[英]Substitute a Sympy matrix into a polynomial
I have a Sympy matrix A and a polynomial expression P, and I would like to compute P(A).我有一个 Sympy 矩阵 A 和一个多项式表达式 P,我想计算 P(A)。
Here is an example:下面是一个例子:
x = Symbol('x')
P = x**2 - 3*x + 5
A = Matrix([ [1,3], [-1,2] ])
P.subs(x, A)
I expect Sympy to compute A**2 - 3*A + 5*eye(2)
(in the example, the result is the zero matrix).我希望 Sympy 计算
A**2 - 3*A + 5*eye(2)
(在示例中,结果是零矩阵)。
But this fails with an error message:但这失败并显示错误消息:
AttributeError: ImmutableMatrix has no attribute as_coeff_Mul.
Is there any way to obtain what I want ?有什么办法可以得到我想要的吗?
Edit: I've tried to convert P
into Sympy's polynomial class, and substitute afterwards, but the result is useless:编辑:我尝试将
P
转换为 Sympy 的多项式类,然后替换,但结果无用:
Poly(P).subs(A)
Poly(Matrix([ [ 1, 3], [-1, 2]])**2 - 3*Matrix([ [ 1, 3],
[-1, 2]]) + 5, Matrix([ [ 1, 3], [-1, 2]]), domain='ZZ')
I can get the correct result with the following function :我可以使用以下函数获得正确的结果:
def poly_matrix(P, A):
coeffs = Poly(P).all_coeffs()[::-1]
res = zeros(A.rows)
for i in range(len(coeffs)):
res += coeffs[i]*(A**i)
return res
But I'm still looking for a more efficient built-in option.但我仍在寻找更高效的内置选项。
Matrix expressions are still a bit buggy, hopefully they'll get fixed in the future.矩阵表达式仍然有一些问题,希望它们在未来得到修复。
Anyway, here's an alternative way to perform the substitution:无论如何,这是执行替换的另一种方法:
In [1]: x = MatrixSymbol('x', 2, 2)
In [2]: P = x**2 - 3*x + 5*eye(2)
In [3]: P
Out[3]:
2
⎡5 0⎤ + -3⋅x + x
⎢ ⎥
⎣0 5⎦
In [4]: A = Matrix([ [1,3], [-1,2] ])
In [5]: P.subs(x, A)
Out[5]:
2
⎡5 0⎤ + -3⋅⎡1 3⎤ + ⎛⎡1 3⎤⎞
⎢ ⎥ ⎢ ⎥ ⎜⎢ ⎥⎟
⎣0 5⎦ ⎣-1 2⎦ ⎝⎣-1 2⎦⎠
In [6]: P.subs(x, A).doit()
Out[6]:
2
⎡2 -9⎤ + ⎛⎡1 3⎤⎞
⎢ ⎥ ⎜⎢ ⎥⎟
⎣3 -1⎦ ⎝⎣-1 2⎦⎠
Here it appears that MatPow isn't able to perform the .doit() operation.这里似乎MatPow无法执行.doit()操作。 It's probably another bug.
这可能是另一个错误。
In output number 5 there's also a bug with the printer ( + -3 should just be -3 ).在输出编号 5 中,打印机也有一个错误( + -3应该只是-3 )。
I really hope that someone will eventually fix all these problems.我真的希望有人最终会解决所有这些问题。
Try to evaluate each term of the polynomial.尝试评估多项式的每一项。
(x**2).subs(x,A) - (3*x).subs(x,A) + 5*(eye(2))
This will evaluate your expression.这将评估您的表达式。
To evaluate P
at A
, you can substitute x
with a Matrix
and convert the constant term by multiplying by eye(2)
:要在
A
处评估P
,您可以用Matrix
替换x
并通过乘以eye(2)
转换常数项:
P_ = Poly(P, x)
(P_ - P_.coeff_monomial(1)).as_expr().subs(x, A) * eye(2) + P_.coeff_monomial(1) * eye(2) # P(A)
The first multiplication by eye(2)
ensures the first term in the sum is a matrix, even if P
is just a constant, ie P_ - P_.coeff_monomial(1) == 0
.第一次乘以
eye(2)
确保总和中的第一项是矩阵,即使P
只是一个常数,即P_ - P_.coeff_monomial(1) == 0
。
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