Linearly independent set

Suppose I have $f\in L(V,V)$ such that $f^n=0$ but $f^k\neq 0\,\,\forall k<n$. Would I be right in thinking that there exists $v\in V$ such that $\{f^k(v)|0\leq k\leq n-1\}$ is linearly independent? If I am right, how could I find this $v$?

Thanks

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Since $f^{n-1}\neq 0$, you can find $v\in V$ such that $f^{n-1}(v)\neq 0$. Now, show that $\{f^k(v)\mid 0\leq k\leq n-1\}$ is linearly independent. – Davide Giraudo Nov 20 '11 at 21:34
Thanks, @DavideGiraudo. Do you mean that for any $v\in V$ such that $f^{n-1}(v)\neq 0$ the set is linearly independent? – ochil Nov 20 '11 at 21:58
Yes, and you can try to prove it, for example assuming that $\sum_{j=0}^{n-1}\alpha_j f^j(v)=0$ for $(\alpha_0,\ldots,\alpha_{n-1})\neq (0,\ldots,0)$ and getting a contradiction. – Davide Giraudo Nov 20 '11 at 22:03
@DavideGiraudo: Thanks, got it now. – ochil Nov 20 '11 at 22:10

If $f^n=0$ but $f^k\neq 0$ for any $k\lt n$, then the minimal polynomial of $f$ is $t^n$; the characteristic polynomial is $t^{\dim (V)}$.

As Davide Giraudo notes, the fact that $f^{n-1}\neq 0$ guarantees the existence of a vector $v\in V$ such that $f^{n-1}(v)\neq\mathbf{0}$. The claim now is that $v$, $f(v),\ldots, f^{n-1}(v)$ are linearly independent.

One way to see this is to note that the $f$-annihilator of $v$ (the polynomial $p(t)$ of least degree such that $p(f)(v) = 0$) must divide the minimal polynomial, and hence must be of the form $t^k$ for some $k\leq n$; but if $f^k(v)=\mathbf{0}$, then $f^{k+i}(v)=\mathbf{0}$ for all $i\geq 0$. Since $f^{n-1}(v)\neq\mathbf{0}$ by choice of $v$, the $f$-annihilator must be $x^n$.

Now, if $\alpha_0,\ldots,\alpha_{n-1}$ are scalars such that $$\alpha_0 v + \alpha_1f(v)+\cdots \alpha_{n-1}f^{n-1}(v)=\mathbf{0},$$ then $p(f)(v)=\mathbf{0}$, where $p(t) = \alpha_0+\alpha_1t+\cdots + \alpha_{n-1}t^{n-1}$. Since this must be a multiple of $t^n$, the only possibility is that $p(t)=0$, so $\alpha_0=\cdots=\alpha_{n-1}=0$, proving that $v$, $f(v),\ldots,f^{n-1}(v)$ are indeed linearly independent.

If you don't have the theory of $f$-annihilators, assume that $$a_0v + \cdots + a_{n-1}f^{n-1}(v)=\mathbf{0}$$ Applying $f$ to both sides $n-1$ times; using the fact that $f^{n+i}(v) = \mathbf{0}$ for all $i\geq 0$ and $f^{n-1}(v)\neq \mathbf{0}$, we get $a_0f^{n-1}(v) = \mathbf{0}$, and so we can deduce that $a_0=0$. This leaves $$a_1f(v) + \cdots + a_{n-1}f^{n-1}(v)=\mathbf{0}.$$ Now apply $f$ again $n-2$ times to conclude $a_1=0$; proceed in this manner to show $a_0=\cdots=a_{n-2}=0$; this leaves $a_{n-1}f^{n-1}(v)=\mathbf{0}$, and since $f^{n-1}(v)\neq \mathbf{0}$, then $a_{n-1}=0$. This proves linear independence.

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