# Prior-conditioning

## introduction

Pre-conditioning is a trick to solve linear systems of equations \(Ax=b\) for iterative solvers. Typically, a simple-to-compute and non-singular matrix \(P\) is chosen to change the system matrix into

\[APy=b\]where \(y = P^{-1}x\) is the new unknown. It is also possible to apply conditioning matrix from the left side, i.e.,

\[PAx = Pb.\]The goal here is to reduce the condition number of \(A\) such that the iterative method converges faster. One also implicitly assumes that \(A\) is full rank such that the solution is unique. In this case pre-conditioning will not affect the solution.

In this post, I would like to discuss the situation when \(A\) is rank-deficient. The iterative solver I picked is conjugate gradient (CG) and model system is

\[A = [1, 1], \quad b=1 \\ P=\begin{bmatrix}1&0\\0&\alpha\end{bmatrix}.\]Obviously, this system does NOT have a unique solution. But **does
CG know that**?

## observations

Using the following python code, we ended up with the solution

\[\tilde x = [0.5, 0.5]^T\]```
import scipy.sparse.linalg as linalg
import numpy as np
A = np.array([[1,1],[1,1]])
b = np.array([1,1])
x = linalg.cg(A,b)
```

Here I augmented the \(A\) matrix to \(2\times2\) so that scipy does not complain that \(A\) is not a square matrix. This augmentation can be viewed either as copying the row of \(A\) as a new row or solving the normal equation

\[A^TA x = A^T b.\]Interestingly, this solution is the matrix pseudo-inverse for matrix with linearly independent rows, i.e.,

\[\tilde x = A^T(AA^T)^{-1} b.\]*(Be very careful about the ordering of \(A\) and \(A^T\) here.)*

This formula can be derived from the minimization problem

\[\min_x\|x\|^2\quad s.t. Ax=b\]The derivation is included at the end of this post. In other words, it is the minimal norm solution for the under-determined system.

With the pre-conditioner \(P\), the normal equation becomes

\[\begin{bmatrix}1&\alpha\\\alpha&\alpha^2\end{bmatrix}x=\begin{bmatrix}1\\\alpha\end{bmatrix}.\]And again the CG result matches with the minimum norm solution

\[\tilde x = \frac{1}{1+\alpha^2}\begin{bmatrix}1\\\alpha^2\end{bmatrix}.\]It seems to hold for randomly generated \(A\) as well. You can check it out with the following code. Feel free to change the matrix dimension as well.

```
A = np.random.rand(3,4)
b = np.random.rand(3,1)
x = linalg.cg(np.dot(A.T,A), np.dot(A.T,b))
print x
print np.dot(A.T, np.dot(np.linalg.inv(np.dot(A,A.T)),b))
```

## caveats

It seems that we have a nice result that CG converges to minimum norm solution. However, there are two caveats to this observation

- The initial condition is set to 0.
- The normal equation is used.

With other initial conditions, the solutions vary. Also, if I copy rows of \(A\) to itself to make it square, CG does not converge to minimum norm solution.

## summary

So far we have numerically demonstrated that CG converges to the minimum norm solution for a rank-deficient system, given 0 initial condition and normal equation are used.

The more interesting observation is that the solution depends on the pre-conditioner \(P\). In some sense, it provides a systematic way to ‘manipulate’ the answer. In the literature, it seems to have the name ‘prior-conditioning’, which refers to confining the structure of \(x\) with \(P\) using some prior information.

In our simple example, the pre-conditioner \(P\) rescales the components of \(x\). It may be motivated by arguing that \(x_2\) is expected to be greater than \(x_1\) thus a pre-conditioner \(P\) with \(\alpha>1\) makes the unknowns more ‘equal’. Looking at the minimum norm solution, it seems to do the trick, i.e., it causes \(x_2>x_1\).

In practice, the system may be so big (hello medical imaging) that it is not possible to explicitly check whether \(A\) is rank-deficient, or even express \(A\) explicitly. Then one needs to be cautious about the stopping criteria, initial condition, and the pre-conditioner of the iterative solvers.

## appendix

### derivation of the pseudo-inverse

The Lagrangian is defined as

\[\mathcal L = x^Tx + \lambda^T(Ax-b)\]where the vector \(\lambda\) is the Lagrangian multiplier.

The stationary condition gives rise to

\[\frac{\partial \mathcal L}{\partial x}=0\rightarrow x=\frac{A^T\lambda}{2}.\]Plugging into the linear equation \(Ax=b\), we get

\[\lambda = 2(AA^T)^{-1}b\]Thus the pseudo-inverse is

\[x = A^T(AA^T)^{-1}b.\]