Using SVD to solve a set of equations.

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The discussion focuses on using Singular Value Decomposition (SVD) to solve the equation Az=0, where A is rank deficient by one. The solution involves finding the SVD of A, represented as [U,S,V], and identifying the right eigenvector corresponding to the zero eigenvalue, which is then normalized. The user confirms that this method works in practice, having tested it in Matlab, but seeks clarification on why the eigenvector associated with the smallest eigenvalue yields a solution. The conversation references additional resources on total least squares and related concepts for further understanding. Insight into the mathematical reasoning behind this approach is requested.
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Homework Statement



This is not a homework problem. I encountered this while working with total least squares for the first time. Ultimately a point is reached where Az=0 must be solved. z is of the form [x,1]^{T}. Let A be nxm, z be mx1.

Suppose A is rank deficient by one. So the SVD of A has one non zero singular value. Then to find z, what i need to do is simply find the SVD of A,

[U,S,V] = svd(A).

and the solution to Az=0 is the right eigenvector corresponding to the 0 eigenvalue, normalized so that the last element equals -1.


Now i have tested it and this works. (Did examples in Matlab). However, i don't know why this is true. Why does the eigenvector corresponding to the smallest eigenvector give you a solution (i'm assuming it gives you a solution to within a scalar multiple).

Any insight would be greatly appreciated.
 
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