video01 matrix det riemann manifold A vector is a set of scalar variables Theorema Egregium 绝妙定理, This matrix is singular, the determinant is zero,

https://tel.archives-ouvertes.fr/tel-00768768/document

http://wwwx.cs.unc.edu/~mn/classes/comp875/doc/diffeomorphisms.pdf
det=把矩陣 matrix 的方括號以細長的垂直線取代

It seems like a homeomorphism is all we want. Because it • prevents folding and • does not allow for tearing either. What else could we ask for? The function f : x 7→ x 3 , f −1 : x 7→ x 1 3 , x ∈ R, is homeomorphic. But the derivative of f −1 is not defined at 0. −1 −0.5 0 0.5 1 −1 −0.5 0 0.5 1 −1 −0.5 0 0.5 1 −1 −0.5 0 0.5 1 Will show later that this is not a diffeomorphism.

Introduction to Smooth Manifolds

By John M. Lee

A new graph density - Self Journal of Science

sjscience.org/article?id=205

Apr 8, 2015 - A graph volume is a function from set of all graphs to : ... When we add a new edge between two existed vertices, the new volume (after edge ...

http://image.diku.dk/hauberg/sommeretal10.pdf

4.3 Human Poses In this experiment, we consider human poses obtained using tracking software. A consumer stereo camera4 is placed in front of a test person, and the tracking software described in [10] is invoked in order to track the pose of the persons upper body. The recorded poses are represented by the human body end-effectors; the end-points of each bone of the skeleton. The placement of each end-effector is given spatial coordinates so that an entire pose with k end-effectors can be considered a point in R 3k . To simplify the representation, only the end-effectors of a subset of the skeleton are included, and, when two bones meet at a joint, their end-points are considered one end-effector. Figure 5 shows a human pose with 11 end-effectors marked by thick dots. −1 −1.5 0 −0.5 1 0.5 2 1.5 2.5 −0.5 0 0.5 0 0.5 1 1.5 2 2.5 Fig. 5. Camera output superimposed with tracking result (left) and a tracked pose with 11 end-effectors marked by thick dots (right). The fact that bones do not change length in short time spans gives rise to a constraint for each bone; the distance between the pair of end-effectors must be constant. We incorporate this into a pose model with b bones by restricting the allowed poses to the preimage F −1 (0) of the map F : R 3k → R b given by F i (x) = kei1 − ei2 k 2 − l 2 i , (12) where ei1 and ei2 denote the spatial coordinates of the end-effectors and li the constant length of the ith bone. In this way, the set of allowed poses constitute a 3k − b-dimensional implicitly represented manifold. We record 26 poses using the tracking setup, and, amongst those, we make 20 random choices of 8 poses and perform linearized PGA and exact PGA. For each experiment, τSvˆ , ˜τSvˆ , ρ, and σ are computed and plotted in Figure 6. The 

A Riemannian manifold is a general d-dimensional manifold with an inner product on the tangent space, i.e., the Riemannian metric. This inner product is defined at each point and vary smoothly from point to point. The Riemannian metric allows one to define the notion of angles, length of the curves, surface area, and volume. We are more concerned with differential geometry of 2-dimensional manifold surfaces as majority of 3D shapes can be treated as the discrete version of a continuous 2D manifold surface embedded in a 3-dimensional Euclidean space. 


2.3.1.2 Tangent Space & Tangent Plane The concept of tangent space in differential geometry is analogous to the idea of linear approximation of a surface in the vector calculus. For a given manifold M embedded in R d , a linear subspace can be associated with each point x ∈ M . This linear subspace of R d is called the tangent space and comprises all the tangent vectors of x. The tangent space is the best linear approximation of the local manifold surface within a small neighborhood around x. In case of a manifold being a parametric 2D-surface patch embedded in R 3 , the tangent space around x is called a tangent plane and is typically depicted as TxM . Figure 2.2 depicts the tangent space of a 2D manifold surface embedded in R 3 


2.3.2 Mapping between Surfaces Surface mapping is an interesting problem in differential geometry. The idea is to seek a (bijective) mapping function that allows one-to-one correspondence between two manifold surfaces which are embedded differently in the Euclidean space. One classical example of surface mapping problem is to find a flat map of the earth’s surface. An important result from Gauss proved that it is impossible to do so without admitting deformations. This significant theorem from Gauss, known as the Theorema Egregium, laid the foundation of modern differential geometry which can deal with intrinsic surface mappings. The surface mapping has many important applications, e.g., texture mapping, dense registration, animation transfer, MRI analysis, etc. Figure 2.8 illustrates surface mapping scenario in R 3 .

