Linear Map, T^p(x)=0, Show Linear Independence

[joypav]

Problem:

Suppose V is a complex vector space of dimension n, and T is a linear map from V to V. Suppose $x \in V$, and p is a positive integer such that $T^p(x)=0$ but $T^{p-1}(x)\ne0$.

Show that $x, Tx, T^2x, ... , T^{p-1}x$ are linearly independent.During class my professor said it was "a fact" that

If V is a complex vector space of dimension n, and T is a linear map from V to V such that $T^n(x)=0$ but $T^{n-1}(x)\ne0$,
then there exists an x such that $x, Tx, T^2x, ... , T^{n-1}x$ are linearly independent.

Is this what I should use for this problem? If so, do I need to know how I can determine such an x?

 

[HOI]

What you assert your professor said and the theorem you state are not quite the same. You say your professor said "If V is a complex vector space of dimension n, and T is a linear map from V to V such that T^n(x)= 0 but T^{n-1}(x) is not 0" then there exist x such that ...". If the conclusion is "there exist x" then what is that x in the hypothesis?

In any case, given that, for some linear transformation there exist vector x and integer n such that T^n(x)= 0 but that T^{n-1}(x)\ne 0 (from which it follows that T^m(x)\ne 0 for any m< n), the suppose to the contrary that x, T(x), T^2(x), ..., T^{n-1}(x) are NOT linearly independent. Then there exist \{a_n\}, not all 0, such that a_0x+ a_1T(x)+\cdot\cdot\cdot+ a_{n-2}T^{n-2}(x)+ a_{n-}T^{n-1}(x)= 0. Apply T to both sides: a_0T(x)+ a_1T^2(x)+ \cdot\cdot\cdot+ a_{n-1}T^{n-1}(x)+ a_nT^n(x)= a_0T(x)+ a_1T^2(x)+ \cdot\cdot\cdot+ a_{n-1}T^{n-1}(x)= 0.

Applying T n-1 times, and repeatedly using the fact that T^n(x)= 0, we arrive at a_0T^{n-1}(x)= 0. If a_0\ne 0 it follows that T^{n-1}(x)= 0, a contradiction. If a_0= 0, we only need to apply T n-2 times to arrive at a_1T^{n-1}(x)= 0, etc.