On Thursday, June 4, 2015 at 3:11:59 PM UTC-7, Michael Orlitzky wrote:
>
> > 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
>
In some cases, Sage's built-in linear algebra functions might be more
efficient:
sage: A = matrix( [[1,2],[3,4]] )
sage: v = vector([1,2])
sage: def is_cyclic_vector( v, A ):
return not A.iterates( v, A.ncols() ).is_singular()
....:
sage: is_cyclic_vector( v, A )
True
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