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how to properly substitute in a matrix?

asked 2011-06-23 06:12:36 -0500

roah gravatar image

updated 2011-06-23 06:15:25 -0500

kcrisman gravatar image

Let's consider a modified example from Sage reference manual

f(x,y)=x^2*y+y^2+y
solutions=solve(list(f.diff()),[x,y])
the_solution=solutions[2]
H=f.diff(2);  # Hessian matrix

How can i properly substitute the_solution into H?

I have tried:

H(x,y).subs(the_solution) - does not work.

This will work for H(x,y).subs(x==0)

H(x,y).subs_expr(*the_solution) - does not work.

This will work for f(x,y).subs_expr(*the_solution)

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answered 2011-06-23 06:17:37 -0500

kcrisman gravatar image

updated 2011-06-23 06:18:42 -0500

Try this:

sage: solutions=solve(list(f.diff()),[x,y],solution_dict=True)
sage: solutions
[{y: 0, x: -I}, {y: 0, x: I}, {y: -1/2, x: 0}]
sage: H.subs(solutions[2])
[(x, y) |--> -1  (x, y) |--> 0]
[ (x, y) |--> 0  (x, y) |--> 2]

Or this:

sage: H(x,y).subs(solutions[2])
[-1  0]
[ 0  2]

I'm not sure exactly what output you are looking for.

The documentation for solve has some more information about this keyword.

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Thanks, kcrisman! Maybe it is possible to include these lines into Sage Reference Manual? In the manual one enters the point manually... I don't know how to do this...

roah gravatar imageroah ( 2011-06-24 00:54:23 -0500 )edit

This is now http://trac.sagemath.org/sage_trac/ticket/11541. It would be a good beginner ticket to try :)

kcrisman gravatar imagekcrisman ( 2011-06-24 03:50:17 -0500 )edit
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answered 2011-06-23 06:25:22 -0500

niles gravatar image

How about this?

sage: soln = [x.rhs() for x in the_solution]; soln
[0, -1/2]
sage: matrix([f(*soln) for f in H])
[-1  0]
[ 0  2]
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Thanks, niles!

roah gravatar imageroah ( 2011-06-24 00:52:04 -0500 )edit

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Asked: 2011-06-23 06:12:36 -0500

Seen: 384 times

Last updated: Jun 23 '11