Let's say we have a 4x4 nilpotent matrix $V$ with an index of 4 and a vector $\vec{x}$ which is an element of $\mathcal{R}^4$, matrix $V$ and $\vec{x}$ are non-zero. How would we show that $V\vec{x}$ and $\vec{x}$ are linearly independent? The following is what I did but I personally think that my answer is somewhat iffy.
$$ \lambda_1\vec{x} + \lambda_2V\vec{x} = \vec{0} \\ \lambda_1(\vec{x_1}+\vec{x_2}+\vec{x_3}+\vec{x_4)} + \lambda_2V(\vec{x_1}+\vec{x_2}+\vec{x_3}+\vec{x_4}) = \vec{0} $$ Factorizing it according to the vectors $\vec{x_1}$ to $\vec{x_4}$: $$ (\lambda_1 + \lambda_2V)\vec{x_1}+(\lambda_1 + \lambda_2V)\vec{x_2}+(\lambda_1 + \lambda_2V)\vec{x_3}+(\lambda_1 + \lambda_2V)\vec{x_4} = \vec{0} $$ Solving this would give, all coefficients as 0 since they're all the same. I think I messed up somewhere in my understanding of linear independence or something but please help and explain it to me. Thank you in advance!
As given, the statement you're asking to be proved is not necessarily true, because $\ V\vec{x}\ $ could be zero, even if both $\ V\ $ and $\ \vec{x}\ $ are non-zero, and if $\ V\vec{x}=\vec{0}\ $, then $\ V\vec{x}\ $ and $\ \vec{x}\ $ are not linearly independent.
Since $\ V\ $ is nilpotent, then there must be a smallest power, $\ V^n\ $, of $\ V\ $ that annihilates $\ \vec{x}\ $ (that is, $\ V^n\vec{x}=\vec{0}\ $ and $\ V^{n-1}\vec{x}\ne\vec{0}\ $), and if $\ V\vec{x}\ne\vec{0}\ $, then $\ n\ge2\ $. Suppose, in this case, that $$ \lambda_1 \vec{x}+ \lambda_2 V\vec{x}= \vec{0}\ . $$ Multiplying this equation by $\ V^{n-1}\ $ gives $$ \vec{0}=\lambda_1 V^{n-1} \vec{x}+ \lambda_2 V^n\vec{x}= \lambda_1 V^{n-1} \vec{x}\ . $$ Since $\ V^{n-1}\vec{x}\ne\vec{0}\ $, it follow that $\ \lambda_1=0\ $ and then that $\ \lambda_2 V\vec{x}= \vec{0}\ $. Since $\ V\vec{x}\ne\vec{0}\ $ it also follows that $\ \lambda_2=0\ $. Thus if $\ V\vec{x} \ne\vec{0}\ $ then $\ \vec{x}\ $ and $\ V\vec{x}\ $ are linearly independent.