I'm struggling with this question for some time. I would appreciate if you'd check what I did and correct me if I'm wrong:
I am requested to check whether this claim is correct or wrong:
Let ${v_1,...,v_n}$ is a basis of $V$ and $T:V→V$ satisfies: $T{v_1}=0$, $T{v_i}=v_{i−1} (2\leq i\leq n)$, then there exists $T^k=0$ for some $1\leq k<n.$
What I did:
Because of $T^k$, I was quite sure that this question is about nilpotent matrix. However, in a nilpotent matrix the only eigenvalue should be $0$, and the minimal polynomial should be $T^k$ where $k \leq n$(because we're speaking of a linear transformation) . Basically, from what I understand, the transformation moves the main diagonal to the left, making it the trace. Because it's linearly independent, $\sum_{k=1}^n \gamma_k v_k \neq 0$, which means that the minimal polynomial is $k=n-1$, which makes it a true claim.
Am I correct? If not please correct me so I can learn and improve.
Thank you very much!
When given a question like this, you can always perform a quick sanity check. Let's take $V$ to be $2$-dimensional, generated by $v_1$ and $v_2$. Then:
$T^0(v_1) = v_1, T^0(v_2) = v_2$
$T^1(v_1) = 0, T^1(v_2) = v_1$
$T^2(v_1) = 0, T^2(v_2) = 0$
Does this satisfy the statement in the question or not? (If so, what is $k$? If not, why not?)