manifold gu A manifold of dimension is a connected Hausdorff space ✁ for which every point has a neighborhood ✂ that is homeomorphic to an open subset ✄ of ☎✝✆ . Such a homeomorphism ✞✠✟✡✂☞☛✌✄ is called a coordinate chart.

http://www3.cs.stonybrook.edu/~gu/publications/pdf/2005/riemann_iccv05.pdf

A manifold of dimension is a connected Hausdorff space ✁ for which every point has a neighborhood ✂ that is homeomorphic to an open subset ✄ of ☎✝✆ . Such a homeomorphism ✞✠✟✡✂☞☛✌✄ is called a coordinate chart.

How to calculate the eigenvector corresponding to zero eigenvalue

up vote4down votefavorite 3

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

activeoldestvotes

up vote8down voteaccepted

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

up vote5down vote

A matrix A$A$ has eigenvalue λ$\lambda$ if and only if there exists a nonzero vector x$\mathbf{x}$ such that Ax=λx$A\mathbf{x}=\lambda \mathbf{x}$. This is equivalent to the existence of a nonzero vector x$\mathbf{x}$ 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, det(A)=∣∣∣−3−445∣∣∣+2∣∣∣2345∣∣∣+3∣∣∣23−3−4∣∣∣=(−15+16)+2(10−12)+3(−8+9)=1−4+3=0,$$\begin{array}{rl}det(A)& =\left|\begin{array}{rr}-3& \phantom{-}4\\ -4& 5\end{array}\right|+2\left|\begin{array}{cc}2& 4\\ 3& 5\end{array}\right|+3\left|\begin{array}{rr}\phantom{-}2& -3\\ 3& -4\end{array}\right|\\ & =(-15+16)+2(10-12)+3(-8+9)\\ & =1-4+3=0,\end{array}$$ 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.