> On 19 Oct 2016, at 19:12, Mike meyer <

[hidden email]> wrote:

>

>> From my reading of the above cited document I get the impression that the algorithm

> (algorithm 3.16, p27) can easily be adapted to handle the case m<n.

> In this case the Jacobian Jf(x) is mxn and the matrix A(x)=Jf(x)'Jf(x) is nxn and rank deficient.

>

> This seems to be a problem since the search direction h at iterate x is computed from

> (A(x)+mu*I)h=-J(x)'f(x)

> where mu -> 0 and so the system becomes ill conditioned.

>

> Why can we not get around this as follows: as soon as mu is below some threshold

> we solve instead the seemingly worse A(x)h=-J(x)'f(x) which however is of the form

>

> J(x)'J(x)h=-J(x)'f(x)

>

> and can be solved by solving J(x)h=-f(x) and the condition for this to be solvable

> is that J(x) have full rank. In fact this is the Gauss-Newton step

Oh?

You are solving a system

Ax=b where A is a matrix mxn with m<n with full rank (m).

Let Aplus be the generalized inverse of A.

Let z <- runif(nrow(Aplus))

The general solution of that linear system is Aplus %*% b + (diag(nrow(Aplus)) - Aplus %*% A) %*% z

where the minimum norm solution is Aplus %*% b.

Since z is any random vector there are many solutions.

So which one is the Gauss-Newton step?

Berend

> and to switch to that in the case of small mu is exactly in keeping with the idea of the

> LM-algorithm.

>

> If h is _any_ solution we have h'F'(x)=-||J(x)h||²<=0 (document, p21, eq(3.10))

> and is a descent direction if this is in fact < 0.

> If it is equal to zero, then f(x)=J(x)h=0 and we have a global minimum at x.

>

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>

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