[sage-support] Re: solving matrix equations
Thanks to both of you!
Mladen
On Feb 7, 12:19 pm, Jason Grout wrote:
> Craig Citro wrote:
> > Hi,
>
> > Yep, there are definitely easy ways of doing this. Here's one way:
>
> > sage: var('a b c d')
> > (a, b, c, d)
> > sage: A = matrix(2,[2,1,1,1])
> > sage: B = matrix(2,[a,b,c,d])
> > sage: C = A*B - B*A
>
> > sage: [ e == 0 for e in C.list() ]
> > [c - b == 0, d + b - a == 0, -d - c + a == 0, b - c == 0]
>
> > sage: solve([ e == 0 for e in C.list() ], N.list())
> > [[a == r2 + r1, b == r2, c == r2, d == r1]]
>
> Just to point out for those that need a guide through the above, Craig
> is basically creating a list of equations from each entry in the matrix.
> He doesn't show it, but I suppose that N is a matrix of the variables.
> You could also do:
>
> solve([ e == 0 for e in C.list() ], C.variables())
>
> That said, I think it would be a great thing if solve could recognize
> matrices and that two matrices are equal if each entry is equal. I
> believe MMA does this (but it's easier there; matrices are nothing more
> than nested lists). It'd certainly make certain things I do more
> natural if I could do:
>
> solve(matrixA==matrixB)
>
> and that was equivalent to:
>
> solve([i==j for i,j in zip(matrixA.list(), matrixB.list())])
>
> if the matrices were of the same dimensions.
>
> Okay, so now that I've written my piece, I suppose the next step is to
> open a trac ticket, write a patch to implement it, and post it for
> review :).
>
> The ticket ishttp://trac.sagemath.org/sage_trac/ticket/5201
>
> I won't cry if someone submits a patch before I get to it :).
>
> Jason
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[sage-support] Re: solving matrix equations
Craig Citro wrote:
> Hi,
>
> Yep, there are definitely easy ways of doing this. Here's one way:
>
> sage: var('a b c d')
> (a, b, c, d)
> sage: A = matrix(2,[2,1,1,1])
> sage: B = matrix(2,[a,b,c,d])
> sage: C = A*B - B*A
>
> sage: [ e == 0 for e in C.list() ]
> [c - b == 0, d + b - a == 0, -d - c + a == 0, b - c == 0]
>
> sage: solve([ e == 0 for e in C.list() ], N.list())
> [[a == r2 + r1, b == r2, c == r2, d == r1]]
Just to point out for those that need a guide through the above, Craig
is basically creating a list of equations from each entry in the matrix.
He doesn't show it, but I suppose that N is a matrix of the variables.
You could also do:
solve([ e == 0 for e in C.list() ], C.variables())
That said, I think it would be a great thing if solve could recognize
matrices and that two matrices are equal if each entry is equal. I
believe MMA does this (but it's easier there; matrices are nothing more
than nested lists). It'd certainly make certain things I do more
natural if I could do:
solve(matrixA==matrixB)
and that was equivalent to:
solve([i==j for i,j in zip(matrixA.list(), matrixB.list())])
if the matrices were of the same dimensions.
Okay, so now that I've written my piece, I suppose the next step is to
open a trac ticket, write a patch to implement it, and post it for
review :).
The ticket is http://trac.sagemath.org/sage_trac/ticket/5201
I won't cry if someone submits a patch before I get to it :).
Jason
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[sage-support] Re: solving matrix equations
Hi,
Yep, there are definitely easy ways of doing this. Here's one way:
sage: var('a b c d')
(a, b, c, d)
sage: A = matrix(2,[2,1,1,1])
sage: B = matrix(2,[a,b,c,d])
sage: C = A*B - B*A
sage: [ e == 0 for e in C.list() ]
[c - b == 0, d + b - a == 0, -d - c + a == 0, b - c == 0]
sage: solve([ e == 0 for e in C.list() ], N.list())
[[a == r2 + r1, b == r2, c == r2, d == r1]]
-cc
On Sat, Feb 7, 2009 at 9:45 AM, mb wrote:
>
> Hi,
>
> Say I want to compute the centralizer of a matrix.
>
> sage: P.=PolynomialRing(QQ)
> sage: A=matrix(P,2,2,[2,1,1,1])
> sage: B=matrix(P,2,2,[a,b,c,d])
> sage: C=A*B-B*A
> sage: C
>
> [-b + c -a + b + d]
> [ a - c - d b - c]
> sage: var('a b c d')
> (a, b, c, d)
> sage: solve([-b + c==0,-a + b + d==0,a - c - d==0,b - c==0],[a,b,c,d])
> [[a == r2 + r1, b == r2, c == r2, d == r1]]
>
> Here I had to manually copy the entries of C into solve, which is a
> problem for larger matrices. Ideally, something like
>
> solve(C==0,[a,b,c,d])
>
> should work, but of course it doesn't. Is there any way of doing this?
>
> Mladen
> >
>
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