Hi Florent,
I usually define
def prod_m(x,y):
z=x*y
z.set_immutable()
return z
and then pass this new product to the monoid/semigroup I am using!
Best,
Anne
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Hi Vincent,
> Having immutability by default would be particuarly annoying in the
> following case
>
> sage: a = identity_matrix(ZZ, 3)
> sage: b = a + a # immutable
> sage: b[0,1] = 2 # niet
Not that much
sage: a = identity_matrix(ZZ, 3)
sage: b = copy(a + a)
sage: b[0,1] = 2 # da
F
True. I was thinking about
sage: M = MatrixSpace(ZZ,3)
sage: a = M.one()
sage: a.is_mutable()
False
On 17/02/16 13:16, Nathann Cohen wrote:
Having immutability by default would be particuarly annoying in the
following case
sage: a = identity_matrix(ZZ, 3)
sage: b = a + a # immutable
sage: b
> Having immutability by default would be particuarly annoying in the
> following case
>
> sage: a = identity_matrix(ZZ, 3)
> sage: b = a + a # immutable
> sage: b[0,1] = 2 # niet
Why would identity_matrix return an immutable matrix ? Or did I miss
something ? All that we need is that the sum/p
Hi Florent,
No. And this is indeed a problem. The only current workaround is to
replace the product with
p = a * b
p.set_immutable()
Note that this is more flexible than having some immutability by default
for the product. Though, as in your example it is a barrier for being
used in any gen
Hi there,
Is there a way to convince sage that the product of two mutable matrices are
immutable ?
I need to consider matrix over a finite ring as a monoid and to take the
monoid algebra (like a group algebra) of it. It works when defined but I can't
compute any product in It because group
