@Lee Worden
Your code is deciding the linear independence of the set
{v,v*A,v*A^2,...,v*A^{dim(A)}}
But I want to decide the linear independence of the set {v,
A*v,..,,A^{dim(A)-1}*v}
On Fri, Jun 5, 2015 at 1:20 PM, Lee Worden <worden....@gmail.com> wrote:
>
>
> 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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