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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