I'm solving an eigenvalue problem when the matrix and the eigenvectors are time dependent. The matrix has dimension 8x8 and is hermitian. The time dependent matrix has the form:
import sympy as sp
t, lbd = sp.symbols(r't,\lambda', real=True)
Had = ...
print(repr(Had))
Matrix([[2*t,0, 0, 0, 0, 0, 0,0],
[ 0,-2*t, 2*t*(1 - t), 0, 0, 0,0,0],
[0, 2*t*(1 - t),0,0, 2 - 2*t, 0,0,0],
[0,0,0,0, 0, 2 - 2*t, 0,0],
[0,0,2 - 2*t,0,0,0,0,0],
[0,0,0, 2 - 2*t,0,0, 2*t*(1 - t),0],
[0,0,0,0,0, 2*t*(1 - t),-2*t,0],
[0,0,0,0,0,0,0,2*t]])
Now the characteristic polynomial has the following for:
P = p.simplify(sp.collect(sp.factor(Had.charpoly(lbd).as_expr()),lbd))
and get
Then I choose the second term and find the solution for lambda:
P_list = sp.factor_list(P)
a,b = P_list[1]
eq,exp = sp.simplify(b)
sol = sp.solve(eq)
With that I get the roots in a list:
r_list = []
for i in range(len(sol)):
a = list(sol[i].values())
r_list.append(a[0])
Solving the problem using sp.eigenvecs:
val_mult_vec = Had.eigenvects()
e_vals = []
mults = []
e_vecs = []
for i in range(len(val_mult_vec)):
val, mult, [vec_i, vec_j] = val_mult_vec[i]
e_vals.append(val)
e_vals.append(val)
mults.append(mult)
e_vecs.append(vec_i)
e_vecs.append(vec_j)
Solving the eigenvectors I get complicated expressions like this:
But I know that this complicated expression can be expressed in terms of the solution of the second term in the characteristic polynomial something like this:
Where r1 are one of the roots of that equation. With the solution to the characteristic polynomial how can I rewrite the eigenvectors in a simplified way like the last image using sympy? rewrite e_vec[i] in terms of r_list[j]
Seems like you want to obtain a compact version of the eigenvectors.
Recepy:
We can create as many symbols as the number of eigenvalues. Each symbol represents an eigenvalue.
Loop over the eigenvectors and for each of its elements substitute the long eigenvalue expression with the respective symbol.
r_symbols = symbols("r0:%s" % len(e_vals))
final_evecs = []
for vec, val, s in zip(e_vecs, e_vals, r_symbols):
final_evecs.append(
vec.applyfunc(lambda t: t.subs(val, s))
)
final_evecs is a list containing eigenvectors in a compact notation.
Let's test one output:
final_evecs[7]
Related
Example: let
M = Matrix([[1,2],[3,4]]) # and
p = Poly(x**3 + x + 1) # then
p.subs(x,M).expand()
gives the error :
TypeError: cannot add <class'sympy.matrices.immutable.ImmutableDenseMatrix'> and <class 'sympy.core.numbers.One'>
which is very plausible since the two first terms become matrices but the last term (the constant term) is not a matrix but a scalar. To remediate to this situation I changed the polynomial to
p = Poly(x**3 + x + x**0) # then
the same error persists. Am I obliged to type the expression by hand, replacing x by M? In this example the polynomial has only three terms but in reality I encounter (multivariate polynomials with) dozens of terms.
So I think the question is mainly revolving around the concept of Matrix polynomial:
(where P is a polynomial, and A is a matrix)
I think this is saying that the free term is a number, and it cannot be added with the rest which is a matrix, effectively the addition operation is undefined between those two types.
TypeError: cannot add <class'sympy.matrices.immutable.ImmutableDenseMatrix'> and <class 'sympy.core.numbers.One'>
However, this can be circumvented by defining a function that evaluates the matrix polynomial for a specific matrix. The difference here is that we're using matrix exponentiation, so we correctly compute the free term of the matrix polynomial a_0 * I where I=A^0 is the identity matrix of the required shape:
from sympy import *
x = symbols('x')
M = Matrix([[1,2],[3,4]])
p = Poly(x**3 + x + 1)
def eval_poly_matrix(P,A):
res = zeros(*A.shape)
for t in enumerate(P.all_coeffs()[::-1]):
i, a_i = t
res += a_i * (A**i)
return res
eval_poly_matrix(p,M)
Output:
In this example the polynomial has only three terms but in reality I encounter (multivariate polynomials with) dozens of terms.
