Can someone give a proof that all $n\times n$ matrices that are nilpotent of order $n$ are similar?
A matrix $A$ is called nilpotent if there exists some positive integer $k$ such that $A^k$ is the $0$-matrix. The order of nilpotency of a matrix is the smallest $k$ with this property. That is, the question is about matrices such that $A^n=0$ yet $A^{n-1} \neq 0$.
I am new in linear algebra and know up to linear transformation and isomorphisms and matrix representation of transformations. The same question is answered in this site using Jordan Matrix and other things which I do not know. So can anyone give a proof using elementary things which I know.
Best Answer
If $A$ is an $n\times n$ matrix that is nilpotent of order $n$, then there is a non-zero vector $x$ such that $A^nx =0$ but $A^{n-1}x \ne 0$. So $\{ x,Ax,A^2x,\cdots,A^{n-1}x\}$ is a basis, and the representation of $A$ with respect to this basis has a standard form. This standard form is the same for any such matrix, which makes any two such matrices similar.