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When E is a minimum, the square root of its value becomes the least RMSD (lRMSD) between and . Being an orthonormal rotation matrix, needs to satisfy the orthonormality property , where is the identity matrix. The orthonormality contraint ensures that the rows and columns are mutually orthogonal, and that their length (as vectors) is one. Any orthonormal matrix represents a rigid orientation (transformation) in space. The only problem with this approach as is, is that all orthonormal matrices encode a rigid transformation, but if the rows/columns of the matrix do not constitute a right handed system , then the rotation is said to be improper . In an improper rotation, one of the three directions may be "mirrored". Fortunately, this case can be detected easily by computing the determinant of the matrix , and if it is negative, correcting the matrix. Denoting as , and moving the constant factor N to the left, the formula for the error becomes:
An alternative way to represent the two point sets, rather than a one-dimensional vector or as separate atom coordinates, is using two 3xN matrices (N atoms, 3 coordinates for each). Using this scheme, is represented by the matrix and is represented by the matrix . Note that column in these matrices stands for point (atom) and , respectively. Using this new representation, we can write:
where and stands for the trace of matrix A, the sum of its diagonal elements. It is easy to see that that the trace of the matrix to the right amounts precisely to the sum on the left (simply carrying out the multiplication of the first row/column should convince the reader). The right-hand side of the equation can be expanded into:
Which follows from the properties of the trace operator, namely: , , and . Furthermore, the first two terms in the expansion above represent the sum of the squares of the components and , so it can be rewritten as:
Note that the components do not need to be primed (i.e., ) since the rotation around the origin does not change the length of . Note that the summation above does not depend on , so minimizing E is equivalent to maximizing . For this reason, the rest of the discussion focuses on finding a proper rotation matrix that maximizes . Remembering that , the quantity to maximize is then . From the property of the trace operator, this is equivalent to . Since is a square 3x3 matrix, it can be decomposed through the Singular Value Decomposition technique (SVD) into , where and are the matrices of left and right eigenvectors (which are orthonormal matrices), respectively, and is a diagonal 3x3 matrix containing the eigenvalues , , in decreasing order. Again from the properties of the trace operator, we obtain that:
If we introduce the 3x3 matrix as the product , we can rewrite the above expression as:
Since is the product of orthonormal matrices, it is itself an orthonormal matrix and . This means that the absolute value of each element of this matrix is no more than one, from where the last equality follows. It is obvious that the maximum value of the left hand side of the equation is reached when the diagonal elements of are equal to 1, and since it is an orthonormal matrix, all other elements must be zero. This results in . Moreover, since , we can write that , and because and are orthonormal, and . Multiplying by to the left and to the right yields a solution for :
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