微分幾何中關於曲面的曲率的重要定理，由高斯發現。這定理說曲面的高斯曲率可以從曲面上的長度和角度的測量完全決定，無需理會曲面如何嵌入三維空間內。換言之，高斯曲率是曲面的內蘊不變量。用現代術語可表述為：

高斯曲率在局部等距變換下不變。

Manifold Learning can be largely classified as linear and non-linear methods. The linear methods try to find a reduced dimensional subspace R d ⊂ R D. On the other hand, the nonlinear methods seek a global parametrization of a manifold Y ⊂ R d . The non-linear methods can further be divided into purely global methods and methods recovering the global manifold structure from local information. Here, we present an overview of the existing manifold learning techniques

How to calculate the eigenvector corresponding to zero eigenvalue

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How can the eigenvector corresponding to zero eigenvalue be found out? I was trying with the following simple matrix in Matlab: A=⎡⎣⎢123−2−3−4345⎤⎦⎥.$$A=\left[\begin{array}{ccc}1& -2& 3\\ 2& -3& 4\\ 3& -4& 5\end{array}\right].$$ In matlab computations, the matrix seemed nearly singular with one of the eigenvalues very close to zero (3e-15). That means the usual shifted inverse power methods for finding out the unit eigenvector corresponding to an eigenvalue won't work. But Matlab returns an eigenvector corresponding to 0. How? Basically, I would like to develop a program to compute this eigenvector given any singular matrix. What algorithm should I use? Edit: (1) Edited to reflect that the 'nearly singular' comment was corresponding to Matlab calculation. (2) Edited to specify the actual question. linear-algebra matrices eigenvalues-eigenvectors shareciteimprove this question edited Mar 24 '11 at 18:34 asked Mar 24 '11 at 5:09 Samik R 148118

1 If all the elements of a matrix are integers, as here, then the determinant is an integer. So 'nearly singular' is impossible! – TonyK Mar 24 '11 at 14:14  add a comment

2 Answers

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This matrix is singular, the determinant is zero, so it has an eigenvector for eigenvalue 0$0$. Nothing mysterious there -- you might want to check the calculation that made you think it was only nearly singular. As for how to find eigenvectors with eigenvalue 0$0$: They are just the solutions of the homogeneous system of linear equations corresponding to this matrix, Ax=0$Ax=0$, so you can use e.g. Gaussian elimination. shareciteimprove this answer answered Mar 24 '11 at 6:03 joriki 126k9127245

Thanks for the comment. I mentioned "nearly singular" because, in matlab the eigenvalue seemed to 3e-15 with long formatting. I of course realize that the matrix is completely singular, but I was wondering why Matlab had it that way which made it seem nearly singular as opposed to completely singular. I will rephrase the original question to reflect that. – Samik R Mar 24 '11 at 18:22 Just edited OP as mentioned above, and change the question to "I would like to develop a program to compute this eigenvector given any singular matrix. What algorithm should I use?" Thanks again for your response. – Samik R Mar 24 '11 at 18:35 @Samik: "in MATLAB the eigenvalue seemed to 3e-15 with long formatting." - because MATLAB uses floating point arithmetic; thus, don't expect supposedly zero results to be *exactly* zero. – J. M. Apr 11 '11 at 2:20 add a comment

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A matrix A$A$ has eigenvalue λ$\lambda$ if and only if there exists a nonzero vector x$x$ such that Ax=λx$A\mathbf{x}=\lambda \mathbf{x}$. This is equivalent to the existence of a nonzero vector x$x$ such that (A−λI)x=0$(A-\lambda I)\mathbf{x}=\mathbf{0}$. This is equivalent to the matrix A−λI$A-\lambda I$ having nontrivial nullspace, which in turn is equivalent to A−λI$A-\lambda I$ being singular (determinant equal to 0$0$). In particular, λ=0$\lambda =0$ is an eigenvector if and only if det(A)=0$det(A)=0$. If the matrix is "nearly singular" but not actually singular, then λ=0$\lambda =0$ is not an eigenvalue. As it happens, $$\begin{array}{rl}det(A)& =\left|\begin{array}{rr}-3& \phantom{-}4\\ -4& 5\end{array}\right|+2\left|\begin{array}{cc}2& 4\\ 3& 5\end{array}\right|+3\left|\begin{array}{rr}\phantom{-}2& -3\\ 3& -4\end{array}\right|\\ & =(-15+16)+2(10-12)+3(-8+9)\\ & =1-4+3=0,\end{array}$$ so the matrix is not "nearly singular", it is just plain singular. The eigenvectors corresponding to λ$\lambda$ are found by solving the system (A−λI)x=0$(A-\lambda I)\mathbf{x}=\mathbf{0}$. So, the eigenvectors corresponding to λ=0$\lambda =0$ are found by solving the system (A−0I)x=Ax=0$(A-0I)\mathbf{x}=A\mathbf{x}=\mathbf{0}$. That is: solve x2x3x−−−2y3y4y+++3z4z5z===000.$$\begin{array}{rcrcrcl}x& -& 2y& +& 3z& =& 0\\ 2x& -& 3y& +& 4z& =& 0\\ 3x& -& 4y& +& 5z& =& 0.\end{array}$$ The solutions (other than the trivial solution) are the eigenvectors. A basis for the solution space (the nullspace of A$A$) is a basis for the eigenspace Eλ$E_{\lambda }$. Added. If you know a square matrix is singular, then finding eigenvectors corresponding to 0$0$ is equivalent to solving the corresponding system of linear equations. There are plenty of algorithms for doing that: Gaussian elimination, for instance (Wikipedia even has pseudocode for implementing it). If you want numerical stability, you can also use Grassmann's algorithm.