Help with Nilpotent Transformation Proof

linear algebralinear-transformationsnilpotence

I have the following linear algebra textbook question and I'm not really sure where to begin with it:

Let $V$ be a (possibly infinite-dimensional) vector space. We say a linear map $T: V \rightarrow V$ is nilpotent if there is a positive integer $k$ such that $T^k = 0$. If $T$ is a nilpotent map, then there is a unique positive integer $k$ such that $T^{k} = 0$ and $T^{k−1} \neq 0$.

Let $T : V \rightarrow V$ be a nilponent linear map. Suppose the
nilpotency index of T is $k$. Show that if $0 ≤ i < k$, then

  1. $\operatorname{Im}(T^{i+1}) \subsetneq \operatorname{Im}(T)$, and
  2. $\ker T^i \subsetneq T^{i+1}$.

(The notation $X \subsetneq Y$ means $X ⊂ Y$ and $X \neq Y$ , i.e. that $X$ is a proper subset of $Y$).

Hint: It should be easy to
see that $\operatorname{Im}(T^{i+1}) \subset \operatorname{Im}(T)$ and $\ker(T^i) \subset \ker(T^{i+1})$. To prove the properness of the inclusions, first argue that
$T(\operatorname{Im}(T^i)) = \operatorname{Im}(T^{i+1})$ and $\operatorname{Im}(T^i) \subset \ker(T^{k-i})$.

My thought process so far is the following: since T is nilpotent, then for some integer $k$, $T^k = 0$. Up until that point, we know that $T^i \neq 0$. Supposing then that $i=k-1$, then $T^{i+1}=T^k=0$, which means all vectors under the linear transformation $T^k$ would go to zero. Hence, $ker(T^k)$ consists of all possible input vectors. Accordingly $Im(T^k)$ would have dimension 0, implying that $Im(T^k) \subsetneq Im(T^{k-1})$. I'm not sure if this is the right direction or how I can then generalize it for any value of $i$

Best Answer

With these sorts of questions it's best to argue by contradiction. We have that $T^i = T^{i-1}T$ and so $Im(T^i) \subseteq Im(T^{i-1})$. Suppose we had $$ Im(T^i) = T^i(V) = T^{i+1}(V) = Im(T^{i-1}) \neq \{0\} $$ Then $$ T^{i+2}(V) = T(T^{i+1}(V)) = T(T^{i}(V)) = T^{i+1}(V) \neq 0 $$ and so by induction, $T^{i + n}(V) \neq 0$ for all $n$. This is a contradiction. In particular we have that $T^{i}(V)$ is properly contained in $T(V)$ for $i > 1$. For the kernels you can apply the hint, if $T^i(v) = 0$ then $T^{i+1}(v) = 0$ and so $$ \ker T^{i} \subseteq \ker T^{i+1} $$ If ker $T^i = \ker T^{i+1}$ then we'd have that $Im(T^{k-(i+1)}) \subseteq \ker T^{i+1} = \ker T^{i}$ and hence $T^i(T^{k-(i+1)}(V)) = T^{k-1}(V) = 0$. This is a contradiction on the nilpotency of $T$.

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