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proof of cyclic vector theorem

Major Section: 
Reference
Type of Math Object: 
Proof

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Given some linear operator
T, I'm having trouble understanding what a cyclic vector is and how to find one.
For example; given some matrix representing a linear operator on say R^2 how can I determine if it has a cyclic vector?

Hi, I am not sure what you mean by a cyclic vector, I suppose $v$ is cyclic if there is a natural number $k$ such that $A^kv=v$ where $A$ is the matrix of your transformation (is it?).

If so, you need to find $k$ such that the kernel of $A^k-Id$ is non-zero.

I hope that helps,
Alvaro

Usually, the term cyclic vector refers to a vector v such that the set {v, Av, A^2 v, A^3 v, ...} spans the vector space. (In the infinite dimensional case, one usually interprets this to mean that the span of the set is dense.)

I have not yet seen anyone use the term "cyclic vector" in the sense which is A. L. R. suggested in his reply to your question.

To make the job of looking for a cyclic vector easier, one could simplify the transformation using a similarity transform. In the case of a finite-dimensional matrix, one could put it into Jordan normal form.

Let us see what happens for 2x2 matrices. There are three possibilities: 1) The matrix is a multiple of the identity. 2) The matrix has two distinct eigenvalues. 3) The matrix has two equal eigenvalues but is not the identity matrix.

Case I. There can be no cyclic vector --- if we multiply any vector v by A, all we obtain are multiples of v. The span of these is a one-dimensional subspace, so we cannot span the whole vector space by repeatedly applying A to a vector.

Case II. The matrix can be diagonalized. Call the two distinct eigenvalues be p and q (one of which may be zero). Then the matrix looks like
p 0
0 q
I claim that the vector v = (1,1) is cyclic. Note that A v = (p,q). Since p is not equal to q, the vectors (1,1) and (p,q) are linearly independant, hence they span the two-dimensional vector space.

Case III. The matrix can be put in the Jordan form
x 1
0 x
Again, consider the vector v = (1,1). Then Av = (x+1,x). Since the set {(1,1), (1+x,x)} is linearly independant for any possible value of x, the vector (1,1) is cyclic.

Hope this helps,
Ray

Thanks,
That really helps clear it up for me. My text book is really ambigous as is my instructor. I'm just tying to understand it.
Melinda

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