# How do I prove these vectors are linearly independent? [closed]

Let $T : V → V$ be a linear map, where $\dim(V ) = n$, and suppose that $T^n = 0$ and that there exists a vector $v ∈ V$ with $T^{n−1}(v) \neq 0$.

Prove that the vectors $v, T(v), T^2(v), . . . , T^{n−1}(v)$ are linearly independent and that the nullity of $T$ is $1$.

## closed as off-topic by The Chaz 2.0, GNUSupporter 8964民主女神 地下教會, Jonas Meyer, Leucippus, NamasteMar 29 '17 at 16:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – The Chaz 2.0, GNUSupporter 8964民主女神 地下教會, Jonas Meyer, Leucippus, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

Suppose to the contrary that $v, T(v), T^2(v), \dots, T^{n-1}(v)$ are linearly dependent. Then there exist coefficients $a_i$ such that $$b_0v + b_1T(v) + \cdots + b_{n-1} T^{n-1}(v) = 0$$ Let $i_0$ denote the lowest $i$ for which $b_i \neq 0$. Then we have $$T^{i_0}v = \sum_{i = i_0 + 1}^n a_iT^iv$$ for some coefficients $a_i$. It follows that $$T^{n-1} v = T^{n-1-i_0}T^{i_0}v = T^{n - 1 - i_0}\sum_{i=i_0+1}^n a_i T^iv = \\ \sum_{i=i_0+1}^n a_i T^{n + (i-(i_0+1))}v = \sum_{j=0}^{k-1} a_{j+1+i_0} \underbrace{T^{n + j}}_{=0}v = 0$$ Which condradicts how we defined $v$.
For the second part, note that $\{T(v),T^2(v),\dots,T^{n-1}(v)\}$ are a linearly independent subset of the image of $T$.