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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 $\operatorname{range}(T^n)=\operatorname{range}(T^{n+1}).$

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  • $\begingroup$ Wait a minute, what about nilpotetent operators? Wouldn't the ranges be different? $\endgroup$
    – abnry
    Oct 18, 2015 at 14:38

3 Answers 3

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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.

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By contradiction, suppose that there exists $v\in\ker(T^{n+1})$ such that $T^nv\neq0$. Then the vectors $v,Tv,...,T^nv$ are all nonzero, and you can see that they are in fact linearly independent by evaluating powers of $T$ on a linear combination $\sum_{k=0}^n\alpha_k\,T^kv=0$. But this is absurd as $\dim V=n$.

Hint for the second statement: write $v=T^nw$ and decompose $w$ as a sum of a vector in $\ker (T^n)$ and a vector in the range of $T$. Second hint: proceeding as in the first point you can show that $\ker (T^n)\cap\operatorname{range}(T^n)=\{0\}$ (see also the answers to this this question).

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The argument in MooS's answer is very nice, but here's an alternative proof via Cayley-Hamilton. Let $p(x)\in F[x]$ be the characteristic polynomial of $T$, and write $p(x)=x^mq(x)$, where $x$ does not divide $q(x)$. Since $p(x)$ has degree $\leq n$, we have $m\leq n$. Write $q(x)=xr(x)+c$ where $c\in F$; by definition of $q$, $c\neq 0$. Now suppose $v\in \ker(T^{n+1})$. We have $$q(T)T^nv=p(T)T^{n-m}v=0$$ since $m\leq n$ and $p(T)=0$ by Cayley-Hamilton, and $$q(T)T^nv=r(T)T^{n+1}v+cT^nv=cT^nv$$ since $T^{n+1}v=0$. Since $c\not=0$, this implies $T^nv=0$. Thus $\ker(T^{n+1})\subseteq \ker(T^n)$.

The statement about ranges can be proven similarly. Suppose $v\in\operatorname{range}(T^n)$; write $v=T^nw$. Then as above, we have $$0=q(T)T^nw=r(T)T^{n+1}w+cT^nw.$$ Rearranging this, we get $v=T^nw=T^{n+1}(-c^{-1}r(T)w)$. Thus $v\in\operatorname{range}(T^{n+1})$.

(Alternatively, the statement about ranges actually follows from applying the statement about kernels to the dual map $T^*:V^*\to V^*$.)

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