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How to enumerate the generalised inverse of given matrix?

asked 2016-03-13 00:38:53 -0500

daviddgl gravatar image

Let $A$ be a given singular matrix over a finite field. Then how to enumerate the solutions of the system of equation

$$Ax=b, b \in R(A), \text{range of } A $$

Since, $A$ is singular matrix I can find the generalized inverse $G$ (i.e., AGA=A) of $A$ . Note that, $Gb$ is one of the solution. And so the set of all solution can be written as $${Gb+(I-GA)u: u \text{ is arbitrary }}$$

Now, how to incorporate these ideas in sagemath?

I am a newbie to sagemath. So, the answerers kindly give me with the details.

Thanks in advance.

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answered 2016-03-13 06:46:20 -0500

tmonteil gravatar image

updated 2016-03-13 06:53:38 -0500

There is an important mathematical notion for dealing with such problems : the kernel of a matrix, and i strongly suggest you to learn about it. If $K$ is the (right) kernel of $A$, if $x_0$ is a solution of the equation $Ax=b$, then the set of solutions of the equation is the set $x_0+K = \{x_0+k \mid k \in K\}$.

So, you can do the following:

Setting:

sage: A = random_matrix(GF(3),4,6) ; A
[2 1 2 2 0 2]
[2 2 0 1 2 0]
[0 2 2 1 2 2]
[2 0 0 1 0 2]
sage: b = vector(GF(3),(2,2,0,2)) ; b
(2, 2, 0, 2)

Find parameters $x_0$ and $K$:

sage: x0 = A.solve_right(b)
sage: x0
(1, 0, 0, 0, 0, 0)

sage: K = A.right_kernel() ; K
Vector space of degree 6 and dimension 2 over Finite Field of size 3
Basis matrix:
[1 0 1 1 0 0]
[0 1 2 1 0 1]

Enumerate solutions:

sage: solutions = [x0 + k for k in K] ; solutions
[(1, 0, 0, 0, 0, 0),
 (2, 0, 1, 1, 0, 0),
 (0, 0, 2, 2, 0, 0),
 (1, 1, 2, 1, 0, 1),
 (2, 1, 0, 2, 0, 1),
 (0, 1, 1, 0, 0, 1),
 (1, 2, 1, 2, 0, 2),
 (2, 2, 2, 0, 0, 2),
 (0, 2, 0, 1, 0, 2)]

You can check:

sage: for i in solutions:
....:     print A*i
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
(2, 2, 0, 2)
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Asked: 2016-03-13 00:30:30 -0500

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Last updated: Mar 13 '16