Linearly independent family from a nilpotent map

Question

Let $V$ be a vector space and let $\psi: V \to V$ be a linear map that $ψ^k \neq 0$ and $ψ^{k+1}=0$ (where $k>0$). To prove that there is an element $x \in V$ such that $\{x, ψ (x), \dots, ψ^k (x)\}$ is linearly independent.

Answer 1

Hint: Any $x$ that isn't in the kernel of $\psi^k$ will do. Consider taking the equation $\sum_{i=0}^k a_i\psi^i x = 0$ and applying an appropriate power of $\psi$ to both sides.

Answer 2

Let $x\in V\setminus {0\ } |\ \psi^k(x)\neq0 \Longrightarrow \psi^i(x)\neq 0 \forall i=0,...,k$, because $Ker(\psi^i)\subseteq Ker(\psi^{i+1})$.

Let $a_0,...,a_k\in K\ | \ \sum_{i=0}^ka_i\psi^i(x)=0, $ where $\psi^0=id_V.$

$\Longrightarrow \psi^k(\sum_{i=0}^ka_i\psi^i(x))=0\ \Longrightarrow a_0x=0$ or $a_0=0$.

$\Longrightarrow \sum_{i=0}^ka_i\psi^i(x)=\sum_{i=1}^ka_i\psi^i(x)=0 \ \Longrightarrow \psi^{k-1}(\sum_{i=0}^ka_i\psi^i(x))=0\ \Longrightarrow a_1=0.$

. . .

So you can show $a_i=0 \forall i=0,...,k$ or the set is linearly independent.

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