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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