I have to show that all $n \times n$ nilpotent matrices of order n are similar. My initial approach was to show that for all nilpotent matrices their minimun characteristic polynomial is of the form: $$\lambda^n$$
Is this sufficient?
Can someone show me a formal approach to this problem?
Thanks!
Best Answer
Examples of nilpotent matrices of the same order that are not similar
$$ A=\begin{pmatrix} 0 & & & \\ & 0 & & \\ & & 0 & 1 \\ & & & 0 \end{pmatrix}, \ \ B=\begin{pmatrix}0 & 1& & \\ & 0 & & \\ & & 0 & 1 \\ & & & 0 \end{pmatrix}. $$ They both have $$ A\neq 0, \ B\neq 0 , \ A^2 = B^2 = 0. $$ However, these matrices $A$ and $B$ are not similar.
Proof that the nilpotent $n\times n$ matrices of order $n$ are similar
For nilpotent $n\times n$ matrix of order $n$, there is only one possible Jordan form. Since it is nilpotent, it has only $0$ as an eigenvalue. Since it is nilpotent of order $n$, it must be similar to the following Jordan block: $$ J(0, n) = \begin{pmatrix} 0 & 1 & & \cdots & \\ & 0 & 1 & \cdots & \\ &\cdots & \\ & & \cdots & 0 & 1 \\ & & \cdots & & 0 \end{pmatrix}. $$ If the matrix has eigenvalue all zeros, and does not have Jordan form as above, the the nilpotency order is less than $n$.