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

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

3 Answers 3


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.


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).


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 $T^*:V^*\to V^*$.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .