Suppose $T$ is a diagonalizable linear operator on a finite dimensional vector space $V$. Prove $V$ is T-cyclic subspace of itself iff every characteristic subspace of it is one-dimensional.

It means there's a $v\in V$ such that $V=\operatorname{span}\langle v,T(v),T^2(v),\dots,T^k(v)\rangle$. How should I proceed?


1 Answer 1


Write every $T^mv$ in the diagonal basis. This gives a matrix of coefficients. The vector $v$ is cyclic for $T$ if and only if this matrix is non-singular. The form of the matrix should ring a bell.

  • $\begingroup$ What does it mean to write $T^m(v)$ in the diagonal basis? could you please explain more? $\endgroup$ Jun 11, 2016 at 8:47
  • $\begingroup$ I mean a basis in which $T$ is diagonal. Being a basis, $v$ can be expressed as a linear combination of vectors in that basis. Apply $T$ and see what happens. $\endgroup$
    – WimC
    Jun 11, 2016 at 8:54
  • $\begingroup$ Alright I was able to prove it all, but I can't prove one thing, why that matrix of coefficients in non-singular? $\endgroup$ Jun 12, 2016 at 6:38

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