Let $1\leq n\in \mathbb{N}$, let $\mathbb{K}$ be a field, $V$ a $\mathbb{K}$-vector space with $\dim_{\mathbb{K}}V=n$ and let $\phi:V\rightarrow V$ be linear.
Let $0\neq v\in V$ and $1\leq m\in \mathbb{N}$, such that $\phi^{m-1}(v)\neq 0=\phi^m(v)$. Show that $\phi^0(v)=v, \phi (v), \ldots ,\phi^{m-1}(v)$ are linearly independent.
Hint : Use induction for $m$ and for the case $m\geq 2$ consider $w=\phi (v)$.
Let $v,m$ as above. Let $B=(b_1, \ldots , b_n)$ be a basis of $V$, with $b_i=\phi^{i-1}(v)$ for all $1\leq i\leq m$. Describe the first $m-1$ columns of $M_B(\phi)$.
I have done the following:
Base case: If $m=1$ the we have the element $\phi^0(v)=v$. This element is linearly independent.
Inductive hypothesis: We assume that $\phi^0(v)=v, \phi (v), \ldots ,\phi^{m-1}(v)$ are linearly independent. (IH)
Inductive step: We want to show that $\phi^0(v)=v, \phi (v), \ldots ,\phi^{m-1}(v), \phi^m(v)$ is linearly independent.
\begin{align*}&\lambda_0\phi^0(v)+\lambda_1\phi (v)+ \ldots +\lambda_{m-1}\phi^{m-1}(v)+\lambda_m \phi^m(v)=0 \\ & \Rightarrow \lambda_m \phi^m(v)=-\lambda_0\phi^0(v)-\lambda_1\phi (v)- \ldots -\lambda_{m-1}\phi^{m-1}(v)\end{align*}
We know that the elements of the right side are linearly independent. How can we use that information here? Could you give me a hint?
i) $v\geq 0$ on a field is non-sense.
ii) I didn't correctly read your equality. In fact $\phi_m(v)=0$. Then it's very easy.
Calculate $\phi^k(\sum_{i=1}^m a_i\phi^{i-1}(v))$.