solving homogeneous system of linear equations

Anonymous

kcrisman
12222 ●41 ●135 ●254

The system is written on the form Ax=0. I know this can be done by using, for example,

solve([eq1==0,eq2==0],x1,x2)

But this is somewhat complex. I wonder if the system can be directly solved by A.solve_right or some other simpler notation?

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kcrisman
12222 ●41 ●135 ●254

Is this what you mean? I'm not sure, because should the zero vector be a solution for you?

sage: A = matrix([[1,2],[2,4]])
sage: A.solve_right(vector([0,0]))
(0, 0)
sage: A\vector([1,2])
(1, 0)
sage: A*vector([1,0])
(1, 2)

Or maybe you wanted this.

sage: A.right_kernel()
Free module of degree 2 and rank 1 over Integer Ring
Echelon basis matrix:
[ 2 -1]

I hope I'm not misunderstanding something here. See this Linear Algebra quickref card for a lot more information.

the right_kernel is exactly what I want. Thanks very much for your answer and the resources provided!

lainme ( 2011-12-14 23:45:04 +0200 )edit

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Asked: 2011-12-14 09:53:03 +0200

Seen: 3,695 times

Last updated: Dec 14 '11