# How quickly minimizing $M*x-v$ (numerically) ?

Let $v$ in $R^m$ and let $M$ be a matrix from $R^n$ to $R^m$, with $m>n$ big numbers.
I want to compute a vector $x$ in $R^n$ such that the norm of $M*x-v$ is minimal.

One way is to compute the projection $w$ of $v$ on the image of $M$.
For so, we can compute the projection $p$ on the image of $M$, as follows :

MTGS=M.transpose().gram_schmidt()[0]  # it's orthogonalization, not orthonormalization
l=MTGS.rank()
U=[]
for i in range(l):
v=MTGS[i]
u=v/(v.norm())
L=list(u)
U.append(L)
N=matrix(m,l,U)
p=N.transpose()*N


Then:

w=p*v
x=M.solve_right(w)


This vector $x$ minimizes the norm of $M*x-v$, but this method is very expensive in time, because it computes $p$ and $w$, while I just need $x$.

Is there another method, less expensive in time, for computing $x$ ?

Remark : I'm ok with numerical methods.

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If I am understanding your question correctly - if you put the additional constraint of getting $x$ with the minimum norm, then you are asking for the Moore-Penrose pseudoinverse. In Sage you can explicitly call the pseudoinverse function of numpy

sage: MN = M.numpy()
sage: import numpy
sage: x = matrix(numpy.linalg.pinv(MN))*v


Edit: Even the command

sage: x = M \ v


works for a nonsquare matrix M. But I am not sure what this is actually doing for a nonsquare matrix. The documentation does not seem to explain. The solution obtained is not the same as obtained from the pseudoinverse.

more

Thank you very much ppurka ! Do you know if we can compute x directly, without computing the pseudo-inverse before ?

( 2013-10-22 09:26:26 -0600 )edit

Well, that's what the Sage command M \ v does - it does not compute the inverse or pseudoinverse. Except, I am not sure exactly what it is computing for nonsquare matrices. Maybe you can put up a question in the sage-support mailing list.

( 2013-10-22 15:27:10 -0600 )edit

Thank you ppurka. I see a difference between "x = matrix(numpy.linalg.pinv(MN))*v" and "x = M \ v" : if v is not in the image of M then the first proposes a pseudo-solution and the second gives "ValueError: matrix equation has no solutions".

( 2013-10-24 00:00:44 -0600 )edit