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Definition of $T$-cyclic subspace: Let $T$ be an linear operator on $V$. Take $v \in V$.

The subspace generated by $\text{span}(\{v,T(v),T^2(v),\cdots\})$ is called $T$-cyclic subspace generated by $v$.

My question is: for a subspace $W$ on $V$, does there exists a linear operator $T$, such that $W$ is $T$-cyclic subspace for some $v \in W$?.

If my question is foolish one. Sorry in advance.

bb_823
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Ganesh Karthi
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    Yes if $W$ has countable dimension: pick a basis $\beta=[w_1,w_2,\ldots]$ for $W$; extend to a basis $\gamma$ for $V$. Then define $T\colon V\to V$ as $T(w_i)=w_{i+1}$ for $w_i\in\beta$, and arbitrarily on elements of $\gamma$ taht are not in $\beta$. If $W$ is the zero subspace, then just take $v=\mathbf{0}$ and $T$ arbitrary. – Arturo Magidin Aug 02 '23 at 18:27
  • (if $\dim(W)=n$, then you just need to define $T(v_{n})$ as any vector in $W$) – Arturo Magidin Aug 02 '23 at 20:26

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