Let $V$ be an $n$-dimensional vector space over a field $F$ and $T$ an operator on $V.$ Prove $\ker(T^n)=\ker(T^{n+1})$ and $\operatorname{range}(T^n)=\operatorname{range}(T^{n+1}).$
Suppose $v \in \ker(T^n).$ Then $T^n(v) = 0,$ implying that $T^{n+1}(v)=T(T^n(v))=T(0)=0.$ Thus $v \in \ker(T^{n+1})$ and so $\ker(T^n) \subset \ker(T^{n+1}).$
Question: How do I prove $\ker(T^n) \supset \ker(T^{n+1})?$ I would also like some hints on proving $\operatorname{range}(T^n)=\operatorname{range}(T^{n+1}).$
Best Answer
Look at the Lemma in this answer: https://math.stackexchange.com/a/1447742/211913
It immediately proves your statement.
Some more details: With the notation of the cited answer: Assume the kernels of $T^n$ and $T^{n+1}$ do not coincide. We then have $k_{n+1}-k_n \geq 1$. Since the sequence of differences is decreasing, we obtain $k_{j+1}-k_j \geq 1$ for all $j = 0, \dotsc, n$. Of course this implies $k_{n+1} \geq k_0+n+1 = n+1$, which is an obvious contradiction.
One should note that the statement about the ranges obviously follows from the statement about the kernels from dimension formula. No need to show this.