue and one small eigenvalue. The large eigenvector is in the direction of fastest change (a direction perpendicular to the ridge), while the small eigenvector is in the direction of the ridge.

  • $\begingroup$ I'm not so familiar with terms like certainty and change. What do you mean by uncertainty? Is it the same as variance? And what change are we talking about? $\endgroup$
    – nglinh
    Commented Dec 3, 2014 at 8:49
  • $\begingroup$ @nglinh I updated my answer to hopefully explain things better. Think of change as the image gradient: a big change in a certain direction means a big gradient in that direction. Uncertainty tells us how accurately we can line up a patch with itself: if we can shift an image patch and still get it to line up with itself then we have uncertainty in that direction. For example a patch with an edge would still mostly line up with itself if you shifted it in the direction of the edge. A patch with a corner wouldn't have this problem. $\endgroup$ Commented Dec 3, 2014 at 19:52
  • $\begingroup$ I think my question was not very clear. What I want to ask is that why does Figure 4.6 F $\endgroup$
    – nglinh
    Commented Dec 30, 2014 at 7:39

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