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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.