Let V be a vector space of dimension n and T be a linear operator on V. Suppose that T is diognalizable. If T has n distinct eigenvalues and {a1,...,an} be a basis of eigenvectors of T, then show that a=a1+ ...+an is a cyclic vector for T.

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- Jul 26th 2013, 12:39 AMxixiCyclic vector for a linear operator
Let V be a vector space of dimension n and T be a linear operator on V. Suppose that T is diognalizable. If T has n distinct eigenvalues and {a1,...,an} be a basis of eigenvectors of T, then show that a=a1+ ...+an is a cyclic vector for T.

- Jul 26th 2013, 07:30 AMHallsofIvyRe: Cyclic vector for a linear operator
Saying that " T is diognalizable. If T has n distinct eigenvalues and {a1,...,an} be a basis of eigenvectors of T" means that if B is the matrix having the vectors {a1,..., an} as columns then $\displaystyle B^{-1}TB= D$ where D is the diagonal matrix having the eigenvalues of T on its diagonal. Now, what is the

**definition**of "cyclic vector"?