I understood what's going on. The variables a0, a1, a2, a3 are a
distraction since they are not to appear in the solution. What matters is
their coefficients in b, which naturally form a matrix.
>>> sp.Matrix([[eq.coeff(v) for v in a] for eq in b])
Matrix([
[0, 0, 0, 1],
[1, 1, 1, 1],
[0, 0, 1, 0],
[3, 2, 1, 0]])
So that's what is really given. And the desired result is the inverse of
that matrix.
>>> sp.Matrix([[eq.coeff(v) for v in a] for eq in b]).inv()
Matrix([
[ 2, -2, 1, 1],
[-3, 3, -2, -1],
[ 0, 0, 1, 0],
[ 1, 0, 0, 0]])
This is the best way to get the result with SymPy, because the best way to
find the inverse of a matrix is to invert that matrix. It's often said
(correctly) that solving a linear system is more efficient than inverting
its matrix. But here, the task was really to find the inverse of a 4 by 4
matrix.
On Wednesday, March 14, 2018 at 7:21:27 AM UTC-4, Matthias Geier wrote:
>
> Thanks a lot for your answer, Leonid!
>
> I think I'm starting to see the problem ...
> When I'm solving the system of equations, I get many solutions, most
> of them containing a0, a1, a2 and a3.
>
> However, I'm just interested in exactly one of those solutions where
> all occurrences of a0, a1, a2 and a3 cancel out and the resulting
> matrix only consists of concrete numbers.
>
> Is there a (simple) way to tell this to SymPy?
>
> In the meantime, I've found a different way to get the desired result.
>
> Let me first repeat the original setup:
>
> >>> import sympy as sp
> >>> a0, a1, a2, a3 = sp.symbols('a:4')
> >>> a = sp.Matrix([a3, a2, a1, a0])
> >>> b = sp.Matrix([a0, a3 + a2 + a1 + a0, a1, 3 * a3 + 2 * a2 + a1])
> >>> M = sp.MatrixSymbol('M', 4, 4)
> >>> sp.Eq(a, M * b)
> Eq(Matrix([[a3], [a2], [a1], [a0]]), M*Matrix([
> [ a0],
> [a0 + a1 + a2 + a3],
> [ a1],
> [ a1 + 2*a2 + 3*a3]]))
>
> I stumbled upon
>
> https://groups.google.com/forum/#!searchin/sympy/linear_eq_to_matrix/sympy/Wqs1OhTBexg/eGoKjrWHBwAJ,
>
>
> which mentions sympy.solvers.solveset.linear_eq_to_matrix().
>
> http://docs.sympy.org/latest/modules/solvers/solveset.html#linear-eq-to-matrix
>
>
> With this I can get my expected solution:
>
> >>> from sympy.solvers.solveset import linear_eq_to_matrix
> >>> linear_eq_to_matrix(list(b), list(a))
> (Matrix([
> [0, 0, 0, 1],
> [1, 1, 1, 1],
> [0, 0, 1, 0],
> [3, 2, 1, 0]]), Matrix([
> [0],
> [0],
> [0],
> [0]]))
> >>> _[0].inv()
> Matrix([
> [ 2, -2, 1, 1],
> [-3, 3, -2, -1],
> [ 0, 0, 1, 0],
> [ 1, 0, 0, 0]])
>
> This still involves the extra step of inverting the result.
>
> Is this already the best way to get this result with SymPy?
>
> cheers,
> Matthias
>
> On Wed, Mar 14, 2018 at 2:33 AM, Leonid Kovalev wrote:
> > You have 4 equations with 16 unknowns, so there are going to be a lot of
> > solutions.
> >
> > Many SymPy functions struggle with M[i, j] construction which creates
> > MatrixElement instead of an ordinary Symbol. I would use a Matrix filled
> > with Symbols:
> >
> > M = sp.Matrix(sp.symarray('M', (4, 4)))
> > sp.solve(a - M*b, list(M))
> >
> > This returns
> >
> > {M_3_0: (-M_3_1*(a0 + a1 + a2 + a3) - M_3_2*a1 - M_3_3*(a1 + 2*a2 +
> 3*a3) +
> > a0)/a0, M_2_0: (-M_2_1*(a0 + a1 + a2 + a3) - M_2_2*a1 - M_2_3*(a1 + 2*a2
> +
> > 3*a3) + a1)/a0, M_1_0: (-M_1_1*(a0 + a1 + a2 + a3) - M_1_2*a1 -
> M_1_3*(a1 +
> > 2*a2 + 3*a3) + a2)/a0, M_0_0: (-M_0_1*(a0 + a1 + a2 + a3) - M_0_2*a1 -
> > M_0_3*(a1 + 2*a2 + 3*a3) + a3)/a0}
> >
> > which, as previously mentioned, is a lot of solutions. You can plug in
> some
> > arbitrary numbers for the free variables here.
> >
> >
> >
> > On Saturday, March 10, 2018 at 7:38:00 AM UTC-5, Matthias Geier wrote:
> >>
> >> Dear list.
> >>
> >> I have this equation:
> >>
> >> a = M * b,
> >>
> >> where a and b are column vectors and M is a 4x4 matrix.
> >> a and b consist of quite simple expressions, M is unknown:
> >>
> >> >>> import sympy as sp
> >> >>> a0, a1, a2, a3 = sp.symbols('a:4')
> >> >>> a = sp.Matrix([a3, a2, a1, a0])
> >> >>> b = sp.Matrix([a0, a3 + a2 + a1 + a0, a1, 3 * a3 + 2 * a2 + a1])
> >> >>> M = sp.MatrixSymbol('M', 4, 4)
> >> >>> sp.Eq(a, M * b)
> >> Eq(Matrix([[a3], [a2], [a1], [a0]]), M*Matrix([
> >> [ a0],
> >> [a0 + a1 + a2 + a3],
> >> [ a1],
> >> [ a1 + 2*a2 + 3*a3]]))
> >>
> >> What's the most straightforward way to solve this for M?
> >>
> >> I have actually found a way, but it seems a bit non-obvious to me:
> >>
> >> >>> def get_value(i, j):
> >> ... return b[i].as_coefficients_dict()[a[j]]
> >> ...
> >> >>> M_inv = sp.Matrix(4, 4, get_value)
> >> >>> M_inv.inv()
> >> Matrix([
> >> [ 2, -2, 1, 1],
> >> [-3, 3, -2, -1],
> >> [ 0, 0, 1, 0],
> >> [ 1, 0, 0, 0]])
> >>
> >> This is the solution I'm looking for, but I was hoping there is a
> >> better way to get it.
> >> Also, the way I did it I had to know that I could look for the inverse
> >> and then undo the inverse afterwards, but I would prefer if there is a
> >> solution without this manual step.
> >>
> >> Thanks in advance!
> >>
> >> cheers,
> >> Matthias
>
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