[0] * i + [1] + [0] * (n-i-1)
On 05/06/15 00:11, Michael Orlitzky wrote:
On 06/04/2015 05:53 PM, Phoenix wrote:
Thanks!
Is there otherwise any standard operation in SAGE to create such vectors?
If the construction isn't too complicated, a "list comprehension"
usually suffices. This will do what you want, I think:
def elem(i,n):
return [ ZZ(i == j) for j in range(0,n) ]
A list comprehension is written much like the usual math notation for
set construction, so nothing to be afraid of.
Given a vector $v$ and a matrix $A$ of dimension $n$, one would say that
$v$ is a cyclic vector of $A$ if the following set is linearly independent
$\{ v,Av,A^2v,..,A^{n-1}v \}$.
Is there a way to test this property on SAGE given a $v$ and a $A$?
Sure, using list comprehensions again. First we construct the list of
A(v), A^2(v), etc. Then we stick those vectors in a big matrix, and ask
for its rank. If the matrix has full rank, it's columns/rows are
independent.
def f(A,v):
M = matrix([ (A^j)*v for j in range(0,len(v)) ])
return M.rank() == len(v)
Note that you will need to pass that function a vector (that you get
from calling vector() on a list), not a list. For example,
sage: A = matrix([[1,2],[3,4]])
sage: v = vector([1,2])
sage: f(A,v)
True
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