Let $V$ be a $K$-vector space, $f: V \rightarrow V$ a linear map. Under some condition, show that $v, f(v),...,f^n(v)$ are linearly independent. [duplicate]

Let $V$ be a $K$-vector space, $f: V \rightarrow V$ a linear map. Let $v \in V$. May a number $n ≥ 0$ exist, so that: $f^n(v) \not= 0$ and $f^{n+1}(v) = 0$. Show that $v, f(v),...,f^n(v)$ are linearly independent.

What i know:

• $f^0 = id, f^1 = f, f^2 = f \circ f, ...$
• $v, f(v),...,f^n(v)$ are linearly independent if:

$a_1v + a_2f(v) + \cdots+a_nf^n(v)=0 \implies a_1 = a_2 =\cdots= a_n = 0$

• $f$ is linear, hence $f(0) = 0$, $af(n) = f(an)$, $f(v_1 + v_2) = f(v_1)+f(v_2)$

My thoughts on it: the definition of linearly independent, in our case, can be rewritten as

$a_1v + f(a_2v) + f(f(a_3v)) + \cdots+f^n(a_nv)=0 \implies a_1 = a_2 =\cdots= a_n = 0$

Taking the function $f$ on both sides we can also obtain

$a_1f(v) + f(f(a_2v)) + f(f(f(a_3v))) + \cdots+f^{n+1}(a_nv)=0 \implies a_1 = a_2 =\cdots= a_n = 0$

... I feel that i am on the right track, but i can't take it further from here.

So i have to show this sum only becomes $0$ when all $a_n$ are $0$. I thought about a proof by contradiction, but i wouldn't know where i can lead this to a contradiction.

Thanks for all help

Yes you are on the right way! Suppose that the $a_n$'s are such that

$$a_0 v + a_1 f(v) +\dots + a_n f^n(v)=0$$

Taking the function times on both side (because of linearity) you obtain that $$a_0 f(v) + a_1 f^2(v) +\dots + a_n f^{n+1}(v)=0$$ and since $f^{n+1}(v)=0$ you get $$a_0 f(v) + a_1 f^2(v) +\dots + a_{n-1} f^{n}(v)=0.$$ Now you can do the same with the equation above and you obtain the relation $$a_0 f^2(v) + a_1 f^3(v) +\dots + a_{n-2} f^{n}(v)=0$$ and so on (for $n$ times). At the end you obtain the relation $$a_0 f^n(v)=0$$ and since $f^n(v) \not= 0$ you can conclude that $a_0=0$. Hence, going back to the equation $$a_0 v + a_1 f(v) +\dots + a_n f^n(v)=0$$ we can conclude that $$a_1 f(v) +\dots + a_n f^n(v)=0.$$ Now we can restart the reasoning above using in place of $v$ the vector $f(v)$, and we arrive to the conclusion that $a_1=0$. Going forward we can conclude that all the $a_n$'s are zero and we are done.

• This is the easiest answer for me to comprehend! thanks May 27, 2015 at 20:51

Yes, you're on the right track.

If $n=0$ the statement is obvious, because it reduces to $v\ne0$.

So suppose we have proved the statement for the pairs $(w,g)$ with $g^{n-1}(w)=0$ and $g^n(w)=0$, for $n>0$.

Let $\alpha_1v+\alpha_2f(v)+\dots+\alpha_nf^{n}(v)=0$. By applying $f$ to this equality and setting $w=f(v)$, we get $$\alpha_1w+\dots+\alpha_{n-1}f^{n-1}(w)=0$$ and, by the induction hypothesis, $\alpha_1=\dots=\alpha_{n-1}=0$. Therefore also $\alpha_n=0$.