The function eval_poly_matrix above can be extended to work for multivariate polynomials by using the .monoms() method to extract monomials with nonzero coefficients, like so:
from sympy import *
x,y = symbols('x y')
M = Matrix([[1,2],[3,4]])
p = poly( x**3 * y + x * y**2 + y )
def eval_poly_matrix(P,*M):
res = zeros(*M[0].shape)
for m in P.monoms():
term = eye(*M[0].shape)
for j in enumerate(m):
i,e = j
term *= M[i]**e
res += term
return res
eval_poly_matrix(p,M,eye(M.rows))
Note: Some sanity checks, edge cases handling and optimizations are possible:
The number of variables present in the polynomial relates to the number of matrices passed as parameters (the former should never be greater than the latter, and if it's lower than some logic needs to be present to handle that, I've only handled the case when the two are equal)
All matrices need to be square as per the definition of the matrix polynomial
A discussion about a multivariate version of the Horner's rule features in the comments of this question. This might be useful to minimize the number of matrix multiplications.
Handle the fact that in a Matrix polynomial x*y is different from y*x because matrix multiplication is non-commutative . Apparently poly functions in sympy do not support non-commutative variables, but you can define symbols with commutative=False and there seems to be a way forward there
About the 4th point above, there is support for Matrix expressions in SymPy, and that can help here:
from sympy import *
from sympy.matrices import MatrixSymbol
A = Matrix([[1,2],[3,4]])
B = Matrix([[2,3],[3,4]])
X = MatrixSymbol('X',2,2)
Y = MatrixSymbol('Y',2,2)
I = eye(X.rows)
p = X**2 * Y + Y * X ** 2 + X ** 3 - I
display(p)
p = p.subs({X: A, Y: B}).doit()
display(p)
Output:
For more developments on this feature follow #18555
As an introduction i want to point out that if one has a matrix A consisting of 4 submatrices in a 2x2 pattern, where the diagonal matrices are square, then if we denote its inverse as X, the submatrix X22 = (A22-A21(A11^-1)A12)^-1, which is quite easy to show by hand.
I was trying to do the same for a matrix of 4x4 submatrices, but its quite tedious by hand. So I thought Sympy would be of some help. But I cannot figure out how (I have started by just trying to reproduce the 2x2 result).
I've tried:
import sympy as s
def blockmatrix(name, sizes, names=None):
if names is None:
names = sizes
ll = []
for i, (s1, n1) in enumerate(zip(sizes, names)):
l = []
for j, (s2, n2) in enumerate(zip(sizes, names)):
l.append(s.MatrixSymbol(name+str(n1)+str(n2), s1, s2))
ll.append(l)
return ll
def eyes(*sizes):
ll = []
for i, s1 in enumerate(sizes):
l = []
for j, s2 in enumerate(sizes):
if i==j:
l.append(s.Identity(s1))
continue
l.append(s.ZeroMatrix(s1, s2))
ll.append(l)
return ll
n1, n2 = s.symbols("n1, n2", integer=True, positive=True, nonzero=True)
M = s.Matrix(blockmatrix("m", (n1, n2)))
X = s.Matrix(blockmatrix("x", (n1, n2)))
I = s.Matrix(eyes(n1, n2))
s.solve(M*X[:, 1:]-I[:, 1:], X[:, 1:])
but it just returns an empty list instead of the result.
I have also tried:
Using M*X==I but that just returns False (boolean, not an Expression)
Entering a list of equations
Using 'ordinary' symbols with commutative=False instead of MatrixSymbols -- this gives an exception with GeneratorsError: non-commutative generators: (x12, x22)
but all without luck.
Can you show how to derive a result with Sympy similar to the one I gave as an example for X22?
The most similar other questions on solving with MatrixSymbols seem to have been solved by working around doing exactly that, by using an array of the inner symbols or some such instead. But since I am dealing with symbolically sized MatrixSymbols, that is not an option for me.
Is this what you mean by a matrix of 2x2 matrices?
>>> a = [MatrixSymbol(i,2,2) for i in symbols('a1:5')]
>>> A = Matrix(2,2,a)
>>> X = A.inv()
>>> print(X[1,1]) # [1,1] instead of [2,2] because indexing starts at 0
a1*(a1*a3 - a3*a1)**(-1)
[You indicated not and pointed out that the above is not correct -- that appears to be an issue that should be resolved.]
I am not sure why this isn't implemented, but we can do the solving manually as follows:
>>> n = 2
>>> v = symbols('b:%s'%n**2,commutative=False)
>>> A = Matrix(n,n,symbols('a:%s'%n**2,commutative=False))
>>> B = Matrix(n,n,v)
>>> eqs = list(A*B - eye(n))
>>> for i in range(n**2):
... s = solve(eqs[i],v[i])[0]
... eqs[i+1:] = [e.subs(v[i],s) for e in eqs[i+1:]]
...
