Linear independence of finite dimentional vector space for a linear transform raised to a power

Question

Let $V$ be a finite-dimensional vector space over $F, T∈L(V)$.

Suppose 0 $\neq v∈V$ and m is a positive integer for which $T^{m-1} v \neq $ 0 but $T^{m} v= 0$.

Show that

S={$v, T v, . . . , T^{m−1}v$} is a linearly independent set?

I love this question but can't see where to start? something to do with it being nilpotent?

Answer 1

When you are asked to prove that a set in linearly independent you should usually start from writing a combination $\lambda_0v+\lambda_1T(v)+...+\lambda_{m-1}T^{m-1}(v)=0$. You have to prove all coefficients $\lambda_0,\lambda_1,...\lambda_{m-1}$ must be zeros. So now using the fact that a linear transformation maps zero to zero we can use $T^{m-1}$ on both sides of the equation and get $\lambda_0T^{m-1}(v)+\lambda_1T^m(v)+...\lambda_{m-1}T^{2m-2}(v)=0$. Easy to see that $T^m(v)=0$ implies $T^k(v)=0$ for all $k\geq m$ and hence we get $\lambda_0T^{m-1}(v)=0$. We know that $T^{m-1}(v)\ne 0$, so it must be $\lambda_0=0$.

Now the linear combination we started from looks like $\lambda_1T(v)+...+\lambda_{m-1}T^{m-1}(v)=0$. Now use $T^{m-2}$ on both sides to get $\lambda_1T^{m-1}(v)+\lambda_2T^m(v)+...+\lambda_{m-1}T^{2m-3}(v)=0$. Once again it can be written shortly as $\lambda_1T^{m-1}(v)=0$ and hence $\lambda_1=0$.

Now the original linear combination looks like $\lambda_2T^2(v)+...+\lambda_{m-1}T^{m-1}(v)=0$. Continue the process I started by induction to get all the coefficients are zeros.

Answer 2

If $m=1$ then there is nothing to prove, otherwise let us assume they are linearly dependent, then we have: $$\alpha_0 v + \dots + \alpha_{m-1} T^{m−1}v =0$$ Applying $T^{m-1}$ on both sides we obtain: $$\alpha_0 T^{m-1} v + T^m v +\dots +T^{2m-2}=\alpha_0 T^{m-1} v= 0$$ Which contradicts the hypothesis.