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I might be wrong. Assume matrix $M$ as a data matrix (e.g., a 2D image). $M$'s rank represents the underlying dimension of the data (or the degree of freedom). For example, $M$ usually isn't full-rank, which means the underlying data lie on some low-dimensional manifold. Now if I rotate the matrix with regard to its centre, even with a tiny angle, the rank of the matrix changes (and the change could be significant). For example, assume $M$ is a matrix of alternating allzero- and allone- columns, like the following:

$$\left(\begin{array}{ccccc} 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0 & 1\\ \end{array}\right)$$ The rank of this matrix is $1$. But even if you rotate it with $0.1$ degree, it becomes full-rank. On the other hand, however, the underlying data hasn't changed during the rotation, which means it still lies on the same low-dimensional manifold. My questions are:

1), Does this conflict that the matrix-rank represents the underlying dimension of the data (or the degree of freedom)?

2), I guess there exist some theory about such matrix-rotation and data dimension, etc. What's the theory? I don't even know what keywords to use for searching the theory. 'matrix rotation' always returns results about 'rotation matrix'..

Thank you all!

M47145
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flyinggip
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  • What do you mean by "rotate the matrix with regard to its centre"? – JiK Jan 18 '14 at 17:37
  • I think user44197's description is more accurate than mine. Basically it can be seen as rotating the target while keeping the camera fixed, or rotating the camera while keeping the target fixed. – flyinggip Jan 19 '14 at 12:45
  • Ok, then what do you think is the "underlying dimension of the data" of a $1000\times1000$ identity matrix? If I understand you correctly, it is an image of a diagonal line but has rank 1000. – JiK Jan 19 '14 at 13:55
  • Yes that's another example of my question. On one hand, the underlying dimension of the I matrix is 1 million since it's full-rank. On the other hand, it's an image of 1 line so its underlying dimension should be low. For example, if you rotate it 45 degrees to change get a 'allones' column, the underlying dimension becomes 1999. – flyinggip Jan 19 '14 at 14:19

1 Answers1

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Note you are not "rotating" the matrix. You are rotating the indices. The matrix you have can be (and is) viewed as a function of $x$ and $y$ (or rows and columns) as

$$ M = M(x,y) = M(p) \text{ where $p = (x,y)^T$ is vector representing the point} $$ When you rotate the image you are doing $$ \hat M(p) = M(R^{-1} p) $$ where $R$ is a rotation matrix.

You are confusing this with rotating the matrix $$ \hat M = R M R^{-1} $$ where $R$ is a rotation matrix.

The two are different. If you view $M$ as a function, the first is rotation in the domain space and the second is rotation in the range space.

user44197
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  • Well, while I think this is more accurate description of my question, is there still my problem? You rotate the coordinates while keeping the camera static, then take a shot. The result is a different image of the same data, however this new image probably has a different rank from the one taken before the rotation. How to explain this? – flyinggip Jan 19 '14 at 12:43
  • Because rank has no meaning. In mathematics you use matrices in a variety of ways. One is to encode data. This is what image processing does. The other is to represent linear transformation. Concept of rank is meaningful only if you use matrices for linear transformation. In the first case it is only a rectangular array for organizing data. If you are trying to use rank for image compression using KLT or similar methods, rotation will introduce very small eigenvalues whose components can be safely ignored. – user44197 Jan 19 '14 at 16:41