>>> s # solution for v[3] which is B22
(-a2*a0**(-1)*a1 + a3)**(-1)
You can change n to 3 and see a modestly complicated expression. Change it to 4 and check the result by hand to give a new definition to the word "tedious" ;-)
The special structure of the equations to be solved can allow for a faster solution, too: the variable of interest is the last factor in each term containing it:
>>> for i in range(n**2):
... c,d = eqs[i].expand().as_independent(v[i])
... assert all(j.args[-1]==v[i] for j in Add.make_args(d))
... s = 1/d.subs(v[i], 1)*-c
... eqs[i+1:] = [e.subs(v[i], s) for e in eqs[i+1:]]
I am running the following in my jupyter-notebook.
from sympy import *
B = IndexedBase('B')
x, L = symbols('x L', real=True)
n = Symbol('n', integer=True)
n = Idx(n, (0, oo))
Bn = Indexed('B', n)
m = Symbol('m', integer=True)
Sum(Piecewise((0, Ne(m, n)), (L, True))*B[n], (n, 0, oo)).doit()
The last expression should evaluate to $L B_n$. I have tried using simplify, doit, limit and evalf methods on it to without success. I also found more similar issues in github but couldn't taylor this to my specific problem.
I also tried fiddling around with the underlying assumptions for integer m but couldn't find anything suitable.
Is there any direct or indirect way to get sympy to simplify infinite sums containing piecewise functions?
Just as an aside this code below works:
p = Piecewise((1, n<5), (0, True))
Sum(p, (n, 1, oo)).evalf()
Motivation. It is well known that generating function for Catalan numbers satisfies quadratic equation. I would like to have first several coefficients of a function, implicitly defined by an algebraic equation (not necessarily a quadratic one!).
Example.
import sympy as sp
sp.init_printing() # math as latex
from IPython.display import display
z = sp.Symbol('z')
F = sp.Function('F')(z)
equation = 1 + z * F**2 - F
display(equation)
solution = sp.solve(equation, F)[0]
display(solution)
display(sp.series(solution))
Question. The approach where we explicitly solve the equation and then expand it as power series, works only for low-degree equations. How to obtain first coefficients of formal power series for more complicated algebraic equations?
Related.
Since algebraic and differential framework may behave differently, I posted another question.
Sympy: how to solve differential equation in formal power series?
I don't know a built-in way, but plugging in a polynomial for F and equating the coefficients works well enough. Although one should not try to find all coefficients at once from a large nonlinear system; those will give SymPy trouble. I take iterative approach, first equating the free term to zero and solving for c0, then equating 2nd and solving for c1, etc.
This assumes a regular algebraic equation, in which the coefficient of z**k in the equation involves the k-th Taylor coefficient of F, and does not involve higher-order coefficients.
from sympy import *
z = Symbol('z')
d = 10 # how many coefficients to find
c = list(symbols('c:{}'.format(d))) # undetermined coefficients
for k in range(d):
F = sum([c[n]*z**n for n in range(k+1)]) # up to z**k inclusive
equation = 1 + z * F**2 - F
coeff_eqn = Poly(series(equation, z, n=k+1).removeO(), z).coeff_monomial(z**k)
c[k] = solve(coeff_eqn, c[k])[0]
sol = sum([c[n]*z**n for n in range(d)]) # solution
print(series(sol + z**d, z, n=d)) # add z**d to get SymPy to print as a series
This prints
1 + z + 2*z**2 + 5*z**3 + 14*z**4 + 42*z**5 + 132*z**6 + 429*z**7 + 1430*z**8 + 4862*z**9 + O(z**10)
I'm trying to use sympy to do some index gymnastics for me. I'm trying to calculate the derivatives of a cost function that looks like
cost = sumi (Mii)2
where M is given by a rotation
Mij = U*ki M0kl Ulj
I've written up a parametrization for the rotation matrix, from which I get the derivatives as products of Kronecker deltas. What I've got so far is
def Uder(p,q,r,s):
return KroneckerDelta(p,r)*KroneckerDelta(q,s) - KroneckerDelta(p,s)*KroneckerDelta(q,r)
from sympy import *
# Matrix size
n = symbols('n')
p = symbols('p');
i = Dummy('i')
k = Dummy('k')
l = Dummy('l')
# Matrix elements
M0 = IndexedBase('M')
U = IndexedBase('U')
# Indices
r, s = map(tensor.Idx, ['r', 's'])
# Derivative
cost_x = Sum(Sum(Sum(M0[i,i]*(Uder(k,i,r,s)*M0[k,l]*U[l,i] + U[k,i]*M0[k,l]*Uder(l,i,r,s)),(k,1,n)),(l,1,n)),(i,1,n))
print cost_x
but sympy is not evaluating the contractions for me, which should reduce to simple sums in terms of r and s, which are the rotation indices. Instead, what I get is
Sum(((-KroneckerDelta(_i, r)*KroneckerDelta(_k, s) + KroneckerDelta(_i, s)*KroneckerDelta(_k, r))*M[_k, _l]*U[_l, _i] + (-KroneckerDelta(_i, r)*KroneckerDelta(_l, s) + KroneckerDelta(_i, s)*KroneckerDelta(_l, r))*M[_k, _l]*U[_k, _i])*M[_i, _i], (_k, 1, n), (_l, 1, n), (_i, 1, n))
I'm using the latest git snapshot 4633fd5713c434c3286e3412a2399bd40fbd9569 of sympy.