Let $T_i:\mathbb R^n\to\mathbb R^n$ be linear transformations, $i=1,...,n$. Assume that $T_i^n=0$ for all $i=1,...,n$, and $T_iT_j=T_jT_i$ for all $i,j=1,...,n$. Prove that $T_1T_2...T_n=0.$
My Attempt:
Suppose the converse, there exists $v\ne 0$, such that $T_1T_2...T_n(v)\ne 0$. Since $T_i$ are commutative we conclude that $v\notin\ker (T_i)$ for every $i$. We also have the sequence: $$\text{Span}\{v\}\nsubseteq\ker (T_i)\subseteq\ker(T_i^2)\subseteq\cdots\subseteq\ker(T_i^n)=\mathbb R^n$$ Then I got stuck. How to get a contradiction?
I know that Let $A$ and $B$ be square matrices of the same size such that $AB = BA$ and $A$ is nilpotent. Show that $AB$ is nilpotent., but given the commutativity of other matrices leading to a much stronger result as claimed in the question.
This can be proven by induction, with $\Bbb R^n$ replaced to any $\le n$ dimensional vector space.
Note that the condition ${T_i}^n=0$ is equivalent to saying $T_i$ is nilpotent.
The base step $n=1$ is obvious.
Assumed the claim for $n-1$, we prove it for $n$.
Let $U:=\mathrm{im} T_n$, the range of $T_n$.
By the commutativity conditions, $U$ is an invariant subspace for each $T_i$.
By $T_n$ being nilpotent, we must have $\dim U\le n-1$
Therefore, we can apply the induction hypothesis with $T_1|_U,\dots, T_{n-1}|_U$ on $U$, yielding $$T_1T_2\dots T_{n-1}(T_nv)=0$$ for all vectors $v$.