If $A$ and $B$ are similar $n\times n$ matrices, then it follows that $B$ is nil-potent if and only if $A$ is nil-potent.
I am not allowed to use Jordan blocks
My attempt is to proof something else but the above lemma follows directly from it.
My attempt:
Suppose that $D$ is a basis for some vector-space $V$, if I can show that $T:V\to V$ is nilpotent if and only if $[T]_D$ (matrix representation with respect to basis $D$) is also nilpotent with identical nilpotency index $k$
Let $u\in V, u\ne0$.
Then it follows that
$T(u)=[T]_Du\ne0$
$T^2(u)=[T]_D^2u\ne0$,
...,
$T^k(u)=[T]_D^k u=0$
It follows that $[T]_D$ is nilpotent of degree $k$.
The other way around:
Let $[T]_D$ be a nil-potent matrix of degree $k$ such that $T(u)=[T]_Du$
Then it follows that $T^k(u)=[T]_d^ku=0$
Have I missed anything? are the notations correctly used?